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1ktct239p | chemistry | solutions | depression-in-freezing-point | 83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ____________ K. (Nearest integer)<br/><br/>[Use : Molal Freezing point depression constant of water = 1.86 K kg mol<sup>$$-$$1</sup>]<br/><br/>Freezing Point of water = 273 K<br/><br/>Atomic masses : C : 12.0 u, O : 16.0 u, H : 1... | [] | null | 269 | k<sub>f</sub> = 1.86 kg/mol<br><br>T$$_f^o$$ = 273 K<br><br>solvent : H<sub>2</sub>O(625 g)<br><br>Solute : 83 g of ethylene glycol<br><br>$$\Rightarrow$$ $$\Delta$$T<sub>f</sub> = k<sub>f</sub> $$\times$$ m<br><br>$$\Rightarrow$$ $$\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}$$<br><br>$$ \Ri... | integer | jee-main-2021-online-26th-august-evening-shift | 3,596 |
1ktedtqil | chemistry | solutions | depression-in-freezing-point | 1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $$-$$4$$^\circ$$C before freezing. The amount of ice (in g) that will be separated out is __________. (Nearest integer)<br/><br/>[Given : K<sub>f</sub>(H<sub>2</sub>O) = 1.86 K kg mol<sup>$$-$$1</sup>] | [] | null | 518 | Let mass of water initially present = x gm<br><br>$$\Rightarrow$$ Mass of sucrose = (1000 $$-$$ x) gm<br><br>$$\Rightarrow$$ moles of sucrose = $$\left( {{{1000 - x} \over {342}}} \right)$$<br><br>$$ \Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {... | integer | jee-main-2021-online-27th-august-morning-shift | 3,597 |
1l57r5qly | chemistry | solutions | depression-in-freezing-point | <p>Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).</p>
<p>Assertion (A) : At 10$$^\circ$$C, the density of a 5 M solution of KCl [atomic masses of K & Cl are 39 & 35.5 g mol<sup>$$-$$1</sup> respectively], is 'x' g ml<sup>$$-$$1</sup>. The solution is c... | [{"identifier": "A", "content": "Both (A) and (R) are true and (R) is the correct explanation of (A)."}, {"identifier": "B", "content": "Both (A) and (R) are true but (R) is not the correct explanation of (A)."}, {"identifier": "C", "content": "(A) is true but (R) is false."}, {"identifier": "D", "content": "(A) is fal... | ["A"] | null | Molality and Mass are temperature Independent so on changing temp., molality and mass remain unchanged. | mcq | jee-main-2022-online-27th-june-morning-shift | 3,600 |
1l6i6acld | chemistry | solutions | depression-in-freezing-point | <p>The elevation in boiling point for 1 molal solution of non-volatile solute A is $$3 \mathrm{~K}$$. The depression in freezing point for 2 molal solution of $$\mathrm{A}$$ in the same solvent is 6 $$K$$. The ratio of $$K_{b}$$ and $$K_{f}$$ i.e., $$K_{b} / K_{f}$$ is $$1: X$$. The value of $$X$$ is [nearest integer]<... | [] | null | 1 | Molality of a solution of non volatile solute $(A)=1$<br/><br/> Elevation in boiling point is given by
<br/><br/>
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}$
<br/><br/>
$$
3=\mathrm{K}_{\mathrm{b}} \times 1
$$ ... (1)<br/><br/>
Molality of $(A)$ in the same ... | integer | jee-main-2022-online-26th-july-evening-shift | 3,601 |
1l6me4949 | chemistry | solutions | depression-in-freezing-point | <p>$$150 \mathrm{~g}$$ of acetic acid was contaminated with $$10.2 \mathrm{~g}$$ ascorbic acid $$\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)$$ to lower down its freezing point by $$\left(x \times 10^{-1}\right)^{\circ} \mathrm{C}$$. The value of $$x$$ is ___________. (Nearest integer)<p>
<p>[Given $$\math... | [] | null | 15 | M.wt. of Acetic acid $=60 \mathrm{~g}$
<br/><br/>
M.wt. of Ascorbic acid $=176 \mathrm{~g}$
<br/><br/>
$$
\Delta T_{f}=K_{f} m
$$
<br/><br/>
$$
\begin{aligned}
\Delta T_{f} &=\frac{3.9 \times 10.2 \times 1000}{176 \times 150} \\\\
\Delta T_{f} &=1.506 \\\\
&=15.06 \times 10^{-1} \\\\
&=15
\end{aligned}
$$ | integer | jee-main-2022-online-28th-july-morning-shift | 3,602 |
1l6rlhdiy | chemistry | solutions | depression-in-freezing-point | <p>$$1.80 \mathrm{~g}$$ of solute A was dissolved in $$62.5 \mathrm{~cm}^{3}$$ of ethanol and freezing point of the solution was found to be $$155.1 \mathrm{~K}$$. The molar mass of solute A is ________ g $$\mathrm{mol}^{-1}$$.</p>
<p>[Given : Freezing point of ethanol is 156.0 K.</p>
<p>Density of ethanol is 0.80 g cm... | [] | null | 80 | Mass of solvent $$=d x v=0.8 \times 62.5=50$$ gram <br/><br/>$$\Delta T_{f}=k_{f} \times m$$<br/><br/>
$$
\begin{aligned}
&0.9=2\left[\frac{1.8 \times 1000}{M_{\text {Solute }} \times 50}\right] \\
&M_{\text {Solute }}=\left(\frac{2 \times 1.8 \times 1000}{0.9 \times 50}\right)=80
\end{aligned}
$$ | integer | jee-main-2022-online-29th-july-evening-shift | 3,603 |
1ldyg764k | chemistry | solutions | depression-in-freezing-point | <p>In the depression of freezing point experiment</p>
<p>A. Vapour pressure of the solution is less than that of pure solvent</p>
<p>B. Vapour pressure of the solution is more than that of pure solvent</p>
<p>C. Only solute molecules solidify at the freezing point</p>
<p>D. Only solvent molecules solidify at the freezi... | [{"identifier": "A", "content": "A and C only"}, {"identifier": "B", "content": "A only"}, {"identifier": "C", "content": "A and D only"}, {"identifier": "D", "content": "B and C only"}] | ["C"] | null | In the depression of freezing point experiment
only solvent molecules solidify and vapour pressure of
solution decreases as some of the surface area is occupied
by solute molecules, so less number of molecules will go in
vapour form. | mcq | jee-main-2023-online-24th-january-morning-shift | 3,604 |
lsap58pv | chemistry | solutions | depression-in-freezing-point | Mass of ethylene glycol (antifreeze) to be added to $18.6 \mathrm{~kg}$ of water to protect the freezing point at $-24^{\circ} \mathrm{C}$ is ________ $\mathrm{kg}$ (Molar mass in $\mathrm{g} ~\mathrm{mol}^{-1}$ for ethylene glycol $62, \mathrm{~K}_f$ of water $=1.86 \mathrm{~K}$ $\mathrm{kg} ~\mathrm{mol}^{-1}$ ) | [] | null | 15 | <p>To determine the mass of ethylene glycol (antifreeze) needed to lower the freezing point of water to $-24^{\circ} \mathrm{C}$, we can use the freezing point depression equation, which is a colligative property given by:</p>
$$ \Delta T_f = i \cdot K_f \cdot m $$
<p>where:</p>
<ul>
<li>$\Delta T_f$ is the depressio... | integer | jee-main-2024-online-1st-february-evening-shift | 3,605 |
1lsg7dshs | chemistry | solutions | depression-in-freezing-point | <p>The solution from the following with highest depression in freezing point/lowest freezing point is</p> | [{"identifier": "A", "content": "$$180 \\mathrm{~g}$$ of acetic acid dissolved in benzene\n"}, {"identifier": "B", "content": "$$180 \\mathrm{~g}$$ of acetic acid dissolved in water\n"}, {"identifier": "C", "content": "$$180 \\mathrm{~g}$$ of benzoic acid dissolved in benzene\n"}, {"identifier": "D", "content": "$$180 ... | ["B"] | null | <p>$$\Delta \mathrm{T}_{\mathrm{f}}$$ is maximum when $$\mathrm{i} \times \mathrm{m}$$ is maximum.</p>
<p>1) $$\mathrm{m}_1=\frac{180}{60}=3, \mathrm{i}=1+\alpha$$</p>
<p>Hence</p>
<p>$$\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}_{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha<<1)$$</p>
<p>2) $... | mcq | jee-main-2024-online-30th-january-evening-shift | 3,606 |
1lsgxwd5m | chemistry | solutions | depression-in-freezing-point | <p>What happens to freezing point of benzene when small quantity of napthalene is added to benzene?</p> | [{"identifier": "A", "content": "Increases\n"}, {"identifier": "B", "content": "Decreases\n"}, {"identifier": "C", "content": "Remains unchanged\n"}, {"identifier": "D", "content": "First decreases and then increases"}] | ["B"] | null | <p>On addition of naphthalene to benzene there is depression in freezing point of benzene.</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 3,607 |
lv2eryip | chemistry | solutions | depression-in-freezing-point | <p>$$2.7 \mathrm{~kg}$$ of each of water and acetic acid are mixed. The freezing point of the solution will be $$-x^{\circ} \mathrm{C}$$. Consider the acetic acid does not dimerise in water, nor dissociates in water. $$x=$$ ________ (nearest integer)</p>
<p>[Given: Molar mass of water $$=18 \mathrm{~g} \mathrm{~mol}^{-... | [] | null | 31 | <p>$$\begin{aligned}
& \text { Molality of acetic acid }=\frac{2700}{60} \times \frac{1}{2.7} \mathrm{~mol} / \mathrm{kg} \\
&=16.667 \\
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times 16.667 \\
&=1.86 \times 16.667 \\
&=31 \mathrm{~K}
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-evening-shift | 3,608 |
lvb2acph | chemistry | solutions | depression-in-freezing-point | <p>When '$$x$$' $$\times 10^{-2} \mathrm{~mL}$$ methanol (molar mass $$=32 \mathrm{~g}$$' density $$=0.792 \mathrm{~g} / \mathrm{cm}^3$$) is added to $$100 \mathrm{~mL}$$. water (density $$=1 \mathrm{~g} / \mathrm{cm}^3$$), the following diagram is obtained.</p>
<p><img src="data:image/png;base64,UklGRpQWAABXRUJQVlA4II... | [] | null | 543 | <p>$$\begin{aligned}
& \Delta T_f=2.5^{\circ} \mathrm{C} \\
& \Delta T_f=i \times k_f \times m \\
& 2.5=1 \times 1.86 \times \frac{n_B}{0.1} \\
& n_B=\frac{2.5 \times 0.1}{1.86}=0.1344 \mathrm{~mol} \\
& \text { mass of methanol }=0.1344 \times 32=4.3 \mathrm{~g} \\
& d=\frac{m}{v} \\
& v=\frac{m}{d} \\
& v=\frac{4.3}{... | integer | jee-main-2024-online-6th-april-evening-shift | 3,610 |
ypqtNICb6KEJG2M3 | chemistry | solutions | elevation-in-boiling-point | A pressure cooker reduces cooking time for food because | [{"identifier": "A", "content": "boiling point of water involved in cooking is increased"}, {"identifier": "B", "content": "the higher pressure inside the cooker crushes the food material"}, {"identifier": "C", "content": "cooking involves chemical changes helped by a rise in temperature"}, {"identifier": "D", "content... | ["A"] | null | <b>NOTE :</b> On increasing pressure, the temperature is also increased. Thus in pressure cooker due to increase in pressure the $$b.p.$$ of water increases. | mcq | aieee-2003 | 3,611 |
