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|---|---|---|---|---|---|---|---|---|---|---|---|
O94vIqwj56Bp3IWojs1kmhv0gf8 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm<sup>$$-$$3</sup>. The molarity of the solution is ____________ mol dm<sup>$$-$$3</sup>. (Round off to the Nearest Integer).<br/><br/>[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u] | [] | null | 9 | $$m = {{1000 \times M} \over {1000 \times d - M \times {M_{solute}}}}$$<br><br>$$6.5 = {{1000 \times M} \over {1890 - M \times 56}}$$<br><br>$$ \Rightarrow $$ $$12285 - 364M = 1000M$$<br><br>$$ \Rightarrow $$ $$1364M = 12285$$<br><br>$$ \Rightarrow $$ $$M = 9$$ | integer | jee-main-2021-online-16th-march-morning-shift | 3,716 |
LnmES6OKMekV2y3cOn1kmitgbw3 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H<sub>3</sub>PO<sub>3</sub> solution and 100 mL of 2 M H<sub>3</sub>PO<sub>2</sub> solution, respectively, are : | [{"identifier": "A", "content": "100 mL and 50 mL"}, {"identifier": "B", "content": "100 mL and 200 mL"}, {"identifier": "C", "content": "100 mL and 100 mL"}, {"identifier": "D", "content": "50 mL and 50 mL"}] | ["B"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3l0kzzj/a9c6d660-3491-4798-86a3-64ab73f568d1/01b6dbf0-dbd9-11ec-adf6-01b3c1852b0a/file-1l3l0kzzk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3l0kzzj/a9c6d660-3491-4798-86a3-64ab73f568d1/01b6dbf0-dbd9-11ec-adf6-01b3c1852b0... | mcq | jee-main-2021-online-16th-march-evening-shift | 3,717 |
uyYL0NOgzLKGlr4wm11kmj8ices | chemistry | some-basic-concepts-of-chemistry | concentration-terms | 15 mL of aqueous solution of Fe<sup>2+</sup> in acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr<sub>2</sub>O$$_7^{2 - }$$. The molarity of the Fe<sup>2+</sup> solution is __________ $$\times$$ 10<sup>-2</sup> M. (Round off to the Nearest Integer). | [] | null | 24 | $$C{r_2}{O_7}^{2 - } + F{e^{2 + }}\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{}} C{r^{3 + }} + F{e^{3 + }}$$<br><br>Valance Factor of $$C{r_2}{O_7}^{2 - }$$ = 6
<br><br>Valance Factor of $$F{e^{2 + }}$$ = 1<br><br>mili eq. of $$C{r_2}{O_7}^{2 - }$$ = mili eq. of Fe<sup>2+</sup><br><br>6(0.03 $$\times$$ 20) = ... | integer | jee-main-2021-online-17th-march-morning-shift | 3,718 |
QasEluCtqV2XkSwQfF1kmj8t8w5 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The mole fraction of a solute in a 100 molal aqueous solution is ___________ $$\times$$ 10<sup>$$-$$2</sup>. (Round off to the Nearest Integer).<br/><br/>[Given : Atomic masses : H : 1.0 u, O : 16.0 u ] | [] | null | 64 | Let weight of H<sub>2</sub>O = 1000 g<br><br>Moles of solute = 100<br><br>(mole)H<sub>2</sub>O = $${{1000} \over {18}}$$<br><br>Mole fraction of solute = $${{mole\,of\,solute} \over {Total\,moles}}$$<br><br>$$ = {{100} \over {100 + {{1000} \over {18}}}} = {{1800} \over {2800}}$$<br><br>$${X_{solute}} = 64 \times {10^{ ... | integer | jee-main-2021-online-17th-march-morning-shift | 3,719 |
1krt7b0pa | chemistry | some-basic-concepts-of-chemistry | concentration-terms | If the concentration of glucose (C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>) in blood is 0.72 g L<sup>$$-$$1</sup>, the molarity of glucose in blood is ____________ $$\times$$ 10<sup>$$-$$3</sup> M. (Nearest integer)<br/><br/>[Given : Atomic mass of C = 12, H = 1, O = 16 u] | [] | null | 4 | [Glucose] = $${{C(gm/l)} \over {M(gm/mol)}} = {{0.72} \over {180}} = 4 \times {10^{ - 3}}$$ M | integer | jee-main-2021-online-22th-july-evening-shift | 3,721 |
1kruv0rrs | chemistry | some-basic-concepts-of-chemistry | concentration-terms | When 10 mL of an aqueous solution of Fe<sup>2+</sup> ions was titrated in the presence of dil H<sub>2</sub>SO<sub>4</sub> using diphenylamine indicator, 15 mL of 0.02 M solution of K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> was required to get the end point. The molarity of the solution containing Fe<sup>2+</sup> ions is... | [] | null | 18 | milli-equivalents of Fe<sup>2+</sup> = milli-equivalents of K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub><br><br>M $$\times$$ 10 $$\times$$ 1 = 0.02 $$\times$$ 15 $$\times$$ 6<br><br>$$ \Rightarrow $$ M = 0.18 = 18 $$\times$$ 10<sup>$$-$$2</sup> M | integer | jee-main-2021-online-25th-july-morning-shift | 3,722 |
1krz1teuy | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The density of NaOH solution is 1.2 g cm<sup>$$-$$3</sup>. The molality of this solution is _____________ m. (Round off to the Nearest Integer)<br/><br/>[Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of H<sub>2</sub>O : 1.0 g cm<sup>$$-$$3</sup>] | [] | null | 5 | Consider 1 litre solution<br><br>mass of solution = (1.2 $$\times$$ 1000)g = 1200 gm<br><br>Neglecting volume of NaOH<br><br>Mass of water = 1000 gm<br><br>$$\Rightarrow$$ Mass of NaOH = (1200 $$-$$ 1000)gm = 200 gm<br><br>$$\Rightarrow$$ Moles of NaOH = $${{200g} \over {50g/mol}} = 5$$ mol<br><br>$$\Rightarrow$$ molal... | integer | jee-main-2021-online-27th-july-morning-shift | 3,723 |
1ktcsp103 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | 100 mL of Na<sub>3</sub>PO<sub>4</sub> solution contains 3.45 g of sodium. The molarity of the solution is _____________ $$\times$$ 10<sup>$$-$$2</sup> mol L<sup>$$-$$1</sup>. (Nearest integer)<br/><br/>[Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u] | [] | null | 50 | Molarity of Na<sub>3</sub>PO<sub>4</sub> Solution = $${{{n_{N{a_3}P{O_4}}}} \over {volume\,of\,solution\,in\,L}}$$<br><br>$$ = {{{1 \over 3} \times {{3.45} \over {23}}mol} \over {0.1\,L}}$$<br><br>= 0.5 = 50 $$\times$$ 10<sup>$$-$$2</sup> | integer | jee-main-2021-online-26th-august-evening-shift | 3,725 |