tDvmievhAkqfTYyd563rsa0w2w9jx0x7b9s | chemistry | solutions | elevation-in-boiling-point | 1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose
ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, $${{\Delta {T_b}(A)} \over {\Delta {T_b}(B)}}$$, is : | [{"identifier": "A", "content": "5 : 1"}, {"identifier": "B", "content": "1 : 0.2"}, {"identifier": "C", "content": "10 : 1"}, {"identifier": "D", "content": "1 : 5"}] | ["D"] | null | Given,
<br><br>$${{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} = {1 \over 5}$$
<br><br>$$\Delta {T_b}\left( A \right) = {k_b}\left( A \right) \times $$ Molality of A
<br><br>$$\Delta {T_b}\left( B \right) = {k_b}\left( B \right) \times $$ Molality of B
<br><br>Here, Molality of A = Molality of B... | mcq | jee-main-2019-online-10th-april-evening-slot | 3,612 |
2Dzc2RBDiHjtenGjEs1kmkjpoc2 | chemistry | solutions | elevation-in-boiling-point | A 1 molal K<sub>4</sub>Fe(CN)<sub>6</sub> solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is __________ u. (Round off to the Nearest Integer). [Density of water = 1.0 g cm<sup>$$-$... | [] | null | 85 | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3hf6b3o/7a43a948-6d10-4eb8-9882-d0d5582ecf46/ef80b740-d9de-11ec-8ea5-5be6b5e3368b/file-1l3hf6b3p.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3hf6b3o/7a43a948-6d10-4eb8-9882-d0d5582ecf46/ef80b740-d9de-11ec-8ea5-5be6b5e3368... | integer | jee-main-2021-online-17th-march-evening-shift | 3,613 |
1krxcofq2 | chemistry | solutions | elevation-in-boiling-point | When 3.00 g of a substance 'X' is dissolved in 100 g of CCl<sub>4</sub>, it raises the boiling point by 0.60 K. The molar mass of the substance 'X' is ______________ g mol<sup>$$-$$1</sup>. (Nearest integer).<br/><br/>[Given K<sub>b</sub> for CCl<sub>4</sub> is 5.0 K kg mol<sup>$$-$$1</sup>] | [] | null | 250 | $$\Delta$$T<sub>b</sub> = K<sub>b</sub> $$\times$$ molality<br><br>0.60 = 5 $$\times$$ $$\left( {{{3/M} \over {100/100}}} \right)$$<br><br>M = 250 | integer | jee-main-2021-online-25th-july-evening-shift | 3,614 |
1l6jkm033 | chemistry | solutions | elevation-in-boiling-point | <p>Boiling point of a $$2 \%$$ aqueous solution of a non-volatile solute A is equal to the boiling point of $$8 \%$$ aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is</p> | [{"identifier": "A", "content": "$$\\mathrm{M}_{\\mathrm{A}}=4 \\mathrm{M}_{\\mathrm{B}}$$"}, {"identifier": "B", "content": "$$\\mathrm{M}_{\\mathrm{B}}=4 \\mathrm{M}_{\\mathrm{A}}$$"}, {"identifier": "C", "content": "$$\\mathrm{M}_{\\mathrm{A}}=8 \\mathrm{M}_{\\mathrm{B}}$$"}, {"identifier": "D", "content": "$$\\math... | ["B"] | null | For $\mathbf{A}: 100 \,\mathrm{gm}$ solution $\rightarrow 2 \,\mathrm{gm}$ solute $\mathrm{A}$<br/><br/> $\therefore$ Molality $=\frac{2 / \mathrm{M}_{\mathrm{A}}}{0.098}$
<br/><br/>
For B : $100 \,\mathrm{gm}$ solution $\rightarrow 8 \,\mathrm{gm}$ solute $\mathrm{B}$<br/><br/>
$$
\begin{aligned}
&\therefore \text { M... | mcq | jee-main-2022-online-27th-july-morning-shift | 3,616 |
1ldr5c521 | chemistry | solutions | elevation-in-boiling-point | <p>A solution containing $$2 \mathrm{~g}$$ of a non-volatile solute in $$20 \mathrm{~g}$$ of water boils at $$373.52 \mathrm{~K}$$. The molecular mass of the solute is ___________ $$\mathrm{g} ~\mathrm{mol}^{-1}$$. (Nearest integer)</p>
<p>Given, water boils at $$373 \mathrm{~K}, \mathrm{~K}_{\mathrm{b}}$$ for water $$... | [] | null | 100 | <p>$$\mathrm{\Delta T_b=K_b.m}$$</p>
<p>(0.52) = (0.52) (m)</p>
<p>$$\mathrm{m=1=\frac{2(1000)}{(mw)(20)}}$$</p>
<p>mw = 100</p> | integer | jee-main-2023-online-30th-january-morning-shift | 3,617 |
1lgp3o9bs | chemistry | solutions | elevation-in-boiling-point | <p>Sea water contains $$29.25 \% ~\mathrm{NaCl}$$ and $$19 \% ~\mathrm{MgCl}_{2}$$ by weight of solution. The normal boiling point of the sea water is _____________ $${ }^{\circ} \mathrm{C}$$ (Nearest integer)</p>
<p>Assume $$100 \%$$ ionization for both $$\mathrm{NaCl}$$ and $$\mathrm{MgCl}_{2}$$</p>
<p>Given : $$\ma... | [] | null | 116 | Amount of solvent = 100 - (29.25 + 19) = 51.75 g
<br/><br/>
Now, we can calculate the boiling point elevation using the given formula:
<br/><br/>
ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52<br/><br/>
ΔTb = 16.075
<br/><br/>
The boiling point of the sea water is the normal boiling... | integer | jee-main-2023-online-13th-april-evening-shift | 3,618 |
1lgystn2c | chemistry | solutions | elevation-in-boiling-point | <p>If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio $$2: 1$$ and their enthalpy of vaporizations are in the ratio $$1: 2$$, then the boiling point elevation constant of $$\mathrm{X}$$ is $$\underline{\mathrm{m}}$$ times the boiling point elevation constant of Y. The value o... | [] | null | 8 | The boiling point elevation constant, also known as the ebullioscopic constant ($K_{b}$), can be determined using the formula:
<br/><br/>
$$K_{b} = \frac{R \cdot T_{b}^{2}}{1000 \cdot \Delta H_{vap}}$$
<br/><br/>
where:<br/><br/>
- $R$ is the universal gas constant (8.31 J mol⁻¹ K⁻¹)<br/><br/>
- $T_{b}$ is the boiling ... | integer | jee-main-2023-online-8th-april-evening-shift | 3,619 |
1lh04z8nh | chemistry | solutions | elevation-in-boiling-point | <p>The vapour pressure vs. temperature curve for a solution solvent system is shown below.</p>
<p><img src="data:image/png;base64,UklGRroaAABXRUJQVlA4IK4aAABwQQGdASoAA9UCP4G812Y2LiwnIRE5isAwCWlu+/7BYVaODPJsjQfz9a/9z4N/4z/oZFfn/+//vp7P+AL5///vujQLNevQL+7v/HoZw87GhS68f/ZhR9rhltEE6ceEfNXgerxhl6uQuGXq5C4ZerkLhl6uQtpn1Xr/aZ... | [] | null | 82 | At any temperature, the vapour pressure of the solution is lower than that of the pure solvent. Hence vapour pressure- temperature curve of solution lies below that of solvent.
<br/><br/>
The more volatile liquid evaporates fast as compared to the less volatile liquid at a low temperature because the volume increases w... | integer | jee-main-2023-online-8th-april-morning-shift | 3,620 |
jaoe38c1lsc54900 | chemistry | solutions | elevation-in-boiling-point | <p>A solution of two miscible liquids showing negative deviation from Raoult's law will have :
</p> | [{"identifier": "A", "content": "increased vapour pressure, increased boiling point\n"}, {"identifier": "B", "content": "increased vapour pressure, decreased boiling point\n"}, {"identifier": "C", "content": "decreased vapour pressure, decreased boiling point\n"}, {"identifier": "D", "content": "decreased vapour pressu... | ["D"] | null | <strong>Negative Deviation from Raoult's Law</strong>
<br/><br/><ul>
<li>Negative deviation means the intermolecular forces of attraction between the molecules of the two liquids (A-B) are stronger than the forces between molecules of the pure liquids (A-A and B-B).</li>
<br/><li>This stronger attraction makes it hard... | mcq | jee-main-2024-online-27th-january-morning-shift | 3,621 |
lv0vyr87 | chemistry | solutions | elevation-in-boiling-point | <p>$$2.5 \mathrm{~g}$$ of a non-volatile, non-electrolyte is dissolved in $$100 \mathrm{~g}$$ of water at $$25^{\circ} \mathrm{C}$$. The solution showed a boiling point elevation by $$2^{\circ} \mathrm{C}$$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of... | [] | null | 707 | <p><ul>
<li>Molal boiling point elevation constant of water ($K_b$) = 0.52 K.kg.mol<sup>-1</sup></li>
<li>1 atm = 760 mm Hg</li>
<li>Molar mass of water = 18 g.mol<sup>-1</sup></li>
</ul>
</p>
<p>First, we calculate the molality (m) of the solution using the boiling point elevation formula:</p>
<p>
<p>$$ \Delta... | integer | jee-main-2024-online-4th-april-morning-shift | 3,622 |
lv5gsxw4 | chemistry | solutions | elevation-in-boiling-point | <p>A solution containing $$10 \mathrm{~g}$$ of an electrolyte $$\mathrm{AB}_2$$ in $$100 \mathrm{~g}$$ of water boils at $$100.52^{\circ} \mathrm{C}$$. The degree of ionization of the electrolyte $$(\alpha)$$ is _________ $$\times 10^{-1}$$. (nearest integer)</p>
<p>[Given : Molar mass of $$\mathrm{AB}_2=200 \mathrm{~g... | [] | null | 5 | <p>To find the degree of ionization $$(\alpha)$$, we need to use the boiling point elevation formula and the van't Hoff factor. The formulas we will use are:</p>
<p>
<p>$$\Delta T_b = i \cdot K_b \cdot m$$</p>
</p>
<p>where:
<p>$$\Delta T_b$$ is the boiling point elevation,</p>
<p>$$i$$ is the van't Hoff factor,<... | integer | jee-main-2024-online-8th-april-morning-shift | 3,623 |
NRMtCtz0O5hwInS10dGrt | chemistry | solutions | henry's-law | The solubility of N<sub>2</sub> in water at 300 K and 500 torr partial pressure is 0.01 g L<sup>−1</sup>. The solubility (in g L<sup>−1</sup>) at 750 torr partial pressure is : | [{"identifier": "A", "content": "0.0075"}, {"identifier": "B", "content": "0.015"}, {"identifier": "C", "content": "0.02"}, {"identifier": "D", "content": "0.005"}] | ["B"] | null | Partial pressure = Mole fraction × Solubility
<br><br>$${{{p_1}} \over {{p_2}}} = {{{s_1}} \over {{s_2}}}$$
<br><br>$$ \Rightarrow $$ $${{500} \over {750}} = {{0.01} \over {{s_2}}}$$
<br><br>$$ \Rightarrow $$ s<sub>2</sub> = 0.015 g L<sup>-1</sup> | mcq | jee-main-2016-online-9th-april-morning-slot | 3,624 |
Gajxr8SuOb34y0PbIcYsL | chemistry | solutions | henry's-law | Which one of the following statements regarding Henry's law is not correct ? | [{"identifier": "A", "content": "Higher the value of K<sub>H</sub> at a given pressure, higher is the solubility of the gas in the liquids"}, {"identifier": "B", "content": "Different gases have different K<sub>H</sub> (Henry's law constant) values at the same temperature."}, {"identifier": "C", "content": "The partial... | ["A"] | null | From Henry's law we know,
<br><br>P<sub>gas</sub> = K<sub>H</sub> x<sub>g</sub>
<br><br>Where, Pgas = Pressure of undissolved gas
<br><br>x<sub>g</sub> = Mole fraction of gas dissolved into the liquid.
<br><br>K<sub>H</sub> = Henry's Carnot.
<br><br>When pressure is constant then,
<br><br>x<sub>g</sub> $$ \propto ... | mcq | jee-main-2019-online-9th-january-morning-slot | 3,625 |
ZcuVrz6CvIekK8ehW8BLx | chemistry | solutions | henry's-law | For the solution of the gases w, x, y and z in
water at 298K, the Henrys law constants (K<sub>H</sub>)
are 0.5, 2, 35 and 40 kbar, respectively. The
correct plot for the given data is :- | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264939/exam_images/acgusoaac1abhyngqgbn.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 8th April Evening Slot Chemistry - Solutions Question 10... | ["A"] | null | Henry's law expression,
<br><br>P<sub>gas</sub> = K<sub>H</sub> $$ \times $$ x<sub>gas</sub>
<br><br>Here, P<sub>gas</sub> = partial pressure of gas
<br><br>K<sub>H</sub> = Henry’s law constant
<br><br>x<sub>gas</sub> = mole fraction of gas
<br><br>$$ \because $$ x<sub>gas</sub> = 1 - x<sub>H<sub>2</sub>O</sub>
<br><br... | mcq | jee-main-2019-online-8th-april-evening-slot | 3,626 |
e75DUB2jBZE6rrFwjx7k9k2k5h15kp6 | chemistry | solutions | henry's-law | A graph of vapour pressure and temperature for three different liquids X, Y, and Z is shown below :
<img src="data:image/png;base64,UklGRjYYAABXRUJQVlA4ICoYAABwfwCdASr0AfoAPm02l0ikIyKhInQ6aIANiWlu/FXCU6QecPCvoyUffDv8x/JXwH/pH5Iei/4n8w/bv7H+zX9g9jH+R6UD98/iP7YeEN71/Ef7B/yv6/82Pz3+yfyH+i/3n+3fs77k/AH4+9gL8s/j3+L/i/9U/6H9... | [{"identifier": "A", "content": "(B)"}, {"identifier": "B", "content": "(A)"}, {"identifier": "C", "content": "(C)"}, {"identifier": "D", "content": "(A) and (C)"}] | ["A"] | null | Vapour pressure of a liquid at a given
temperature is inversely proportional to
intermolecular force of attraction. At the same
temperature, vapour pressure of X is higher
than that of Y.
<br><br>Therefore (X) has lower intermolecular
interactions compared to Y. Statement (B) is
correct. | mcq | jee-main-2020-online-8th-january-morning-slot | 3,627 |
jo1LfOJz8I06ykYcybjgy2xukf2bn9id | chemistry | solutions | henry's-law | Henry’s constant (in kbar) for four gases $$\alpha $$, $$\beta $$, $$\gamma $$ and $$\delta $$ in water at 298 K is given below :
<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px... | [{"identifier": "A", "content": "solubility of $$\\gamma $$ at 308 K is lower than at 298 K"}, {"identifier": "B", "content": "The pressure of a 55.5 molal solution of $$\\delta $$ is 250 bar"}, {"identifier": "C", "content": "$$\\alpha $$ has the highest solubility in water at a given pressure"}, {"identifier": "D", "... | null | null | (1) On increasing temperature solubility of gases decreases.
<br><br>(2) 55.5 molal solution of
$$\gamma $$ = 55.5 mol of
<br><br>P<sub>$$\gamma $$</sub>
= K<sub>H</sub>X<sub>$$\gamma $$</sub>
<br><br>= 2 $$ \times $$ 10<sup>-5</sup> $$ \times $$ $${{55.5} \over {55.5 + {{1000} \over {18}}}}$$
<br><br>= 10<sup>–5</s... | mcqm | jee-main-2020-online-3rd-september-morning-slot | 3,628 |
nQQWklFtMUwr7Mrh8t1kmj8z2gy | chemistry | solutions | henry's-law | The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is __________ $$\times$$ 10<sup>$$-$$5</sup> mol dm<sup>$$-$$3</sup>. (Round off to the Nearest Integer).<br/><br/>[Given : Henry's law constant = K<sub>H</sub> = 8.0 $$\times$$ 10<sup>4<... | [] | null | 25 | Partial pressure (P<sub>g</sub>) $$ \propto $$ Solubility
<br><br>Partial pressure (P<sub>g</sub>) = K<sub>H</sub>$$ \times $$ Solubility
<br><br>20 × 10<sup>3</sup> = [8.0 × 10<sup>4</sup> × 10<sup>3</sup>] × Solubility
<br><br>Solubility = $${{20 \times {{10}^3}} \over {8.0 \times {{10}^7}}}$$ = 25 $$ \times $$ 10<su... | integer | jee-main-2021-online-17th-march-morning-shift | 3,629 |
1krutq9fh | chemistry | solutions | henry's-law | CO<sub>2</sub> gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO<sub>2</sub> exerts a partial pressure of 0.835 bar then x m mol of CO<sub>2</sub> would dissolve in 0.9 L of water. The value of x is ____________. (Nearest integer)<br/><br/>(Henry's law constant for CO<sub>2</sub> at... | [] | null | 25 | From Henry's law<br><br>P<sub>gas</sub> = K<sub>H</sub>.X<sub>gas</sub><br><br>0.835 = 1.67 $$\times$$ 10<sup>3</sup> $$\times$$ $${{n(C{O_2})} \over {{{0.9 \times 1000} \over {18}}}}$$<br><br>$$ \Rightarrow $$ n(CO<sub>2</sub>) = 0.025<br><br>Millimoles of CO<sub>2</sub> = 0.025 $$\times$$ 1000 = 25 | integer | jee-main-2021-online-25th-july-morning-shift | 3,630 |
1l5bdteim | chemistry | solutions | henry's-law | <p>A company dissolves 'x' amount of CO<sub>2</sub> at 298 K in 1 litre of water to prepare soda water. X = __________ $$\times$$ 10<sup>$$-$$3</sup> g. (nearest integer)</p>
<p>(Given : partial pressure of CO<sub>2</sub> at 298 K = 0.835 bar.</p>
<p>Henry's law constant for CO<sub>2</sub> at 298 K = 1.67 kbar.</p>
<p>... | [] | null | 1221 | According to Henry's law, partial pressure of a gas is given by
<br/><br/>
$P_{g}=\left(K_{H}\right) X_{g}$
<br/><br/>
where $X_{g}$ is mole fraction of gas in solution
<br/><br/>
$0.835=1.67 \times 10^{3}\left(\mathrm{X}_{\mathrm{CO}_{2}}\right)$
<br/><br/>
$\mathrm{X}_{\mathrm{CO}_{2}}=5 \times 10^{-4}$
<br/><br/>
Ma... | integer | jee-main-2022-online-24th-june-evening-shift | 3,631 |
1l6p8shcy | chemistry | solutions | henry's-law | <p>If $$\mathrm{O}_{2}$$ gas is bubbled through water at $$303 \mathrm{~K}$$, the number of millimoles of $$\mathrm{O}_{2}$$ gas that dissolve in 1 litre of water is __________. (Nearest Integer)</p>
<p>(Given : Henry's Law constant for $$\mathrm{O}_{2}$$ at $$303 \mathrm{~K}$$ is $$46.82 \,\mathrm{k}$$ bar and partial... | [] | null | 1 | $$\mathrm{P}=\mathrm{K}_{\mathrm{H}} \times \mathrm{X}$$
<br/><br/>
$0.920 \mathrm{bar}=46.82 \times 10^{3}\, \mathrm{bar} \times \frac{\mathrm{mol} \,\mathrm{of} \,\mathrm{O}_{2}}{\mathrm{~mol} \text { of } \mathrm{H}_{2} \mathrm{O}}$
<br/><br/>
$$0.920=46.82 \times 10^{3} \times \frac{\text { mol of O}_{2}}{1000 / 18... | integer | jee-main-2022-online-29th-july-morning-shift | 3,633 |
tZRwmMZwPTq9I1Doun3rsa0w2w9jx8yalrl | chemistry | solutions | osmotic-pressure | A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol<sup>–1</sup>) and 1.8 g of glucose (molar
mass = 180 g mol<sup>–1</sup>) in 100 mL of water at 27<sup>o</sup>C. The osmotic pressure of the solution is :
<br/>(R = 0.08206 L atm K<sup>–1</sup>
mol<sup>–1</sup>) | [{"identifier": "A", "content": "8.2 atm"}, {"identifier": "B", "content": "2.46 atm"}, {"identifier": "C", "content": "4.92 atm"}, {"identifier": "D", "content": "1.64 atm"}] | ["C"] | null | We know, osmotic pressure ($$\pi $$) = (Resultant molarity)$$ \times $$R$$ \times $$T
<br><br>Resultant molarity = $${{Total\,moles\,of\,all\,solutions} \over {Total\,volume(in\,lit)}}$$
<br>= $${{{{0.6} \over {60}} + {{1.8} \over {180}}} \over {0.1}}$$ = 0.2 mol/lit
<br><br>$$ \therefore $$ Osmotic pressure ($$\pi $$)... | mcq | jee-main-2019-online-12th-april-evening-slot | 3,636 |
a74MKCDwo8d8xFT0j0jgy2xukeyapmgh | chemistry | solutions | osmotic-pressure | The size of a raw mango shrinks to a much
smaller size when kept in a concentrated salt
solution. Which one of the following processes
can explain this? | [{"identifier": "A", "content": "Osmosis"}, {"identifier": "B", "content": "Reverse osmosis"}, {"identifier": "C", "content": "Diffusion"}, {"identifier": "D", "content": "Dialysis"}] | ["A"] | null | Raw mango shrink in salt solution due to net transfer of water molecules from mango to salt solution due to
phenomenon of osmosis. | mcq | jee-main-2020-online-2nd-september-evening-slot | 3,637 |
Nzt3ZgqFSlbHmzGHYpjgy2xukf3nq87t | chemistry | solutions | osmotic-pressure | If 250 cm<sup>3</sup>
of an aqueous solution containing 0.73 g of a protein A is isotonic with one litre of
another aqueous solution containing 1.65 g of a protein B, at 298 K, the ratio of the molecular
masses of A and B is ______ × 10<sup>–2</sup> (to the nearest integer). | [] | null | 177 | Let molar mass of protein A = x g/mol
<br><br>Let molar mass of protein B = y g/mol
<br><br>$$\pi $$<sub>A</sub> = osmotic pressure of protein A = $${{0.73} \over x} \times {{1000} \over {250}} \times RT$$
<br><br>$$\pi $$<sub>B</sub> = osmotic pressure of protein B = $${{1.65} \over y} \times {1 \over 1} \times RT$$
<... | integer | jee-main-2020-online-3rd-september-evening-slot | 3,638 |
LFGTUcTtTbDoLQ6U0cjgy2xukfch6jyt | chemistry | solutions | osmotic-pressure | The osmotic pressure of a solution of NaCl is
0.10 atm and that of a glucose solution is
0.20 atm. The osmotic pressure of a solution
formed by mixing 1 L of the sodium chloride
solution with 2 L of <br/>the glucose solution is
x $$ \times $$ 10<sup>–3</sup> atm. x is _____. (nearest integer) | [] | null | 167 | For NaCl : $$\pi $$<sub>1</sub> = 0.1 atm, v<sub>1</sub> = 1 L
<br><br>For Glucose: $$\pi $$<sub>2</sub> = 0.2 atm, v<sub>2</sub> = 2 L
<br><br>$$\pi $$<sub>mix</sub> = $${{{\pi _1}{v_1} + {\pi _2}{v_2}} \over {{v_1} + {v_2}}}$$
<br><br>= $${{0.1 \times 1 + 0.2 \times 2} \over {1 + 2}}$$
<br><br>= $${{0.5} \over 3}$$ =... | integer | jee-main-2020-online-4th-september-evening-slot | 3,639 |