1ktfu24md | chemistry | some-basic-concepts-of-chemistry | concentration-terms | 100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x $$\times$$ 10<sup>$$-$$2</sup>. The value of x is ____________. (Nearest integer)<br/><br/>[Atomic weight : H = 1.008; C = 12.00; O = 16.00] | [] | null | 19 | C<sub>3</sub>H<sub>8(g)</sub> + 5O<sub>2(g)</sub> $$\to$$ 3CO<sub>2(g)</sub> + 4H<sub>2</sub>O<sub>(l)</sub><br><br><table>
<thead>
<tr>
<th>t = 0</th>
<th>2.27 mole</th>
<th>31.25 mol</th>
<th></th>
<th></th>
</tr>
</thead>
<tbody>
<tr>
<td>t = $$\infty $$</td>
<td>0</td>
<td>19.9... | integer | jee-main-2021-online-27th-august-evening-shift | 3,726 |
1ktifhx0l | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H<sub>2</sub>C<sub>2</sub>O<sub>4</sub>.2H<sub>2</sub>O) in 250 mL of water in mol L<sup>$$-$$1</sup> is x $$\times$$ 10<sup>$$-$$2</sup>. The value of x is _____________. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0] | [] | null | 20 | $$[{H_2}{C_2}{O_4}.2{H_2}O] = {{weight/{M_w}} \over {V(L)}}$$<br><br>$$ \Rightarrow x \times {10^{ - 2}} = {{6.3/126} \over {250/1000}}$$<br><br>$$x = 20$$ | integer | jee-main-2021-online-31st-august-morning-shift | 3,727 |
1ktn2mwbz | chemistry | some-basic-concepts-of-chemistry | concentration-terms | If 80 g of copper sulphate CuSO<sub>4</sub> . 5H<sub>2</sub>O is dissolved in deionised water to make 5L of solution. The concentration of the copper sulphate solution is x $$\times$$ 10<sup>$$-$$3</sup> mol L<sup>$$-$$1</sup>. The value of x is _____________.<br/><br/>[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : ... | [] | null | 64 | Given, mass of CuSO<sub>4</sub> . 5H<sub>2</sub>O = 80 g<br/><br/>The concentration of copper sulphate solution is x $$\times$$ 10<sup>$$-$$3</sup> mol/L.<br/><br/>Molarity = $${{Number\,of\,moles\,of\,solute} \over {Volume\,of\,solution(L)}}$$ ..... (i)<br/><br/>Molar mass of CuSO<sub>4</sub> . 5H<sub>2</sub>O = 63.54... | integer | jee-main-2021-online-1st-september-evening-shift | 3,729 |
1l5c7c1o1 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>A 0.166 g sample of an organic compound was digested with conc. H<sub>2</sub>SO<sub>4</sub> and then distilled with NaOH. The ammonia gas evolved was passed through 50.0 mL of 0.5 N H<sub>2</sub>SO<sub>4</sub>. The used acid required 30.0 mL of 0.25 N NaOH for complete neutralization. The mass percentage of nitrogen... | [] | null | 63 | Millimoles of used acid $=\frac{30 \times 0.25}{2}$
<br/><br/>
Millimoles of $\mathrm{NH}_{3}=30 \times 0.25=7.5$
<br/><br/>
Mass $\%$ of nitrogen $=\frac{7.5}{0.166} \times 10^{-3} \times 14 \times 100 \simeq 63 \%$ | integer | jee-main-2022-online-24th-june-morning-shift | 3,731 |
1l6e0ioba | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>$$\mathrm{SO}_{2} \mathrm{Cl}_{2}$$ on reaction with excess of water results into acidic mixture</p>
<p>$$\mathrm{SO}_{2} \mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}$$</p>
<p>16 moles of $$\mathrm{NaOH}$$ is required for the complete neutralisation of the res... | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $$\mathrm{SO}_{2} \mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}$$
<br/><br/>
Moles of $$\mathrm{NaOH}$$ required for complete neutralization of resultant acidic mixture $$=16$$ moles
<br/><br/>
And 1 mole of $$\mathrm{SO}_{2} \mathrm{Cl}_{2}$$ produced 4 moles of ... | mcq | jee-main-2022-online-25th-july-morning-shift | 3,732 |
1l6gs2ci9 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>Chlorophyll extracted from the crushed green leaves was dissolved in water to make $$2 \mathrm{~L}$$ solution of Mg of concentration $$48\, \mathrm{ppm}$$. The number of atoms of $$\mathrm{Mg}$$ in this solution is $$x \times 10^{20}$$ atoms. The value of $$x$$ is ___________. (Nearest Integer)</p>
<p>(Given : Atomi... | [] | null | 24 | In $$2 \mathrm{L} \rightarrow 96 \,\mathrm{mg}$$ of $$\mathrm{Mg}$$
<br/><br/>
Number of atoms of $$\mathrm{Mg}=\frac{96 \times 10^{-3}}{24} \times \mathrm{N}_{\mathrm{A}}$$
<br/><br/>
$$
\begin{aligned}
&=4 \times 10^{-3} \times 6 \times 10^{23} \\
&=24 \times 10^{20}
\end{aligned}
$$
<br/><br/>
$$x=24$$ | integer | jee-main-2022-online-26th-july-morning-shift | 3,733 |
1l6gsd3s4 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is x $$\times$$ 10<sup>$$-$$2</sup> M. (Nearest integer)</p>
<p>(Molar mass of nitric acid is 63 g mol<sup>$$-$$1</sup>)</p> | [] | null | 54 | $$\mathrm{m}$$ moles of $$\mathrm{HNO}_{3}=800 \times 0.5$$
<br/><br/>
Moles of $$\mathrm{HNO}_{3}=400 \times 10^{-3} = 0.4 \text { moles }$$
<br/><br/>
Weight of $$\mathrm{HNO}_{3}=0.4 \times 63 \mathrm{~g}
=25.2 \mathrm{~g}
$$
<br/><br/>
Remaining acid $$=25.2-11.5
=13.7 \mathrm{~g}
$$
<br/><br/>
$$
\begin{aligned}
... | integer | jee-main-2022-online-26th-july-morning-shift | 3,734 |
1l6jk7xj0 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>$$250 \mathrm{~g}$$ solution of $$\mathrm{D}$$-glucose in water contains $$10.8 \%$$ of carbon by weight. The molality of the solution is nearest to</p>
<p>(Given: Atomic Weights are, $$\mathrm{H}, 1 \,\mathrm{u} ; \mathrm{C}, 12 \,\mathrm{u} ; \mathrm{O}, 16 \,\mathrm{u}$$)</p> | [{"identifier": "A", "content": "1.03"}, {"identifier": "B", "content": "2.06"}, {"identifier": "C", "content": "3.09"}, {"identifier": "D", "content": "5.40"}] | ["B"] | null | Weight of D-glucose in water $=250 \mathrm{~g}$