1krz2avqn | chemistry | solutions | osmotic-pressure | 1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 $$\times$$ 10<sup>$$-$$3</sup> bar.<br/><br/>The molar mass of the biopolymer is _____________ $$\times$$ 10<sup>4</sup> g mol<sup>$$-$$1</sup>. (Round off to the Nearest Integer)<br/><br/>[Use : R = 0.083 L bar mol<sup>$$-$... | [] | null | 15 | $$\pi$$ = CRT;<br><br>$$\pi$$ = osmotic pressure<br><br>C = molarity<br><br>T = Temperature of solution<br><br>let the molar mass be M gm/mol<br><br>2.42 $$\times$$ 10<sup>$$-$$3</sup> bar = $${{\left( {{{1.46g} \over {Mgm/mol}}} \right)} \over {0.1l}} \times \left( {{{0.083l - bar} \over {mol - K}}} \right) \times (30... | integer | jee-main-2021-online-27th-july-morning-shift | 3,640 |
1l5c74iui | chemistry | solutions | osmotic-pressure | <p>The osmotic pressure of blood is 7.47 bar at 300 K. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in gL<sup>$$-$$1</sup> is _____________.</p>
<p>(Molar mass of glucose = 180 g mol<sup>$$-$$1</sup>, R = 0.083 L bar K<sup>$$-$$1</sup> mol<sup>$$-... | [] | null | 54 | $7.47=\mathrm{C} \times 0.083 \times 300$
<br/><br/>
$(\pi=\mathrm{CRT})$
<br/><br/>
(Where C represents the concentration of glucose solution and $\pi$ represents osmotic pressure)
<br/><br/>
$\mathrm{C}=\frac{7.47}{0.083 \times 300}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)$
<br/><br/>
which in $\mathrm{gm} / \mathr... | integer | jee-main-2022-online-24th-june-morning-shift | 3,641 |
1ldpq3zg8 | chemistry | solutions | osmotic-pressure | <p>At $$27^{\circ} \mathrm{C}$$, a solution containing $$2.5 \mathrm{~g}$$ of solute in $$250.0 \mathrm{~mL}$$ of solution exerts an osmotic pressure of $$400 \mathrm{~Pa}$$. The molar mass of the solute is ___________ $$\mathrm{g} \mathrm{~mol}^{-1}$$ (Nearest integer)</p>
<p>(Given : $$\mathrm{R}=0.083 \mathrm{~L} \m... | [] | null | 62250 | $$\pi = CRT$$
<br/><br/>$$
\begin{aligned}
400 & =\frac{2.5}{\mathrm{m_w}} \times 4 \times\left(.083 \times 10^5\right) \times 300 \\\\
\mathrm{m_w} & =\frac{10 \times 0.083 \times 3}{4} \times 10^5 \\\\
& =62250
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-morning-shift | 3,642 |
1ldults02 | chemistry | solutions | osmotic-pressure | <p>The osmotic pressure of solutions of PVC in cyclohexanone at 300 K are plotted on the graph.</p>
<p>The molar mass of PVC is ____________ g mol$$^{-1}$$ (Nearest integer)</p>
<p><img src="data:image/png;base64,UklGRkILAABXRUJQVlA4IDYLAADwrgCdASoAA+UBP4HA2WW2MKynITHpUsAwCWlu7mBzqI9f8RvP03+pEgf7d5f4NHh2/e4iXf+CMA+exFO... | [] | null | 41500 | $\begin{aligned} & \pi=\mathrm{M}^{\prime} \mathrm{RT}=\left(\frac{\mathrm{W} / \mathrm{M}}{\mathrm{V}}\right) \mathrm{RT} \\\\ & \Rightarrow \ \pi=\left(\frac{\mathrm{W}}{\mathrm{V}}\right)\left(\frac{1}{\mathrm{M}}\right) \mathrm{RT}=\mathrm{C}\left(\frac{\mathrm{RT}}{\mathrm{M}}\right) \\\\ & \Rightarrow... | integer | jee-main-2023-online-25th-january-morning-shift | 3,643 |
1lgvv57yh | chemistry | solutions | osmotic-pressure | <p>An aqueous solution of volume $$300 \mathrm{~cm}^{3}$$ contains $$0.63 \mathrm{~g}$$ of protein. The osmotic pressure of the solution at $$300 \mathrm{~K}$$ is 1.29 mbar. The molar mass of the protein is ___________ $$\mathrm{g} ~\mathrm{mol}^{-1}$$</p>
<p>Given : R = 0.083 L bar K$$^{-1}$$ mol$$^{-1}$$</p> | [] | null | 40535 | <p>The concept we are utilizing to solve this problem is osmotic pressure. Osmotic pressure is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. For dilute solutions, osmotic pressure behaves similarly to an ideal gas, hence the formula we use is similar to the ideal ga... | integer | jee-main-2023-online-10th-april-evening-shift | 3,644 |
1lh338j26 | chemistry | solutions | osmotic-pressure | <p>Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is / are ___________.</p>
<p>A. $$1 ~\mathrm{M}$$ aq. $$\mathrm{NaCl}$$ and $$2 ~\mathrm{M}$$ aq. urea</p>
<p>B. $$1 ~\mathrm{M}$$ aq. $$\mathrm{CaCl}_{2}$$ and $$1.5 ~\mathrm{M}$$ aq. $$\mathrm{... | [] | null | 4 | <p>To find the number of pairs of solutions that are isotonic with each other, we need to consider the van 't Hoff factor ($i$) which represents the number of particles a substance dissociates into in solution. We then use this to find the osmotic pressure of each solution in the pairs, using the formula:</p>
<p>$ ... | integer | jee-main-2023-online-6th-april-evening-shift | 3,645 |
jaoe38c1lsfk8a0m | chemistry | solutions | osmotic-pressure | <p>The osmotic pressure of a dilute solution is $$7 \times 10^5 \mathrm{~Pa}$$ at $$273 \mathrm{~K}$$. Osmotic pressure of the same solution at $$283 \mathrm{~K}$$ is _________ $$\times 10^4 \mathrm{Nm}^{-2}$$.</p> | [] | null | 73 | <p>$$\begin{aligned}
& \pi=\text { CRT } \\
& \Rightarrow \frac{\pi_1}{\pi_2}=\frac{\mathrm{T}_1}{\mathrm{~T}_2} \\
& \Rightarrow \pi_2=\frac{\pi_1 \mathrm{~T}_2}{\mathrm{~T}_1}=\frac{7 \times 10^5 \times 283}{273} \\
& =72.56 \times 10^4 \mathrm{Nm}^{-2}
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift | 3,646 |
luz2v4b2 | chemistry | solutions | osmotic-pressure | <p>$$0.05 \mathrm{M} \mathrm{~CuSO}_4$$ when treated with $$0.01 \mathrm{M} \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7$$ gives green colour solution of $$\mathrm{Cu}_2 \mathrm{Cr}_2 \mathrm{O}_7$$. The two solutions are separated as shown below :
[SPM : Semi Permeable Membrane]</p>
<p><img src="data:image/png;base64,UklG... | [{"identifier": "A", "content": "Molarity of $$\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$$ solution is lowered.\n"}, {"identifier": "B", "content": "Green colour formation observed on side Y.\n"}, {"identifier": "C", "content": "Molarity of $$\\mathrm{CuSO}_4$$ solution is lowered.\n"}, {"identifier": "D", "content": ... | ["C"] | null | <p>Osmosis leads to the net movement of water molecule from low osmotic pressure to high osmotic pressure. Since the iM value of $$0.05 \mathrm{~M} \mathrm{~CuSO}_4$$ solution is higher hence it has higher osmotic pressure so the water moves towards $$\mathrm{CuSO}_4$$ solution leads to drop of its molarity. The solute... | mcq | jee-main-2024-online-9th-april-morning-shift | 3,647 |
lv7v47rj | chemistry | solutions | osmotic-pressure | <p>An artificial cell is made by encapsulating $$0.2 \mathrm{~M}$$ glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a $$0.05 \mathrm{~M}$$ solution of $$\mathrm{NaCl}$$ at $$300 \mathrm{~K}$$ is ________ $$\times 10^{-1}$$ bar. (nearest integer).... | [] | null | 25 | <p>$$\begin{aligned}
& \pi=\left(\mathrm{i_1 C_1-i_2 C_2}\right) \mathrm{R T}=(1 \times 0.2-2 \times 0.05) 0.083 \times 300 \\
& =2.5 \text { bar }=25 \times 10^{-1} \text { bar }
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-morning-shift | 3,648 |
tdNQsECygguVLg9X | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | In a mixture of A and B, components show negative deviation when : | [{"identifier": "A", "content": "$$\\Delta V_{mix}$$ > 0, $$\\Delta S_{mix}$$ > 0"}, {"identifier": "B", "content": "$$\\Delta V_{mix}$$ = 0, $$\\Delta S_{mix}$$ > 0"}, {"identifier": "C", "content": "A - B interaction is weaker than A - A and B - B interaction"}, {"identifier": "D", "content": "A - B interact... | ["D"] | null | <p>In the context of solutions and mixtures, a negative deviation from Raoult's law occurs when the interaction between different molecules (A-B interaction) is stronger than the interaction between similar molecules (A-A and B-B interactions). This is because the molecules prefer to interact with each other rather... | mcq | aieee-2002 | 3,649 |
LWh53htJcfZZITWZ | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | Which of the following liquid pairs shows a positive deviation from Raoult’s law? | [{"identifier": "A", "content": "Water \u2013 hydrochloric acid "}, {"identifier": "B", "content": "Acetone \u2013 chloroform"}, {"identifier": "C", "content": "Water \u2013 nitric acid "}, {"identifier": "D", "content": "Benzene \u2013 methanol "}] | ["D"] | null | <b>NOTE :</b> Positive deviations are shown by such solutions in which solvent-solvent and solute-solute interactions are stronger than the solvent interactions. In such solution, the interactions among molecules becomes weaker. Therefore their escaping tendency increases which results in the increase in their partial ... | mcq | aieee-2004 | 3,650 |
9yNGNtOML1UDdjFp | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | For which of the following parameters the structural isomers C<sub>2</sub>H<sub>5</sub>OH and CH<sub>3</sub>OCH<sub>3</sub> would
be expected to have the same values? (Assume ideal behaviour) | [{"identifier": "A", "content": "Heat of vaporization "}, {"identifier": "B", "content": "Gaseous densities at the same temperature and pressure"}, {"identifier": "C", "content": "Boiling points"}, {"identifier": "D", "content": "Vapour pressure at the same temperature "}] | ["B"] | null | Gaseous densities of ethanol and dimethyl ether would be same at same temperature and pressure. The heat of vaporisation , $$V.P.$$ and $$b.pts$$ will differ due to $$H$$-bonding in ethanol. | mcq | aieee-2004 | 3,651 |
1u1t3ZHohgiyEWLX | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | Benzene and toluene form nearly ideal solutions. At 20 <sup>o</sup>C, the vapour pressure of
benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of
benzene at 20 <sup>o</sup>C for a solution containing 78 g of benzene and 46 g of toluene in
torr is | [{"identifier": "A", "content": "50"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "53.5"}, {"identifier": "D", "content": "37.5"}] | ["A"] | null | Given, Vapour pressure of benzene $$ = 75\,\,torr$$
<br><br>Vapour pressure of toluene $$ = 22\,\,torr$$
<br><br>mass of benzene in $$=78g$$
<br><br>hence moles of benzene $$ = {{78} \over {78}} = 1\,mole$$
<br><br>(mol. wt of benzene $$=78$$)
<br><br>mass of toluene in solution $$=46g$$
<br><br>hence moles of tolue... | mcq | aieee-2005 | 3,652 |
kkM4jRkZl5GzAZIJ | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour
pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure
(in mm) at the same temperature will be | [{"identifier": "A", "content": "360"}, {"identifier": "B", "content": "350"}, {"identifier": "C", "content": "300"}, {"identifier": "D", "content": "700"}] | ["B"] | null | $$P_A^ \circ = ?,\,\,$$
<br><br>Given $$P_B^ \circ = 200mm,\,\,{x_A} = 0.6$$
<br><br>$${x_B} = 1 - 0.6 = 0.4,\,\,P = 290$$
<br><br>$$P = {P_A} + {P_B} = P_A^ \circ {x_A} + P_B^ \circ {x_B}$$
<br><br>$$ \Rightarrow 290 = P_A^ \circ \times 0.6 + 200 \times 0.4$$
<br><br>$$\therefore$$ $$\,\,\,\,\,\,\,\,$$ $$P_A^ \cir... | mcq | aieee-2007 | 3,654 |
ssn7tyGt7YEUBPqG | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | At 80<sup>o</sup>C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80<sup>o</sup>C and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg) | [{"identifier": "A", "content": "52 mol percent "}, {"identifier": "B", "content": "34 mol percent "}, {"identifier": "C", "content": "48 mol percent"}, {"identifier": "D", "content": "50 mol percent"}] | ["D"] | null | At $$1$$ atmospheric pressure the boiling point of mixture is $${80^ \circ }C.$$
<br><br>At boiling point the vapour pressure of mixture, $${P_T} = 1$$
<br><br>atmosphere $$ = 760\,mm\,Hg.$$
<br><br>Using the relation,
<br><br>$${P_T} = P_A^ \circ {X_A} + P_B^ \circ {X_B},\,\,$$ we get
<br><br>$${P_T} = 520{X_A} + 100... | mcq | aieee-2008 | 3,655 |
5uJrBAVb3j7ItlHp | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | The vapour pressure of water at 20<sup>o</sup>C is 17.5 mm Hg. If 18 g of glucose (C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>) is added to 178.2 g of water at 20<sup>o</sup>C, the vapour pressure of the resulting solution will be | [{"identifier": "A", "content": "17.675 mm Hg "}, {"identifier": "B", "content": "15.750 mm Hg "}, {"identifier": "C", "content": "16.500 mm Hg "}, {"identifier": "D", "content": "17.325 mm Hg"}] | ["D"] | null | <b>NOTE :</b> On addition of glucose to water, vapour pressure of water will decrease. The vapour pressure of a solution of glucose in water can be calculated using the relation
<br><br>$${{{P^ \circ } - {P_S}} \over {{P_S}}} = {{Moles\,\,of\,\,glu\cos e\,\,in\,\,solution} \over {moles\,\,of\,\,water\,\,in\,\,solution}... | mcq | aieee-2008 | 3,656 |
ILX6KRkuWYKgMWov | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following
statements is correct regarding the behaviour of the solution ? | [{"identifier": "A", "content": "The solution formed is an ideal solution "}, {"identifier": "B", "content": "The solution is non-ideal, showing +ve deviation from Raoult\u2019s law. "}, {"identifier": "C", "content": "The solution is non-ideal, showing \u2013ve deviation from Raoult\u2019s law. "}, {"identifier": "D",... | ["B"] | null | For this solution intermolecular interactions between $$n$$-heptane and ethanol aare weaker than $$n$$-heptane -$$n$$- heptane & ethanol-ethanol interactions hence the solution of $$n$$-heptane and ethanol is non-ideal and shows positive deviation from Raoult's law. | mcq | aieee-2009 | 3,657 |
MyqgZzr0SGZfhccA | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol
of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this
solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X
and Y in their pure sta... | [{"identifier": "A", "content": "200 and 300 "}, {"identifier": "B", "content": "300 and 400"}, {"identifier": "C", "content": "400 and 600 "}, {"identifier": "D", "content": "500 and 600 "}] | ["C"] | null | $${P_{total}} = P_A^ \circ {X_A} + P_B^ \circ {X_B};$$
<br><br>$$\,\,\,\,\,\,\,\,$$ $$550 = P_A^ \circ \times {1 \over 4} + P_B^ \circ \times {3 \over 4}$$
<br><br>$$P_A^ \circ + 3P_B^ \circ = 550 \times 4\,\,...\left( i \right)$$
<br><br>In second case
<br><br>$${P_{total}} = P_A^ \circ \times {1 \over 5} + P_B^ ... | mcq | aieee-2009 | 3,658 |
ixCL3HUWCKtYjHb6 | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two
liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of
the solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol<sup>–1</sup> and o... | [{"identifier": "A", "content": "72.0 kPa"}, {"identifier": "B", "content": "36.1 kPa"}, {"identifier": "C", "content": "96.2 kPa"}, {"identifier": "D", "content": "144.5 kPa"}] | ["A"] | null | $${P_{Total}} = {P_A}^ \circ {x_A} + {P_B}^ \circ {x_B}$$
<br><br>$$ = {P^ \circ }_{Hep\tan e}\,{x_{hep\tan e}} + {P^ \circ }_{Oc\tan e}\,{x_{Oc\tan e}}$$
<br><br>$$ = 105 \times {{25/100} \over {{{25} \over {100}} + {{35} \over {114}}}} + 45 \times {{35/114} \over {{{25} \over {100}} + {{35} \over {114}}}}$$
<br><br>$... | mcq | aieee-2010 | 3,659 |
1xgTTVc3tihLuKOr | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | The vapour pressure of acetone at 20<sup>o</sup>C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in
100 g of acetone at 20<sup>o</sup>C, its vapour pressure was 183 torr. The molar mass (g mol<sup>-1</sup>) of the substance is: | [{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "128"}, {"identifier": "C", "content": "488"}, {"identifier": "D", "content": "32"}] | ["A"] | null | Using relation,
<br><br>$${{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{w_2}{M_1}} \over {{w_1}{M_2}}}$$
<br><br>where $${w_1},$$ $${M_1} = $$ mass in $$g$$
<br><br>and mol. mass of solvent
<br><br>$${w_2},{M_2} = $$ mass in $$g$$
<br><br>and mol. mass of solute
<br><br>Let $${M_2} = x$$
<br><br>$${P^ \circ } = 185\,\,... | mcq | jee-main-2015-offline | 3,660 |
w3wJEpPsCM8YS4Ez | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | 18 g glucose (C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is : | [{"identifier": "A", "content": "76.0 "}, {"identifier": "B", "content": "752.4"}, {"identifier": "C", "content": "759.0"}, {"identifier": "D", "content": "7.6"}] | ["B"] | null | According to Raoult's Law
<br><br>$${{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{W_B} \times {M_A}} \over {{M_B} \times {W_A}}}\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>Here $${P^ \circ } = $$ Vapour pressure of pure solvent,
<br><br>$${P_s} = $$ Vapour pressure of solution
<br><br>$${W_B} = $$ Mass of solute,
... | mcq | jee-main-2016-offline | 3,661 |
RdC7mcDXSuzFMaYlE74U3 | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | A solution is prepared by mixing 8.5 g of CH<sub>2</sub>Cl<sub>2</sub> and 11.95 g of CHCl<sub>3</sub> . If vapour pressure of CH<sub>2</sub>Cl<sub>2</sub>
and CHCl<sub>3</sub> at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl<sub>3</sub> in vapour form is : (Molar mass of Cl = 35.5 g mol<sup>−1... | [{"identifier": "A", "content": "0.162"}, {"identifier": "B", "content": "0.675"}, {"identifier": "C", "content": "0.325"}, {"identifier": "D", "content": "0.486"}] | ["C"] | null | <p>Mole fractions can be calculated as</p>
<p>$${n_{C{H_2}C{l_2}}} = {{8.5} \over {8.5}} = 0.1$$ and $${n_{CHC{l_3}}} = {{11.9} \over {119.5}} = 0.1$$</p>
<p>We know that</p>
<p>$${p_{total}} = p_{C{H_2}C{l_2}}^o \times \,{x_{C{H_2}C{l_2}}} + p_{CHC{l_3}}^o \times \,{x_{CHC{l_3}}}$$</p>
<p>$$ = 415 \times 0.10 + 200 \t... | mcq | jee-main-2017-online-9th-april-morning-slot | 3,662 |
kN5RXVNCwkUaqSfOsf1BU | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are M<sub>x</sub> and M<sub>y</sub>, respectively where M<sub>x</sub> = $${3 \over 4}$$ My. The relative lowering of vapor pressure of the solution in X is ''m... | [{"identifier": "A", "content": "$${4 \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["B"] | null | Relative lowering of vapour pressure,
<br><br> $${{\Delta P} \over P}$$ = $${{{n_2}} \over {{n_1}}}$$
<br><br>n<sub>2</sub> = Number of moles of solute
<br><br>n<sub>1</sub> = Number of moles of solvent.
<br><br>Given that,
<br><br>Here is 5 molal solution, mean... | mcq | jee-main-2018-online-15th-april-evening-slot | 3,663 |
NWznHMOmrbIAE97rJtiNP | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol<sup>-1</sup> ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is : | [{"identifier": "A", "content": "37.5 g"}, {"identifier": "B", "content": "75 g"}, {"identifier": "C", "content": "150 g"}, {"identifier": "D", "content": "50 g"}] | ["C"] | null | Molar mass of octane
<br><br>(C<sub>8</sub> H<sub>18</sub>) = 8 $$ \times $$ 12 + 18 = 114 g/mol
<br><br>Let, $$\omega $$ is the mass of solute.
<br><br>Relative lowering vapour pressure,
<br><br>$${{\Delta P} \over P}$$ = $${{{\omega \over {50}}} \over {{\omega \over {50}} + {{114} \over {114}}}}$$
<br><br>$$ \Ri... | mcq | jee-main-2018-online-16th-april-morning-slot | 3,664 |
KWklP8kxZBhWq5Gzjo3rsa0w2w9jwvebznx | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the
vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be.