<br/><br/>
$$
\begin{aligned}
\therefore \text { Weight of carbon in D-glucose } &=\frac{250}{180} \times 72 \\\\
&=100 \mathrm{~g}
\end{aligned}
$$
<br/><br/>
$\%$ of carbon in the aqueous solution of glucose is $=10.8 \%$
<br/><br/>
$\therefore$ Weight of the solution i... | mcq | jee-main-2022-online-27th-july-morning-shift | 3,735 |
1l6jmimfy | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>$$20 \mathrm{~mL}$$ of $$0.02 \,\mathrm{M} \,\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$$ solution is used for the titration of $$10 \mathrm{~mL}$$ of $$\mathrm{Fe}^{2+}$$ solution in the acidic medium.</p>
<p>The molarity of $$\mathrm{Fe}^{2+}$$ solution is __________ $$\times \,10^{-2}\, \mathrm{M}$$. (Nearest ... | [] | null | 24 | Applying the law of equivalence, milliequivalents of $\mathrm{Fe}^{2+}=$ milliequivalents of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
<br/><br/>
$$
\begin{aligned}
&10 \times 1 \times M=20 \times 6 \times .02 \\\\
&M=24 \times 10^{-2} M
\end{aligned}
$$
<br/><br/>
$\therefore $ Answer will be 24 | integer | jee-main-2022-online-27th-july-morning-shift | 3,736 |
1l6nxecai | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>2L of 0.2M H<sub>2</sub>SO<sub>4</sub> is reacted with 2L of 0.1M NaOH solution, the molarity of the resulting product Na<sub>2</sub>SO<sub>4</sub> in the solution is _________ millimolar. (Nearest integer) </p> | [] | null | 25 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7vya06e/57a2bec7-f6bd-48ad-b27f-de2cadb545d6/7a5ca060-310d-11ed-9c98-ad39d53b642b/file-1l7vya06f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7vya06e/57a2bec7-f6bd-48ad-b27f-de2cadb545d6/7a5ca060-310d-11ed-9c98-ad39d53b642b/fi... | integer | jee-main-2022-online-28th-july-evening-shift | 3,737 |
1ldo46li6 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>The molality of a $$10 \%(\mathrm{v} / \mathrm{v})$$ solution of di-bromine solution in $$\mathrm{CCl}_{4}$$ (carbon tetrachloride) is '$$x$$'. $$x=$$ ____________ $$\times 10^{-2} ~\mathrm{M}$$. (Nearest integer)</p>
<p>[Given : molar mass of $$\mathrm{Br}_{2}=160 \mathrm{~g} \mathrm{~mol}^{-1}$$</p>
<p>atomic mass... | [] | null | 139 | Mass of $10 \mathrm{~mL}$ of $\mathrm{Br}_2=10 \times 3.2=32 \mathrm{gm}$
<br/><br/>Mass of $90 \mathrm{~mL}^{\mathrm{L}} \mathrm{CCl}_4=90 \times 1.6=144 \mathrm{gm}$
<br/><br/>Molality of $\mathrm{Br}_2$ solution in $\mathrm{CCl}_4$
<br/><br/>$$
\begin{aligned}
& =\frac{32 \times 1000}{160 \times 144} \\\\
& =1.39 \m... | integer | jee-main-2023-online-1st-february-evening-shift | 3,738 |
ldqxivpf | chemistry | some-basic-concepts-of-chemistry | concentration-terms | $1 \mathrm{~L}, 0.02 \mathrm{M}$ solution of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right]$ Br is mixed with $1 \mathrm{~L}, 0.02 \mathrm{M}$ solution of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}$. The resulting solution is divided into two equal p... | [{"identifier": "A", "content": "$0 .01,0.01$"}, {"identifier": "B", "content": "$0.01,0.02$"}, {"identifier": "C", "content": "$0.02,0.01$"}, {"identifier": "D", "content": "$0.02,0.02$"}] | ["A"] | null | <p>On mixing both $$\mathrm{\left[ {Co{{(N{H_5})}_5}S{O_4}} \right]Br}$$ and $$\mathrm{\left[ {Co{{(N{H_3})}_5}Br} \right]S{O_4}}$$ becomes 0.01 molar.</p>
<p>$$\therefore$$ Moles of y and z formed will also be 0.01 both.</p> | mcq | jee-main-2023-online-30th-january-evening-shift | 3,740 |
ldqxybyf | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>The strength of 50 volume solution of hydrogen peroxide is ______ $\mathrm{g} / \mathrm{L}$ (Nearest integer).
</p>
Given:
<br/><br/>
Molar mass of $\mathrm{H}_{2} \mathrm{O}_{2}$ is $34 \mathrm{~g} \mathrm{~mol}^{-1}$
<br/><br/>
Molar volume of gas at $\mathrm{STP}=22.7 \mathrm{~L}$ | [] | null | 150 | <p>$$\mathrm{H_2O_2\longrightarrow H_2O+\frac{1}{2}O_2}$$</p>
<p>$$\frac{50}{22.7}$$</p>
<p>$$\therefore$$ Moles of $$\mathrm{H_2O_2}$$ in solution $$=\frac{50}{22.7}\times2$$</p>
<p>$$\therefore$$ Strength $$=\frac{\frac{50\times2}{22.7}\times34}{1}=149.78\approx150$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 3,741 |
1ldr4stxs | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>Some amount of dichloromethane $$\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)$$ is added to $$671.141 \mathrm{~mL}$$ of chloroform $$\left(\mathrm{CHCl}_{3}\right)$$ to prepare $$2.6 \times 10^{-3} \mathrm{M}$$ solution of $$\mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{DCM})$$. The concentration of $$\mathrm{DCM}$$ is __... | [] | null | 148 | <p>Mass of $$\mathrm{CHCl_3=671.141\times1.49=1000}$$ gm</p>
<p>$$2.6 \times {10^{ - 3}} = {{moles\,of\,C{H_2}C{l_2}} \over {0.671141}}$$</p>
<p>$$\Rightarrow$$ moles of $$C{H_2}C{l_2} = 1.74496 \times {10^{ - 3}}$$</p>
<p>mass of $$C{H_2}C{l_2} = 148.32 \times {10^{ - 3}}$$ gm</p>
<p>Composition of $$C{H_2}C{l_2} = {{... | integer | jee-main-2023-online-30th-january-morning-shift | 3,742 |
1ldu0fnza | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>What is the mass ratio of ethylene glycol ($$\mathrm{C_2H_6O_2}$$, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?</p> | [{"identifier": "A", "content": "1 : 2"}, {"identifier": "B", "content": "1 : 1"}, {"identifier": "C", "content": "2 : 1"}, {"identifier": "D", "content": "3 : 1"}] | ["C"] | null | For 500 g of 0.25 molal aqueous solution,
<br/><br/>0.25 = $${{{{{w_1}} \over {62}}} \over {{{500} \over {1000}}}}$$ = $${{2{w_2}} \over {62}}$$ ......(1)
<br/><br/>For 250 mL of 0.25 molar aqueous solution
<br/><br/>$${{{{{w_2}} \over {62}}} \over {{{250} \over {1000}}}} = {{4{w_2}} \over {62}}$$ .......(2)
<br/><br/>... | mcq | jee-main-2023-online-25th-january-evening-shift | 3,743 |
1ldukrf6s | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>'25 volume' hydrogen peroxide means</p> | [{"identifier": "A", "content": "1 L marketed solution contains 75 g of H$$_2$$O$$_2$$."}, {"identifier": "B", "content": "1 L marketed solution contains 250 g of H$$_2$$O$$_2$$."}, {"identifier": "C", "content": "1 L marketed solution contains 25 g of H$$_2$$O$$_2$$."}, {"identifier": "D", "content": "100 mL marketed ... | ["A"] | null | Molarity of $\mathrm{H}_{2} \mathrm{O}_{2}$ soln $=\frac{\text { volume strength }}{11.2}$
<br/><br/>
$=\frac{25}{11.2}=2.23 \mathrm{M}$
<br/><br/>
$\therefore$ amount of $\mathrm{H}_{2} \mathrm{O}_{2}$ in one litre $=2.23 \times 34=75 \mathrm{gm}$ | mcq | jee-main-2023-online-25th-january-morning-shift | 3,744 |
1ldwvoouk | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>The number of units, which are used to express concentration of solutions from the following is _________</p>
<p>Mass percent, Mole, Mole fraction, Molarity, ppm, Molality</p> | [] | null | 5 | Mass percent, mole fraction, molarity, ppm & molality is used to express concentration. So, the number of units = 5 | integer | jee-main-2023-online-24th-january-evening-shift | 3,745 |
1ldyhik44 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _____________</p>
<p>Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol$$^{-1}$$ respectively</p> | [] | null | 180 | Molarity of solution $=\frac{5}{(40)} \frac{(1000)}{(450)}$
<br/><br/>
$\Rightarrow M \times V=500 \times .1$
<br/><br/>
$\Rightarrow \frac{5}{40} \times \frac{1000}{450} \times V=500 \times 0.1$
<br/><br/>
$\mathrm{V}=180 \mathrm{~mL}$ | integer | jee-main-2023-online-24th-january-morning-shift | 3,746 |
lgo0b1p0 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | The volume (in $\mathrm{mL}$) of $0.1 \mathrm{M} ~\mathrm{AgNO}_{3}$ required for complete precipitation of chloride ions present in $20 \mathrm{~mL}$ of $0.01 \mathrm{M}$ solution of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ as silver chloride is __________. | [] | null | 4 | The given solution contains a complex ion $\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2$ which contains one chloride ion per complex. Therefore, when this complex is treated with silver nitrate, each mole of the complex will consume 2 moles of silver nitrate to form two moles... | integer | jee-main-2023-online-15th-april-morning-shift | 3,747 |
1lgsyhrki | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>A solution is prepared by adding $$2 \mathrm{~g}$$ of "$$\mathrm{X}$$" to 1 mole of water. Mass percent of "$$\mathrm{X}$$" in the solution is :</p> | [{"identifier": "A", "content": "20%"}, {"identifier": "B", "content": "10%"}, {"identifier": "C", "content": "2%"}, {"identifier": "D", "content": "5%"}] | ["B"] | null | First, we need to calculate the mass of 1 mole of water (H<sub>2</sub>O). The molecular weight of water is 18 g/mol, so 1 mole of water weighs 18 g.
<br/><br/>The total mass of the solution is the mass of "X" plus the mass of water, which is 2 g + 18 g = 20 g.
<br/><br/>We can calculate the mass percent of "X" in the s... | mcq | jee-main-2023-online-11th-april-evening-shift | 3,749 |
1lgv037r2 | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>A solution of sugar is obtained by mixing $$200 \mathrm{~g}$$ of its $$25 \%$$ solution and $$500 \mathrm{~g}$$ of its $$40 \%$$ solution (both by mass). The mass percentage of the resulting sugar solution is ___________ (Nearest integer)</p> | [] | null | 36 | <p>Let's first calculate the amount of sugar in each of the solutions :</p>
<ol>
<li><p>In the 25% sugar solution, the amount of sugar is 25% of 200 g, which equals 50 g.</p>
</li>
<li><p>In the 40% sugar solution, the amount of sugar is 40% of 500 g, which equals 200 g.</p>
</li>
</ol>
<p>Now, let's mix these ... | integer | jee-main-2023-online-11th-april-morning-shift | 3,750 |
1lgvu3ogf | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>Given below are two statements: one is labelled as Assertion $$\mathbf{A}$$ and the other is labelled as Reason $$\mathbf{R}$$</p>
<p>Assertion A : $$3.1500 \mathrm{~g}$$ of hydrated oxalic acid dissolved in water to make $$250.0 \mathrm{~mL}$$ solution will result in $$0.1 \mathrm{~M}$$ oxalic acid solution.</p>
<p... | [{"identifier": "A", "content": "Both A and R are true but R is NOT the correct explanation of A"}, {"identifier": "B", "content": "A is true but R is false"}, {"identifier": "C", "content": "Both A and R are true and R is the correct explanation of A"}, {"identifier": "D", "content": "A is false but R is true"}] | ["C"] | null | <p>Firstly, let's calculate the molarity of the oxalic acid solution given in the Assertion (A). </p>
<p>The molarity (M) is defined as the number of moles of solute per liter of solution. The number of moles of solute is given by the mass of the solute divided by its molar mass. </p>
<p>So, the number of moles of ... | mcq | jee-main-2023-online-10th-april-evening-shift | 3,751 |