(molar mass of urea = 60 g mol<sup>–1</sup>) | [{"identifier": "A", "content": "0.031 mmHg"}, {"identifier": "B", "content": "0.017 mmHg"}, {"identifier": "C", "content": "0.028 mmHg"}, {"identifier": "D", "content": "0.027 mmHg"}] | ["B"] | null | Given that,
<br><br>w<sub>solute</sub> = w<sub>urea</sub> = 0.6 gm
<br><br>w<sub>solvent</sub> = w<sub>H<sub>2</sub>O</sub> = 360 gm
<br><br>p<sup>o</sup> = 35
<br><br>We know,
<br><br>lowering of vapour pressure
<br><br>$$\Delta $$p = x<sub>solute</sub> $$ \times $$ p<sup>o</sup>
<br><br>= $${{{n_{urea}}} \over {{n_... | mcq | jee-main-2019-online-10th-april-morning-slot | 3,665 |
adPGBqmHlWclT9dmmDRye | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vaapor pressures of pure A and pure B are 7 $$ \times $$ 10<sup>3</sup> Pa and 12 $$ \times $$ 10<sup>3</sup> Pa, respectively . The composition of the vapor in equilibriumwith a solution containing 40 mole percent of A at this temper... | [{"identifier": "A", "content": "x<sub>A</sub> = 0.76; x<sub>B</sub> = 0.24 "}, {"identifier": "B", "content": "x<sub>A</sub> = 0.28; x<sub>B</sub> = 0.72"}, {"identifier": "C", "content": "x<sub>A</sub> = 0.4; x<sub>B</sub> = 0.6"}, {"identifier": "D", "content": "x<sub>A</sub> = 0.37; x<sub>B</sub> = 0.63 "}] | ["B"] | null | $${y_A} = {{{P_A}} \over {{P_{Total}}}} = {{P_A^0{x_A}} \over {P_A^0{X_A} \times P_B^0X{}_B}}$$
<br><br>$$ = {{7 \times {{10}^3} \times 0.4} \over {7 \times {{10}^3} \times 0.4 + 12 \times {{10}^3} \times 0.6}}$$
<br><br>$$ = {{2.8} \over {10}} = 0.28$$
<br><br>$${y_B} = 0.72$$ | mcq | jee-main-2019-online-10th-january-morning-slot | 3,667 |
ScWeujDIfg1TcG9q6G7k9k2k5epvacb | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | Two open beakers one containing a solvent and the other containing a mixture of that solvent
with a non voltatile solute are together selated in a container. Over time :
| [{"identifier": "A", "content": "The volume of the solution increases and the volume of the solvent decreases."}, {"identifier": "B", "content": "The volume of solution does not change and the volume of the solvent decreases."}, {"identifier": "C", "content": "The volumer of the solution and the solvent does not change... | ["A"] | null | There will be lowering in vapour pressure for
solution containing non-volatile solute. So,
there will be transfer of solvent molecules
from pure solvent to solution and hence,
volume of beaker containing solvent (pure) will
decrease and volume of beaker containing
solution will increase. | mcq | jee-main-2020-online-7th-january-evening-slot | 3,670 |
bYYg4U9huWqOqJPBD9jgy2xukf9l0nsx | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | At 300 K, the vapour pressure of a solution
containing 1 mole of n-hexane and 3 moles of
n-heptane is 550 mm of Hg. At the same
temperature, if one more mole of n-heptane is
added to this solution, the vapour pressure of
the solution increases by 10 mm of Hg. What is
the vapour pressure in mm Hg of n-heptane in
its pur... | [] | null | 600 | Let X<sub>1</sub> and P$$P_1^o$$
are the mole fraction and vapour
pressure of n-hexane in solution and X<sub>2</sub> and $$P_2^o$$
are the mole fraction and vapour pressure of
n-heptane in solution then
<br><br>550 = $$P_1^o$$$$ \times $$$${1 \over 4}$$ + $$P_2^o$$$$ \times $$$${3 \over 4}$$
<br><br>$$ \Rightarrow $... | integer | jee-main-2020-online-4th-september-morning-slot | 3,671 |
tTBo7z5xTZntIy4s7Hjgy2xukg3f2u1b | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | A set of solutions is prepared using 180 g of
water as a solvent and 10 g of different nonvolatile solutes A, B and C. The relative
lowering of vapour pressure in the presence of
these solutes are in the order :<br/><br/>[Given, molar
mass of A = 100 g mol<sup>–1</sup>; B = 200 g mol<sup>–1</sup>;
C = 10,000 g mol<sup>... | [{"identifier": "A", "content": "A > C > B"}, {"identifier": "B", "content": "C > B > A"}, {"identifier": "C", "content": "A > B > C"}, {"identifier": "D", "content": "B > C > A"}] | ["C"] | null | Relative lowering in vapour pressure (RLVP)
<br><br>= $${{P - {P_s}} \over P} = {n \over {n + N}}$$
<br><br>n $$ \to $$ moles of solute
<br>N $$ \to $$ moles of solvent
<br><br>$$ \therefore $$ (RLVP)<sub>A</sub> = $${{{{10} \over {100}}} \over {{{10} \over {100}} + {{180} \over {18}}}}$$
<br><br>(RLVP)<sub>B</sub> = $... | mcq | jee-main-2020-online-6th-september-evening-slot | 3,672 |
0OLQ5S531KFDgV60XJ1kmiv4jg3 | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | At 363 K, the vapour pressure of A is 21 kPa and that of B is 18 kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is ___________ kPa. (Round off to the Nearest Integer). | [] | null | 19 | $${X_A} = {1 \over {1 + 2}} = {1 \over 3}$$<br><br>$${X_B} = {2 \over 3}$$<br><br>$$P_A^o = 21kPa$$<br><br>$$P_B^o = 18kPa$$<br><br>$${P_{total}} = P_A^o{X_A} + P_B^o{X_B}$$<br><br>$$ = 21 \times {1 \over 3} + 18 \times {2 \over 3}$$<br><br>$$ = 7 + 12$$<br><br>$$ = 19 kPa$$ | integer | jee-main-2021-online-16th-march-evening-shift | 3,673 |
1krrl0i24 | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | The vapour pressures of A and B at 25$$^\circ$$C are 90 mm Hg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6, then the mole fraction of B in the vapour phase is x $$\times$$ 10<sup>$$-$$1</sup>. The value of x is ___________. (Nearest integer) | [] | null | 1 | x<sub>A</sub> = 0.6<br><br>P<sub>T</sub> = x<sub>A</sub>P<sub>A</sub><sup>o</sup> + x<sub>B</sub>P<sub>B</sub><sup>o</sup><br><br>= 0.6 $$\times$$ 90 + 0.4 $$\times$$ 15<br><br>= 54 + 6 = 60<br><br>x<sub>A</sub>P<sub>A</sub><sup>o</sup> = y<sub>A</sub>P<sub>T</sub><br><br>0.6 $$\times$$ 90 = y<sub>A</sub>(60)<br><br>$$... | integer | jee-main-2021-online-20th-july-evening-shift | 3,675 |
1l56bhnou | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | <p>The vapour pressures of two volatile liquids A and B at 25$$^\circ$$C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is $${x \over {17}}$$. The value of x is ______________.</p> | [] | null | 14 | $$
\begin{aligned}
&\frac{\mathrm{y}_{\mathrm{B}}}{1-\mathrm{y}_{\mathrm{B}}}=\frac{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}\left[\frac{\mathrm{X}_{\mathrm{B}}}{1-\mathrm{X}_{\mathrm{B}}}\right] \\\\
&\Rightarrow \frac{\mathrm{y}_{\mathrm{B}}}{1-\mathrm{y}_{\mathrm{B}}}=\frac{100}{50}... | integer | jee-main-2022-online-28th-june-morning-shift | 3,676 |
1l6kujy4i | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | <p>When a certain amount of solid A is dissolved in $$100 \mathrm{~g}$$ of water at $$25^{\circ} \mathrm{C}$$ to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $$23.76 \,\mathrm{mmHg}$$. The number of moles of solute A added... | [] | null | 6 | $\because$ Diliute solution given:<br/><br/>
$$
\frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0} \sim \frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }}
$$<br/><br/>
$$
\frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text {... | integer | jee-main-2022-online-27th-july-evening-shift | 3,677 |
1l6nxmtwu | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | <p>A gaseous mixture of two substances A and B, under a total pressure of $$0.8$$ atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is $$0.5$$ in the vapour phase and $$0.2$$ in the liquid phase. The vapour pressure of pure liquid $$\mathrm{A}$$ is __________ atm. (Nearest integer)</... | [] | null | 2 | Given that $X_{A}=0.2, Y_{A}=0.5, P_{T}=0.8 \mathrm{~atm}$
<br/><br/>
We know that $P_{A}=Y_{A} \times P_{T}$
<br/><br/>
$$
P_{A}=0.5 \times 0.8=0.4
$$
<br/><br/>
Now $P_{A}=X_{A} \times P_{A}^{\circ} \Rightarrow P_{A}^{\circ}=\frac{0.4}{0.2}=2 \mathrm{~atm}$
| integer | jee-main-2022-online-28th-july-evening-shift | 3,678 |
1ldwvqdbz | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | <p>The total pressure observed by mixing two liquids A and B is 350 mm Hg when their mole fractions are 0.7 and 0.3 respectively.</p>
<p>The total pressure becomes 410 mm Hg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B. The vapour pressure of pure A is __________ mm Hg. (Nearest integer)</p... | [] | null | 314 | Let $V . P$ of pure $A$ be $P_A^0$<br/><br/>
Let $V . P$ of pure $B$ be $P_B^0$<br/><br/>
When $\mathrm{X}_{\mathrm{A}}=0.7 \& \mathrm{X}_{\mathrm{B}}=0.3$<br/><br/>
$$
\begin{aligned}
& \mathrm{P}_{\mathrm{s}}=350 \\\\
& \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.7+\mathrm{P}_{\mathrm{B}}^0 \times 0.3=350
\end{ali... | integer | jee-main-2023-online-24th-january-evening-shift | 3,679 |
lgo0eo44 | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | The vapour pressure of $30 \%(\mathrm{w} / \mathrm{v})$ aqueous solution of glucose is __________ $\mathrm{mm} ~\mathrm{Hg}$ at $25^{\circ} \mathrm{C}$.
<br/><br/>
[Given : The density of $30 \%$ (w/v), aqueous solution of glucose is $1.2 \mathrm{~g} \mathrm{~cm}^{-3}$ and vapour pressure of pure water is $24 \mathrm{~... | [] | null | 23 | The given solution is a $30 \%$ (w/v) aqueous solution of glucose. This means that $30\ \mathrm{g}$ of glucose is dissolved in $100\ \mathrm{mL}$ of the solution. Since the density of the solution is $1.2\ \mathrm{g/mL}$, the weight of $100\ \mathrm{mL}$ of the solution is:
<br/><br/>
$$\mathrm{Wt.~of~solution} = 100\ ... | integer | jee-main-2023-online-15th-april-morning-shift | 3,680 |
1lgsygi7r | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | <p>What weight of glucose must be dissolved in $$100 \mathrm{~g}$$ of water to lower the vapour pressure by $$0.20 \mathrm{~mm} ~\mathrm{Hg}$$ ?</p>
<p>(Assume dilute solution is being formed)</p>
<p>Given : Vapour pressure of pure water is $$54.2 \mathrm{~mm} ~\mathrm{Hg}$$ at room temperature. Molar mass of glucose i... | [{"identifier": "A", "content": "3.69 g"}, {"identifier": "B", "content": "2.59 g"}, {"identifier": "C", "content": "3.59 g"}, {"identifier": "D", "content": "4.69 g"}] | ["A"] | null | The lowering of vapor pressure of a solvent by a nonvolatile solute is given by Raoult's law:
<br/><br/>$$\Delta P = x_{\text{solute}} \cdot P_0$$
<br/><br/>where $\Delta P$ is the change in vapor pressure, $x_{\text{solute}}$ is the mole fraction of the solute, and $P_0$ is the vapor pressure of the pure solvent.