1lh32l8lw | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>The volume of $$0.02 ~\mathrm{M}$$ aqueous $$\mathrm{HBr}$$ required to neutralize $$10.0 \mathrm{~mL}$$ of $$0.01 ~\mathrm{M}$$ aqueous $$\mathrm{Ba}(\mathrm{OH})_{2}$$ is (Assume complete neutralization)</p> | [{"identifier": "A", "content": "7.5 mL"}, {"identifier": "B", "content": "5.0 mL"}, {"identifier": "C", "content": "10.0 mL"}, {"identifier": "D", "content": "2.5 mL"}] | ["C"] | null | <p>The reaction between $\mathrm{HBr}$ and $\mathrm{Ba(OH)_2}$ is a neutralization reaction where one $\mathrm{Ba(OH)_2}$ reacts with two $\mathrm{HBr}$ to form $\mathrm{BaBr_2}$ and $2\mathrm{H_2O}$:</p>
<p>$\mathrm{Ba(OH)_2} + 2\mathrm{HBr} \rightarrow \mathrm{BaBr_2} + 2\mathrm{H_2O}$</p>
<p>We can use the stoichiom... | mcq | jee-main-2023-online-6th-april-evening-shift | 3,752 |
jaoe38c1lscryk9k | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>The quantity which changes with temperature is :</p> | [{"identifier": "A", "content": "Molality"}, {"identifier": "B", "content": "Molarity"}, {"identifier": "C", "content": "Mole fraction"}, {"identifier": "D", "content": "Mass percentage"}] | ["B"] | null | <p>The quantity that changes with temperature from the given options is <strong>Molarity</strong>.</p>
<p>Molarity, which is denoted with a capital M, measures the number of moles of solute dissolved in one liter of solution. It is expressed as:</p>
<p>$$ Molarity (M) = \frac{moles \ of \ solute}{liters \ of \ solution... | mcq | jee-main-2024-online-27th-january-evening-shift | 3,753 |
jaoe38c1lsdaf22k | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>The molarity of $$1 \mathrm{~L}$$ orthophosphoric acid $$\left(\mathrm{H}_3 \mathrm{PO}_4\right)$$ having $$70 \%$$ purity by weight (specific gravity $$1.54 \mathrm{~g} \mathrm{~cm}^{-3}$$) is __________ $$\mathrm{M}$$.</p>
<p>(Molar mass of $$\mathrm{H}_3 \mathrm{PO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1}$$)</p> | [] | null | 11 | <p>Specific gravity (density) $$=1.54 \mathrm{~g} / \mathrm{cc}$$.</p>
<p>Volume $$=1 \mathrm{~L}=1000 \mathrm{~ml}$$</p>
<p>Mass of solution $$=1.54 \times 1000$$</p>
<p>$$=1540 \mathrm{~g}$$</p>
<p>$$\%$$ purity of $$\mathrm{H}_2 \mathrm{SO}_4$$ is $$70 \%$$</p>
<p>So weight of $$\mathrm{H}_3 \mathrm{PO}_4=0.7 \times... | integer | jee-main-2024-online-31st-january-evening-shift | 3,754 |
jaoe38c1lsfk2ayb | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>A solution of $$\mathrm{H}_2 \mathrm{SO}_4$$ is $$31.4 \% \mathrm{H}_2 \mathrm{SO}_4$$ by mass and has a density of $$1.25 \mathrm{~g} / \mathrm{mL}$$. The molarity of the $$\mathrm{H}_2 \mathrm{SO}_4$$ solution is _________ $$\mathrm{M}$$ (nearest integer)</p>
<p>[Given molar mass of $$\mathrm{H}_2 \mathrm{SO}_4=98... | [] | null | 4 | <p>$$\begin{aligned}
& M=\frac{n_{\text {solute }}}{V} \times 1000 \\
& =\frac{\left(\frac{31.4}{98}\right)}{\left(\frac{100}{1.25}\right)} \times 1000 \\
& =4.005 \approx 4
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift | 3,755 |
jaoe38c1lsfpoo1e | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>Molality of 0.8 M H$$_2$$SO$$_4$$ solution (density 1.06 g cm$$^{-3}$$) is ________ $$\times10^{-3}$$ m.</p> | [] | null | 815 | <p>$$\mathrm{m}=\frac{\mathrm{M} \times 1000}{\mathrm{~d}_{\text {sol }} \times 1000-\mathrm{M} \times \text { Molar mass }_{\text {solute }}}$$</p>
<p>$$815 \times 10^{-3} \mathrm{~m}$$</p> | integer | jee-main-2024-online-29th-january-evening-shift | 3,756 |
1lsgyl9gy | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>The mass of sodium acetate $$\left(\mathrm{CH}_3 \mathrm{COONa}\right)$$ required to prepare $$250 \mathrm{~mL}$$ of $$0.35 \mathrm{~M}$$ aqueous solution is ________ g. (Molar mass of $$\mathrm{CH}_3 \mathrm{COONa}$$ is $$82.02 \mathrm{~g} \mathrm{~mol}^{-1}$$)</p> | [] | null | 7 | <p>$$\begin{aligned}
& \text { Moles }=\text { Molarity } \times \text { Volume in litres } \\
& =0.35 \times 0.25 \\
& \text { Mass }=\text { moles } \times \text { molar mass } \\
& =0.35 \times 0.25 \times 82.02=7.18 \mathrm{~g}
\end{aligned}$$</p>
<p>Ans. 7</p> | integer | jee-main-2024-online-30th-january-morning-shift | 3,758 |
lv0vyqoa | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>The Molarity (M) of an aqueous solution containing $$5.85 \mathrm{~g}$$ of $$\mathrm{NaCl}$$ in $$500 \mathrm{~mL}$$ water is : (Given : Molar Mass $$\mathrm{Na}: 23$$ and $$\mathrm{Cl}: 35.5 \mathrm{~gmol}^{-1}$$)</p> | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "0.2"}] | ["D"] | null | <p>Moles $$=0.1$$</p>
<p>Volume $$=0.5 \mathrm{~L}$$</p>
<p>Molarity $$=\frac{0.1}{0.5}=0.2 \mathrm{M}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift | 3,760 |
lv40vb6r | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is ________ (Integer answer) (Given : Molar mass in $$\mathrm{g} \mathrm{~mol}^{-1}$$ Ethyl alcohol : 46 water : 18)</p> | [] | null | 22 | <p>To determine the mass percent of the solute (ethyl alcohol) in the solution, we start by calculating the masses of ethyl alcohol and water using their respective molar masses.</p>
<p>The molar mass of ethyl alcohol (C<sub>2</sub>H<sub>5</sub>OH) is given as 46 g/mol and the molar mass of water (H<sub>2</sub>O) is g... | integer | jee-main-2024-online-8th-april-evening-shift | 3,761 |
lv40vbct | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>Molality of an aqueous solution of urea is $$4.44 \mathrm{~m}$$. Mole fraction of urea in solution is $$x \times 10^{-3}$$, Value of $$x$$ is ________. (Integer answer)</p> | [] | null | 74 | <p>To determine the mole fraction of urea in a solution where the molality is $$4.44 \ \mathrm{m}$$, we first need to understand the definitions and relationships involved.</p>
<p>Molality ($$\mathrm{m}$$) is given by:</p>
<p>$$ \mathrm{m} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} $$</p>
<p>Here,... | integer | jee-main-2024-online-8th-april-evening-shift | 3,762 |
lvb2a2bp | chemistry | some-basic-concepts-of-chemistry | concentration-terms | <p>Molality $$(\mathrm{m})$$ of $$3 \mathrm{M}$$ aqueous solution of $$\mathrm{NaCl}$$ is :
(Given : Density of solution $$=1.25 \mathrm{~g} \mathrm{~mL}^{-1}$$, Molar mass in $$\mathrm{g} \mathrm{~mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5$$)</p> | [{"identifier": "A", "content": "2.90 m"}, {"identifier": "B", "content": "3.85 m"}, {"identifier": "C", "content": "1.90 m"}, {"identifier": "D", "content": "2.79 m"}] | ["D"] | null | <p>To calculate the molality $(\mathrm{m})$ of a 3M (molar) aqueous solution of $\mathrm{NaCl}$, we need to follow certain steps, using the given data and understanding the definitions properly.</p>
<p>Molarity (M) is defined as the number of moles of solute per liter of solution. Molality $(\mathrm{m})$, on the other... | mcq | jee-main-2024-online-6th-april-evening-shift | 3,763 |
29zPo9XaMdIFlyBJ | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen? | [{"identifier": "A", "content": "67.2 L"}, {"identifier": "B", "content": "44.8 L"}, {"identifier": "C", "content": "22.4 L"}, {"identifier": "D", "content": "89.6 L"}] | ["A"] | null | The reaction is
<br>2BCl<sub>3</sub> + 3H<sub>2</sub> $$ \to $$ 2B + 6HCL
<br><br>$$\therefore$$ 3 moles of hydrogen produces 2 moles of boron
<br><br>So 2$$ \times $$10.8 gm = 21.6 gm of boron is produces from 3$$ \times $$22.4 = 67.2 L hydrogen | mcq | aieee-2003 | 3,765 |
scr2lDOm0WB6sdIf | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is | [{"identifier": "A", "content": "urea"}, {"identifier": "B", "content": "benzamide"}, {"identifier": "C", "content": "acetamide"}, {"identifier": "D", "content": "thiourea"}] | ["A"] | null | Initially total H<sub>2</sub>SO<sub>4</sub> present = 100 mL of 0.1 M = $${{{100} \over {1000}} \times 0.1}$$ mole = 0.01 mole
<br><br>2NaOH + H<sub>2</sub>SO<sub>4</sub> $$ \to $$ Na<sub>2</sub>SO<sub>4</sub> + 2H<sub>2</sub>O
<br>Let in this reaction H<sub>2</sub>SO<sub>4</sub> required n mole
<br><br>$$\therefore$$ ... | mcq | aieee-2004 | 3,766 |
iFOCpvHDjmR9VR1X | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of
their molecule is: | [{"identifier": "A", "content": "1: 8"}, {"identifier": "B", "content": "3 : 16"}, {"identifier": "C", "content": "1 : 4"}, {"identifier": "D", "content": "7 : 32"}] | ["D"] | null | Given ratio of mass of O<sub>2</sub> and N<sub>2</sub> = 1 : 4
<br>Let mass of O<sub>2</sub> = w
<br>and mass of N<sub>2</sub> = 4w
<br><br>$$\therefore$$ Number of moles of O<sub>2</sub> = $${w \over {32}}$$
<br>Number of molecules of O<sub>2</sub> = $${w \over {32}}$$$$ \times $$N<sub>A</sub>
<br><br>Number of moles ... | mcq | jee-main-2014-offline | 3,767 |
bdKhAnBfK4pYimHvgfLID | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must
be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid
solution ?
| [{"identifier": "A", "content": "320"}, {"identifier": "B", "content": "325"}, {"identifier": "C", "content": "316"}, {"identifier": "D", "content": "330"}] | ["C"] | null | <p>Let the volume of a monoprotic acid solution in mL be V</p>
<p>$${{V \times 45} \over {100}} + {{(800 - V)20} \over {100}} = {{800 \times 29.875} \over {100}}$$</p>
<p>$${{9V} \over {20}} + 160 - {V \over 5} = 239$$</p>
<p>$${{5V} \over {20}} = 79 \Rightarrow V = 316$$ mL</p> | mcq | jee-main-2017-online-9th-april-morning-slot | 3,768 |
IrGVkdxwdeaOjVy1IFsWX | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO<sub>2</sub> at T = 298.15 K and p = 1 bar. If molar volume of CO<sub>2</sub> is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ?
<br/>[Molar mass of NaHCO<sub>3</sub> = 84 g mol<su... | [{"identifier": "A", "content": "33.6"}, {"identifier": "B", "content": "0.84"}, {"identifier": "C", "content": "8.4"}, {"identifier": "D", "content": "16.8"}] | ["C"] | null | 2NaHCO<sub>3</sub>
+ (COOH)<sub>2</sub> $$ \to $$ (COONa)<sub>2</sub> + 2H<sub>2</sub>O + 2CO<sub>2</sub>
<br><br>$$ \therefore $$ 1 mole of CO<sub>2</sub> is produced by 1 mole of NaHCO<sub>3</sub>.
<br><br>Given, volume of CO<sub>2</sub> produced = 0.25 ml
<br><br>25 L of CO<sub>2</sub> contains 1 mol
<br><br>$$ \th... | mcq | jee-main-2019-online-11th-january-morning-slot | 3,769 |
Bd3kIn7RoGOVqmZN4C7k9k2k5dymla5 | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | Amongst the following statements, that which was not proposed by Dalton was : | [{"identifier": "A", "content": "Matter consists of indivisible atoms all the atoms of a given element have."}, {"identifier": "B", "content": "Chemical reactions involve reorganization of atoms. These are neither created not destroyed in\na chemical reaction."}, {"identifier": "C", "content": "When gases combine or re... | ["C"] | null | Option(3) is according to Avogadro's law of
volume combination. | mcq | jee-main-2020-online-7th-january-morning-slot | 3,770 |
BcmaMdKKD0IqRFuqf37k9k2k5h6tnhh | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | The volume (in mL) of 0.125 M AgNO<sub>3</sub> required to quantitatively precipitate chloride ions in 0.3 g of
[Co(NH<sub>3</sub>)<sub>6</sub>]Cl<sub>3</sub> is ________.