<... | mcq | jee-main-2023-online-11th-april-evening-shift | 3,682 |
1lh29pkyt | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | <p>Mass of Urea $$\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$$ required to be dissolved in $$1000 \mathrm{~g}$$ of water in order to reduce the vapour pressure of water by $$25 \%$$ is _________ g. (Nearest integer)</p>
<p>Given: Molar mass of N, C, O and H are $$14,12,16$$ and $$1 \mathrm{~g} \mathrm{~mol}^{-1}$$... | [] | null | 1111 | <p>Given:</p>
<ul>
<li>Vapor pressure reduction: $25\%$ ($$0.75$$ times the vapor pressure of pure water)</li><br/>
<li>Molar mass of water ($$\text{H}_2\text{O}$$): $$18 \, \text{g/mol}$$</li><br/>
<li>Mass of solvent (water): $$1000 \, \text{g}$$</li>
</ul>
<p>Using Raoult's law:<br/><br/>
$$\frac{P^0 - P_s}{P_s} = \... | integer | jee-main-2023-online-6th-april-morning-shift | 3,683 |
jaoe38c1lse76yow | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | <p>Identify the mixture that shows positive deviations from Raoult's Law</p> | [{"identifier": "A", "content": "$$\\left(\\mathrm{CH}_3\\right)_2 \\mathrm{CO}+\\mathrm{CS}_2$$\n"}, {"identifier": "B", "content": "$$\\left(\\mathrm{CH}_3\\right)_2 \\mathrm{CO}+\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{NH}_2$$\n"}, {"identifier": "C", "content": "$$\\mathrm{CHCl}_3+\\mathrm{C}_6 \\mathrm{H}_6$$\n"}, {"i... | ["A"] | null | <p>$$\left(\mathrm{CH}_3\right)_2 \mathrm{CO}+\mathrm{CS}_2$$ Exibits positive deviations from Raoult's Law</p> | mcq | jee-main-2024-online-31st-january-morning-shift | 3,684 |
luy1mwup | chemistry | solutions | relative-lowering-of-vapour-pressure-and-roult’s-law | <p>The vapor pressure of pure benzene and methyl benzene at $$27^{\circ} \mathrm{C}$$ is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapor phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is _________ $$\times 10^{-2}$$... | [] | null | 23 | <p>To find the mole fraction of methyl benzene in the vapor phase when it is in equilibrium with an equimolar (equal mole) mixture of benzene and methyl benzene at $$27^{\circ}C$$, we can apply Raoult's law for an ideal solution. Given the vapor pressures of pure benzene and methyl benzene are 80 Torr and 24 Torr, resp... | integer | jee-main-2024-online-9th-april-evening-shift | 3,685 |
4XWEvDZgov05rNf5 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | With increase of temperature, which of these changes? | [{"identifier": "A", "content": "molality"}, {"identifier": "B", "content": "weight fraction of solute"}, {"identifier": "C", "content": "molarity"}, {"identifier": "D", "content": "mole fraction"}] | ["C"] | null | Molarity = <span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;">Number of moles of solute</div>
<div style="text-align: center;">Volume of solution (in L)</div>
</span>
<br><br>Molality = <span style="display: inline-block;vertical-align:... | mcq | aieee-2002 | 3,686 |
JLmhW0rwv9npaRo3 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H<sub>3</sub>PO<sub>3</sub>), the volume of 0.1 M aqueous KOH solution required is | [{"identifier": "A", "content": "40 mL"}, {"identifier": "B", "content": "20 mL"}, {"identifier": "C", "content": "10 mL"}, {"identifier": "D", "content": "60 mL"}] | ["A"] | null | H<sub>3</sub>PO<sub>3</sub> + 2KOH $$ \to $$ K<sub>2</sub>HPO<sub>3</sub> + 2H<sub>2</sub>O
<br><br>Let the volume of KOH solution is = V mL = $${V \over {1000}}$$ litre
<br><br>Volume of H<sub>3</sub>PO<sub>3</sub> = 20 mL = $${{20} \over {1000}}$$ litre
<br><br>No of moles of H<sub>3</sub>PO<sub>3</sub> present = $${... | mcq | aieee-2004 | 3,688 |
QjmzkdZvdbblx3Fb | chemistry | some-basic-concepts-of-chemistry | concentration-terms | 6.02 $$\times$$ 10<sup>20</sup> molecules of urea are present in 100 ml of its solution. The concentration of urea solution is <br/>(Avogadro constant, N<sub>A</sub> = 6.02 $$\times$$ 10<sup>23</sup> mol<sup>-1</sup>) | [{"identifier": "A", "content": "0.02 M"}, {"identifier": "B", "content": "0.01 M"}, {"identifier": "C", "content": "0.001 M"}, {"identifier": "D", "content": "0.1 M"}] | ["B"] | null | Moles (n) = $${{molecules} \over {{N_A}}}$$<br><br>
$$ \Rightarrow $$ $${{6.02 \times {{10}^{20}}} \over {6.02 \times {{10}^{23}}}}$$<br> <br>
$$ \Rightarrow $$ 0.001<br><br>
Concentration (M) = $${{moles} \over {volume}}$$<br><br>
Given volume of urea is 100 ml = 0.1 litre<br><br>
[<b>Note :</b> While calculating C... | mcq | aieee-2004 | 3,689 |
QB3J5rvjWKHvWeza | chemistry | some-basic-concepts-of-chemistry | concentration-terms | Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture? | [{"identifier": "A", "content": "2.70 M"}, {"identifier": "B", "content": "1.344 M"}, {"identifier": "C", "content": "1.50 M"}, {"identifier": "D", "content": "1.20 M"}] | ["B"] | null | <b>Short Cut Method :</b> Molarity, Normality, Average Atomic Mass, Average Molar Mass, Average Density, Average Vapour Pressure all of those values will be between the solutions which are going to be mixed.
<br><br>Here molarity of first solution is 1.2 and molarity of second solution is 1.5.
<br>So the molarity of th... | mcq | aieee-2005 | 3,690 |
844sUB4RFISSFOXR | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The density (in g mL<sup>-1</sup>) of a 3.60 M sulphuric acid solution that is 29% H<sub>2</sub>SO<sub>4</sub> (molar mass = 98 g mol<sup>-1</sup>) by mass will be | [{"identifier": "A", "content": "1.45"}, {"identifier": "B", "content": "1.64"}, {"identifier": "C", "content": "1.88"}, {"identifier": "D", "content": "1.22"}] | ["D"] | null | Important formula of Molarity(M) when % w/w is given
<br><br>M = $${{10 \times \% w/w \times d} \over {{M_{solute}}}}$$
<br><br>Here M = 3.6, % w/w = 29, d = density, M<sub>solute</sub> = 98
<br><br>$$\therefore$$ 3.6 = $${{10 \times 29 \times d} \over {98}}$$
<br><br>$$ \Rightarrow d = 1.22$$ | mcq | aieee-2007 | 3,692 |
x4Gyy58dmTWLt229 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | A 5.2 molal aqueous solution of methyl alcohol, CH<sub>3</sub>OH, is supplied. What is the mole fraction of methyl
alcohol in the solution ? | [{"identifier": "A", "content": "0.190"}, {"identifier": "B", "content": "0.086"}, {"identifier": "C", "content": "0.050"}, {"identifier": "D", "content": "0.100"}] | ["B"] | null | The formula between Mole fraction and Molality is
$$$m = {{{X_{solute}}} \over {{X_{solvent}}}} \times {{1000} \over {{M_{solvent}}}}$$$
<br>$$m$$ = molality, X<sub>solute</sub> = Mole fraction of solute, X<sub>solvent</sub> = Mole fraction of solvent, M<sub>solvent</sub> = Molar mass of solvent
<br><br>Here solute is ... | mcq | aieee-2011 | 3,693 |
gxeMWkrz1ZAzgkNf | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is
1.15 g/mL. The molarity of this solution is : | [{"identifier": "A", "content": "0.50 M"}, {"identifier": "B", "content": "1.78 M"}, {"identifier": "C", "content": "1.02 M"}, {"identifier": "D", "content": "2.05 M"}] | ["D"] | null | We know molarity (M) = $${{no\,of\,moles\,of\,solute} \over {volume\,of\,solution\,in\,litre}}$$
<br><br>Moles of solute = $${{120} \over {60}}$$ = 2
<br><br>Mass of solution = 1000 + 120 = 1120 gm
<br><br/>Density of solution = 1.15 g/mL
<br><br>$$\therefore$$ volume of solution = $${{1120} \over {1.15}}$$ = 973.9 mL ... | mcq | aieee-2012 | 3,694 |
3ERrWbhv3mTCxxp8 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be: | [{"identifier": "A", "content": "1.00 M"}, {"identifier": "B", "content": "1. 75 M "}, {"identifier": "C", "content": "0.975 M "}, {"identifier": "D", "content": "0.875 M"}] | ["D"] | null | The formula for molarity of mixture of two substance is = $${{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}$$
<br><br>Here $${M_1} = 0.5$$, $${V_1} = 750$$, $${M_2} = 2$$, $${V_2} = 250$$
<br><br>$$\therefore$$ Molarity of mixture = $${{0.5 \times 750 + 2 \times 250} \over {750 + 250}}$$ <br><br>= $${{375 + 500} \ove... | mcq | jee-main-2013-offline | 3,695 |
PVKotErnxHw0lyKaAF2oI | chemistry | some-basic-concepts-of-chemistry | concentration-terms | Excess of NaOH (aq) was added to 100 mL of FeCl<sub>3</sub> (aq) resulting into 2.14 g of Fe(OH)<sub>3</sub> . The molarity of FeCl<sub>3</sub> (aq) is :
<br/><br/>(Given molar mass of Fe = 56 g mol<sup>−1</sup> and molar mass of Cl = 35.5 g mol<sub>−1</sub>)
| [{"identifier": "A", "content": "0.2 M"}, {"identifier": "B", "content": "03 M"}, {"identifier": "C", "content": "0.6 M"}, {"identifier": "D", "content": "1.8 M"}] | ["A"] | null | 3 NaOH (aq.) + FeCl<sub>3</sub>(aq) $$ \to $$ Fe(OH)<sub>3</sub>(s)+ 3 NaCl(aq).
<br><br>Moles of Fe(OH)<sub>3</sub> = $${{2.14} \over {107}}$$ = 2 $$ \times $$ 10<sup>$$-$$2</sup>
<br><br>1 mole of Fe(OH)<sub>3</sub> is obtained from = 1 mole of FeCl<sub>3</sub>
<br><br>$$\therefore\,\,\,$$ 2 $$ \times $$ 10<sup>$$-$$... | mcq | jee-main-2017-online-8th-april-morning-slot | 3,696 |
K9SYpSYXNXvIRgks | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The ratio of mass percent of C and H of an organic compound (C<sub>X</sub>H<sub>Y</sub>O<sub>Z</sub>) is 6 : 1. If one molecule of the above compound (C<sub>X</sub>H<sub>Y</sub>O<sub>Z</sub>) contains half as much oxygen as required to burn one molecule of compound C<sub>X</sub>H<sub>Y</sub> completely to CO<sub>2</sub... | [{"identifier": "A", "content": "C<sub>2</sub>H<sub>4</sub>O<sub>3</sub>"}, {"identifier": "B", "content": "C<sub>3</sub>H<sub>6</sub>O<sub>3</sub>"}, {"identifier": "C", "content": "C<sub>2</sub>H<sub>4</sub>O"}, {"identifier": "D", "content": "C<sub>3</sub>H<sub>4</sub>O<sub>2</sub>"}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264447/exam_images/omb7cmyem4cuacaeyca0.webp" loading="lazy" alt="JEE Main 2018 (Offline) Chemistry - Some Basic Concepts of Chemistry Question 178 English Explanation">
<br><br>$$\therefore\,\,\,$$ The ratio of no of atoms of C and... | mcq | jee-main-2018-offline | 3,697 |
Y43xWlFoQQdJAQFKzvGzu | chemistry | some-basic-concepts-of-chemistry | concentration-terms | A solution of sodium sulphate contains 92 g of Na<sup>+</sup> ions per kilogram of water. The molality of Na<sup>+</sup> ions in that solution in mol kg<sup>$$-$$1</sup> is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "16"}] | ["B"] | null | Molality of Na<sup>+</sup> = $${{Moles\,\,of\,\,N{a^ + }} \over {mass\,\,of\,\,{H_2}O\,\,in\,\,kg}}$$
<br><br>Moles of Na<sup>+</sup> = $${{92} \over {23}}$$ = 4
<br><br>Mass of H<sub>2</sub>O = 1 kg
<br><br>$$ \therefore $$ Molality = $${4 \over 1}$$ = 4 | mcq | jee-main-2019-online-9th-january-morning-slot | 3,698 |
u0OEQcFRPlqr9tiY883rsa0w2w9jx4rwcfo | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg<sup>–1</sup>
) of the
aqueous solution is : | [{"identifier": "A", "content": "13.88 \u00d7 10<sup>\u20131</sup>"}, {"identifier": "B", "content": "13.88 \u00d7 10<sup>\u20133</sup>"}, {"identifier": "C", "content": "13.88"}, {"identifier": "D", "content": "13.88 \u00d7 10<sup>\u20132</sup>"}] | ["C"] | null | We know,
<br><br>molality(m) = $${{{x_{solute}}} \over {{x_{solvent}}}} \times {{1000} \over {molar\,weight\,of\,solvent}}$$
<br><br>Here solvent is H<sub>2</sub>O
<br><br>x<sub>solvent</sub> = 0.8
<br><br>$$ \therefore $$ x<sub>solute</sub> = 0.2
<br><br>$$ \therefore $$ Molality = $${{0.2} \over {0.8}} \times {{1000}... | mcq | jee-main-2019-online-12th-april-morning-slot | 3,699 |
ov1PKjoChpAPDQ3efSBqx | chemistry | some-basic-concepts-of-chemistry | concentration-terms | What would be the molality of 20% (mass/
mass) aqueous solution of KI?<br/>
(molar mass of KI = 166 g mol<sup>–1</sup>) | [{"identifier": "A", "content": "1.51"}, {"identifier": "B", "content": "1.35"}, {"identifier": "C", "content": "1.08"}, {"identifier": "D", "content": "1.48"}] | ["A"] | null | 20% (mass/
mass) aqueous solution means,
<br><br>100 g solution contains 20 g KI
<br><br>$$ \therefore $$ Mass of solvent = 100 – 20 = 80 g
<br><br>So, 20 g of KI present in 80 g of solvent.