<br/>M<sub>[Co(NH<sub>3</sub>)<sub>6</sub>Cl<sub>3</sub>]</sub> = 267.46 g/mol
<br/>M<sub>AgNO<sub>3</sub></sub> = 169.87 g/mol | [] | null | 26.80to27.00 | [Co(NH<sub>3</sub>)<sub>6</sub>]Cl<sub>3</sub> + 3AgNO<sub>3</sub> $$ \to $$ 3AgCl
<br><br><span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;">Mole of [Co(NH<sub>3</sub>)<sub>6</sub>]Cl<sub>3</sub></div>
<div style="text-align: center;">... | integer | jee-main-2020-online-8th-january-morning-slot | 3,771 |
Bp8Ev13EnWjZbZQ2KYjgy2xukfcgtpoy | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | Consider the following equations :
<br/>2Fe<sup>2+</sup> + H<sub>2</sub>O<sub>2</sub> $$ \to $$ xA + yB
<br/>(in basic medium)
<br/>2MnO<sub>4</sub><sup>-</sup> + 6H<sup>+</sup> + 5H<sub>2</sub>O<sub>2</sub> $$ \to $$ x'C + y'D + z'E
<br/>(in acidic medium)
<br/>The sum of the stoichiometric coefficients
x, y, x', y'... | [] | null | 19 | 2Fe<sup>2+</sup> + H<sub>2</sub>O<sub>2</sub> $$ \to $$ 2Fe<sup>3+</sup> + 2OH<sup>–</sup>
<br><br>2MnO<sub>4</sub><sup>-</sup> + 6H<sup>+</sup> + 5H<sub>2</sub>O<sub>2</sub> $$ \to $$ 2Mn<sup>2+</sup> + 8H<sub>2</sub>O + 5O<sub>2</sub>
<br><br>$$ \therefore $$ x + y + x' + y' + z' = 19 | integer | jee-main-2020-online-4th-september-evening-slot | 3,772 |
eU8LhsnUGXRFHufLBq1kmhuwbk4 | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | $$2MnO_4^ - + b{C_2}O_4^{2 - } + c{H^ + } \to xM{n^{2 + }} + yC{O_2} + z{H_2}O$$<br/><br/>If the above equation is balanced with integer coefficients, the value of c is ___________. (Round off to the Nearest Integer). | [] | null | 16 | $$2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$$ | integer | jee-main-2021-online-16th-march-morning-shift | 3,773 |
1krq6fn86 | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | 250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ___________ $$\times$$ 10<sup>21</sup>. (Nearest integer)<br/><br/>(N<sub>A</sub> = 6.022 $$\times$$ 10<sup>23</sup>) | [] | null | 226 | We know that, number of moles = V<sub>L</sub> $$\times$$ molarity and number of millimoles = V<sub>mL</sub> $$\times$$ molarity<br><br>So, millimoles of NaOH = 250 $$\times$$ 0.5 = 125<br><br>Millimoles of HCl = 500 $$\times$$ 1 = 500<br><br>Now, reaction is<br><br><img src="https://app-content.cdn.examgoal.net/fly/@wi... | integer | jee-main-2021-online-20th-july-morning-shift | 3,774 |
1l58kj1r9 | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | <p>The moles of methane required to produce 81 g of water after complete combustion is _____________ $$\times$$ 10<sup>$$-$$2</sup> mol. [nearest integer]</p> | [] | null | 225 | $$
\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}
$$<br/><br/>
POAC on $\mathrm{H}$ atom<br/><br/>
$$
\begin{aligned}
&\mathrm{n}_{\mathrm{CH} 4} \times 4=\mathrm{n}_{\mathrm{H} 2 \mathrm{O}} \times 2 \\\\
&\mathrm{n}_{\mathrm{CH}_4}=\frac{81}{18} \times 2 \times \frac{1}{4}=\frac{81}{... | integer | jee-main-2022-online-26th-june-evening-shift | 3,775 |
1l5amw0b1 | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | <p>Number of grams of bromine that will completely react with 5.0 g of pent-1-ene is ___________ $$\times$$ 10<sup>$$-$$2</sup> g. (Atomic mass of Br = 80 g/mol) [Nearest Integer]</p> | [] | null | 1143 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5fiey4l/9f7176f9-a5d6-487c-9f31-6b0fb1a2791b/789bc040-006a-11ed-983d-b1f8143a2272/file-1l5fiey4m.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5fiey4l/9f7176f9-a5d6-487c-9f31-6b0fb1a2791b/789bc040-006a-11ed-983d-b1f8143a2272... | integer | jee-main-2022-online-25th-june-morning-shift | 3,777 |
1l5w6k3gr | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | <p>Blister copper is produced by reaction of copper oxide with copper sulphide.</p>
<p>2Cu<sub>2</sub>O + Cu<sub>2</sub>S $$\to$$ 6Cu + SO<sub>2</sub></p>
<p>When 2.86 $$\times$$ 10<sup>3</sup> g of Cu<sub>2</sub>O and 4.77 $$\times$$ 10<sup>3</sup> g of Cu<sub>2</sub>S are used for reaction, the mass of copper produce... | [] | null | 3810 | moles of $\mathrm{Cu}_{2} \mathrm{O}=\frac{2.86 \times 10^{3}}{143}=20$
<br/><br/>
moles of $\mathrm{Cu}_{2} \mathrm{S}=\frac{4.77 \times 10^{3}}{159}=30$
<br/><br/>
$\mathrm{Cu}_{2} \mathrm{O}$ is limiting reagent
<br/><br/>
$$
\begin{array}{lccc}
2 \mathrm{Cu}_{2} \mathrm{O} & + & \mathrm{Cu}_{2} \mathrm{~S} \longrig... | integer | jee-main-2022-online-30th-june-morning-shift | 3,778 |
1l6f7j8oo | chemistry | some-basic-concepts-of-chemistry | laws-of-chemical-combination | <p>56.0 L of nitrogen gas is mixed with excess hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _________ L.</p> | [] | null | 46 | $$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})$$
<br/><br/>
Since $$\mathrm{H}_{2}$$ is in excess and $$20 \mathrm{~L}$$ of ammonia gas is produced.