<br><br>$$ \therefore $$ $${{20} \over {166}}$$ moles of KI present in 80 g of solvent.
<br><br>We know,
<br><br> Molality (m) = ... | mcq | jee-main-2019-online-9th-april-evening-slot | 3,700 |
bKxsooNQMecIGrtOBKQyP | chemistry | some-basic-concepts-of-chemistry | concentration-terms | 8 g of NaOH is dissolved in 18g of H<sub>2</sub>O. Mole fraction of NaOH in solution and molality (in mol kg<sup>–1</sup>) of the solution respectively are - | [{"identifier": "A", "content": "0.2, 11.11"}, {"identifier": "B", "content": "0.167, 22.20"}, {"identifier": "C", "content": "0.167, 11.11"}, {"identifier": "D", "content": "0.2, 22.20"}] | ["C"] | null | 8 gm of NaOH = $${8 \over {40}}$$ = 0.2 mol of NaOH
<br><br>18 gm of H<sub>2</sub>O = $${18 \over {18}}$$ = 1 mol of H<sub>2</sub>O
<br><br>$$ \therefore $$ Total mole = 1 + 0.2  ... | mcq | jee-main-2019-online-12th-january-evening-slot | 3,701 |
rlNzEv7uCwtb5vrmG7L1P | chemistry | some-basic-concepts-of-chemistry | concentration-terms | 25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solutions ? | [{"identifier": "A", "content": "50 mL"}, {"identifier": "B", "content": "12.5 mL"}, {"identifier": "C", "content": "25 mL"}, {"identifier": "D", "content": "75 mL"}] | ["C"] | null | 2HCl(aq) + Na<sub>2</sub>CO<sub>3</sub>(aq) $$ \to $$ H<sub>2</sub>CO<sub>3</sub> + NaCl
<br><br><span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;">moles of HCl</div>
<div style="text-align: center;">2</div>
</span> = <span style="displ... | mcq | jee-main-2019-online-11th-january-evening-slot | 3,702 |
hFSv53q3s7OTcGsqIre6r | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The amount of sugar (C<sub>12</sub>H<sub>22</sub>O<sub>11</sub>) required to prepare 2L of its 0.1 M aqueous solution is : | [{"identifier": "A", "content": "17.1 g"}, {"identifier": "B", "content": "34.2 g"}, {"identifier": "C", "content": "68.4 g"}, {"identifier": "D", "content": "136.8 g"}] | ["C"] | null | Molarity = $${{{{(n)}_{solute}}} \over {{V_{solution}}(in\,\,lit)}}$$
<br><br>0.1 = $${{wt./342} \over 2}$$
<br><br>wt (C<sub>12</sub>H<sub>22</sub>O<sub>11</sub>) = 68.4 gram | mcq | jee-main-2019-online-10th-january-evening-slot | 3,703 |
z0nMxb9i873GWe6YLXjgy2xukft5pysc | chemistry | some-basic-concepts-of-chemistry | concentration-terms | A solution of two components containing
n<sub>1</sub> moles of the 1<sup>st</sup> component and n<sub>2</sub> moles of
the 2<sup>nd</sup> component is prepared. M<sub>1</sub> and M<sub>2</sub> are
the molecular weights of component 1 and 2
respectively. If d is the density of the solution
in g mL<sup>–1</sup>, C<sub>2<... | [{"identifier": "A", "content": "$${C_2} = {{1000{x_2}} \\over {{M_1} + {x_2}\\left( {{M_2} - {M_1}} \\right)}}$$"}, {"identifier": "B", "content": "$${C_2} = {{1000d{x_2}} \\over {{M_1} + {x_2}\\left( {{M_2} - {M_1}} \\right)}}$$"}, {"identifier": "C", "content": "$${C_2} = {{d{x_2}} \\over {{M_1} + {x_2}\\left( {{M_2... | ["B"] | null | <table class="tg">
<thead>
<tr>
<th class="tg-baqh"></th>
<th class="tg-baqh"><b>1<sup>st</sup> component</b></th>
<th class="tg-baqh"><b>2<sup>nd</sup> component</b></th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh"><b>Mole</b></td>
<td class="tg-baqh">n<sub>1</sub></td>
<td class="tg-... | mcq | jee-main-2020-online-6th-september-morning-slot | 3,704 |
2srajthxeOV333nomnjgy2xukf9l5wrn | chemistry | some-basic-concepts-of-chemistry | concentration-terms | A 20.0 mL solution containing 0.2 g impure
H<sub>2</sub>O<sub>2</sub> reacts completely with 0.316 g of KMnO<sub>4</sub>
in acid solution. The purity of H<sub>2</sub>O<sub>2</sub> (in %) is
_____________
<br/>(mol. wt. of H<sub>2</sub>O<sub>2</sub> = 34; mol. wt. of
KMnO<sub>4</sub> = 158) | [] | null | 85 | 5H<sub>2</sub>O<sub>2</sub> + 2MnO<sub>4</sub><sup>-</sup> + 6H<sup>+</sup> $$ \to $$ 2Mn<sup>2+</sup> + 5O<sub>2</sub> + 8H<sub>2</sub>O
<br><br>Moles of KMnO<sub>4</sub> = $${{0.316} \over {158}}$$ = 2 $$ \times $$ 10<sup>-3</sup>
<br><br>Equivalents of H<sub>2</sub>O<sub>2</sub> = Equivalent of KMnO<sub>4</sub>
<br>... | integer | jee-main-2020-online-4th-september-morning-slot | 3,705 |
sAa1Qc9FeH0mL5N0Ag7k9k2k5llryvo | chemistry | some-basic-concepts-of-chemistry | concentration-terms | 10.30 mg of O<sub>2</sub> is dissolved into a liter of sea
water of density 1.03 g/mL. The concentration
of O<sub>2</sub> in ppm is__________. | [] | null | 10 | 1030 gm of sea water contains = 10.3 × 10<sup>–3</sup> gm
<br><br>10<sup>6</sup>
gm of sea water contains = $${{10.3 \times {{10}^{ - 3}}} \over {1030}} \times {10^6}$$ = 10 ppm | integer | jee-main-2020-online-9th-january-evening-slot | 3,707 |
pSpi4IXasGNW4SUBXGjgy2xukeyi08ck | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The ratio of the mass percentages of ‘C & H’
and ‘C & O’ of a saturated acyclic organic
compound ‘X’ are 4 : 1 and 3 : 4 respectively.
Then, the moles of oxygen gas required for
complete combustion of two moles of organic
compound ‘X’ is ________. | [] | null | 5 | Let the organic compound X is = C<sub>x</sub>H<sub>y</sub>O<sub>z</sub>
<br><br>Here moles of C = x, moles of H = y, moles of O = z
<br><br>Given, $${{{W_C}} \over {{W_H}}} = {4 \over 1}$$
<br><br>$$ \therefore $$ $${x \over y} = {{{{{W_C}} \over {12}}} \over {{{{W_H}} \over 1}}}$$ = $${4 \over 1} \times {1 \over {12}}... | integer | jee-main-2020-online-2nd-september-evening-slot | 3,709 |
cUVTb5iR4DrE7DqM7D7k9k2k5idqbnl | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The hardness of a water sample containing
10<sup>–3</sup> M MgSO<sub>4</sub> expressed as CaCO<sub>3</sub> equivalents
(in ppm) is ______.<br/>
(molar mass of MgSO<sub>4</sub> is 120.37 g/mol) | [] | null | 100 | Equivalance of MgSO<sub>4</sub> = Equivalance of CaCO<sub>3</sub>
<br><br>$$ \Rightarrow $$ 10<sup>-3</sup> $$ \times $$ 2 = [CaCO<sub>3</sub>] $$ \times $$ 2
<br><br>$$ \Rightarrow $$ [CaCO<sub>3</sub>] = 10<sup>-3</sup> M = 10<sup>-3</sup> mol/lit
<br><br>= 10<sup>-3</sup> $$ \times $$ 100 g/lit
<br><br>= 100 mg/lit
... | integer | jee-main-2020-online-9th-january-morning-slot | 3,711 |
IN4NQ5VLOi58nyG0p5jgy2xukf2cs80s | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The volume strength of 8.9 M H<sub>2</sub>O<sub>2</sub>
solution calculated at 273 K and 1 atm is ______. (R = 0.0821 L
atm K<sup>-1</sup> mol<sup>-1</sup>) (rounded off ot the nearest integer) | [] | null | 100 | Volume strength of H<sub>2</sub>O<sub>2</sub> at 1 atm
273 kelvin <br>= M × 11.2 = 8.9 × 11.2 = 99.68 $$ \simeq $$ 100 | integer | jee-main-2020-online-3rd-september-morning-slot | 3,713 |
itCVYu94Chhuq3eClp1klsc9bpt | chemistry | some-basic-concepts-of-chemistry | concentration-terms | In basic medium $$Cr{O_4}^{2 - }$$ oxidises $${S_2}{O_3}^{2 - }$$ to form $$S{O_4}^{2 - }$$ and itself changes into $$Cr{(OH)_4}^ - $$. The volume of 0.154 M $$Cr{O_4}^{2 - }$$ required to react with 40 mL of 0.25 M $${S_2}{O_3}^{2 - }$$ is __________ mL. (Rounded off to the nearest integer) | [] | null | 173 | $$17{H_2}O + 8Cr{O_4} + 3{S_2}{O_3}\buildrel {} \over
\longrightarrow 6S{O_4} + 8Cr{(OH)_4}^ - + 2O{H^ - }$$<br><br>Applying mole-mole analysis<br><br>$${{0.154 \times v} \over 8} = {{40 \times 0.25} \over 3}$$<br><br>v = 173 mL | integer | jee-main-2021-online-25th-february-morning-slot | 3,715 |
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