<br/><br/>
Hence, 2 moles $$\mathrm{NH}_{3} \equiv 1$$ mole $$\mathrm{N}_{2} \quad(v \propto \mathrm{n})$$
<br/><br... | integer | jee-main-2022-online-25th-july-evening-shift | 3,779 |
QzGotHLXODpQD9r6 | chemistry | some-basic-concepts-of-chemistry | mole-concept | If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will | [{"identifier": "A", "content": "be a function of the molecular mass of the substance"}, {"identifier": "B", "content": "remain unchanged"}, {"identifier": "C", "content": "increase two fold"}, {"identifier": "D", "content": "decrease twice"}] | ["D"] | null | Let Relative Atomic Mass is in short <b>R.A.M</b> then<br><br>
R.A.M = $${{mass\,of\,1\,atom} \over {{1 \over {12}} \times mass\,of\,1\,atom\,of\,{C^{12}}}}$$<br><br>
$$ \Rightarrow $$ $${{mass\,of\,1\,atom} \over {mass\,of\,1\,atom\,of\,{C^{12}}}}$$ $$\times$$ 12<br><br>
Now if we take 1/6 instead of 1/12 at first th... | mcq | aieee-2005 | 3,781 |
ml8U6CMJWDAl2xLH | chemistry | some-basic-concepts-of-chemistry | mole-concept | How many moles of magnesium phosphate Mg<sub>3</sub>(PO<sub>4</sub>)<sub>2</sub> will contain 0.25 mole of oxygen atoms? | [{"identifier": "A", "content": "1.25 $$\\times$$ 10<sup>-2</sup>"}, {"identifier": "B", "content": "2.5 $$\\times$$ 10<sup>-2</sup>"}, {"identifier": "C", "content": "0.02"}, {"identifier": "D", "content": "3.125 $$\\times$$ 10<sup>-2</sup>"}] | ["D"] | null | In one Mg<sub>3</sub>(PO<sub>4</sub>)<sub>2</sub> molecule no of oxygen atoms = 8
<br><br>So if there is M mole of Mg<sub>3</sub>(PO<sub>4</sub>)<sub>2</sub> molecules then no of moles of oxygen atoms is = M $$ \times $$ 8
<br><br>According to question,
<br><br>M $$ \times $$ 8 = 0.25
<br><br>$$ \Rightarrow $$ M = 0.03... | mcq | aieee-2006 | 3,782 |
ywVAkLfCch5kY2EFQiWDH | chemistry | some-basic-concepts-of-chemistry | mole-concept | The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H<sub>2</sub>S in the presence of conc. HCl ( assuming 100% conversion) is : | [{"identifier": "A", "content": "0.50 mol "}, {"identifier": "B", "content": "0.25 mol"}, {"identifier": "C", "content": "0.125 mol"}, {"identifier": "D", "content": "0.333 mol"}] | ["C"] | null | 2H<sub>3</sub> As O<sub>4</sub> + 5H<sub>2</sub>S $$\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{HCl}^{conc}} $$ As<sub>2</sub>S<sub>5</sub> + 8H<sub>2</sub>O
<br><br>35.5 g of H<sub>3</sub>AsO<sub>4</sub> = $${{35.5} \over {142}}$$ = 0.25 moles
<br><br>Let, As<sub>2</sub>S<sub>5</sub> prod... | mcq | jee-main-2016-online-9th-april-morning-slot | 3,785 |
BqSFClk0IVTCYAg3sNptI | chemistry | some-basic-concepts-of-chemistry | mole-concept | An organic compound contains C, H and S. The minimum molecular weight of the
compound containing 8% sulphur is :
<br/><br/>(atomic weight of S = 32 amu)
| [{"identifier": "A", "content": "200 g mol<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "400 g mol<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "600 g mol<sup>\u22121</sup>"}, {"identifier": "D", "content": "300 g mol<sup>\u22121</sup>"}] | ["B"] | null | <p>We know that 8% sulphur means 8 g of sulphur present in 100 g of an organic compound.</p>
<p>Thus, 32 g of sulphur present in $${{100} \over 8} \times 32g$$ = 400 g of organic compound.</p>
<p>Therefore, minimum molecular weight of the compound is 400 g mol<sup>$$-$$1</sup>.</p> | mcq | jee-main-2016-online-9th-april-morning-slot | 3,786 |
7CF58UDFv4yg0OcG | chemistry | some-basic-concepts-of-chemistry | mole-concept | 1 gram of a carbonate (M<sub>2</sub>CO<sub>3</sub>) on treatment with excess HCl produces 0.01186 mole of CO<sub>2</sub>. The molar mass of M<sub>2</sub>CO<sub>3</sub> in g mol<sup>–1</sup> is: | [{"identifier": "A", "content": "84.3"}, {"identifier": "B", "content": "118.6"}, {"identifier": "C", "content": "11.86"}, {"identifier": "D", "content": "1186"}] | ["A"] | null | M<sub>2</sub>CO<sub>3</sub> + 2HCl $$\buildrel \, \over
\longrightarrow $$ 2MCl + CO<sub>2</sub> + H<sub>2</sub>O
<br><br>Here weight of M<sub>2</sub>CO<sub>3</sub> = 1 gm
<br><br>Let molar mass of M<sub>2</sub>CO<sub>3</sub> = M
<br><br>$$\therefore\,\,\,$$ No of moles of M<sub>2</sub>CO<sub>3</sub> = $${1 \over M}... | mcq | jee-main-2017-offline | 3,787 |
c1FL4mkVPUwV7MZv | chemistry | some-basic-concepts-of-chemistry | mole-concept | The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon
(22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all
<sup>1</sup>H atoms are replaced by <sup>2</sup>H atoms is: | [{"identifier": "A", "content": "37.5 kg"}, {"identifier": "B", "content": "7.5 kg"}, {"identifier": "C", "content": "10 kg"}, {"identifier": "D", "content": "15 kg"}] | ["B"] | null | Given that weight of human adult is = 75 kg
<br><br>Among those 75 kg, 10% is Hydrogen(<sup>1</sup>H).
<br><br>$$\therefore$$ Mass of <sup>1</sup>H = $$75 \times {{10} \over {100}}$$ = 7.5 kg
<br><br>Now when every <sup>1</sup>H atom is replaced by <sup>2</sup>H atom then weight of every atom is become double. So total... | mcq | jee-main-2017-offline | 3,788 |
4R8sj0h5Qd1wYZP3QAxlC | chemistry | some-basic-concepts-of-chemistry | mole-concept | An organic compound is estimated through Duma's method and was found to evolve 6 moles of CO<sub>2</sub>. 4 moles of H<sub>2</sub>O and 1 mole of nitrogen gas. The formula of the compound is : | [{"identifier": "A", "content": "C<sub>6</sub>H<sub>8</sub>N"}, {"identifier": "B", "content": "C<sub>6</sub>H<sub>8</sub>N<sub>2</sub>"}, {"identifier": "C", "content": "C<sub>12</sub>H<sub>8</sub>N"}, {"identifier": "D", "content": "C<sub>12</sub>H<sub>8</sub>N<sub>2</sub>"}] | ["B"] | null | Molar ratio of C : H : N : : 6 : 8 : 2 i.e., 3 : 4 : 1
<br><br>Thus, the correct formula is C<sub>6</sub>H<sub>8</sub>N<sub>2</sub>.
. | mcq | jee-main-2019-online-11th-january-morning-slot | 3,790 |
qkiVs9dLdkXLEkkmNV3rsa0w2w9jx0xkpa8 | chemistry | some-basic-concepts-of-chemistry | mole-concept | The minimum amount of O<sub>2</sub>(g) consumed per gram of reactant is for the reaction : <br/>(Given atomic mass : Fe =
56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
| [{"identifier": "A", "content": "4Fe(s) + 3O<sub>2</sub>(g) $$ \\to $$ 2Fe<sub>2</sub>O<sub>3</sub>(s)"}, {"identifier": "B", "content": "P<sub>4</sub>(s) + 5O<sub>2</sub>(g) $$ \\to $$ P<sub>4</sub>O<sub>10</sub>(s)"}, {"identifier": "C", "content": "C<sub>3</sub>H<sub>8</sub>(g) + 5O<sub>2</sub>(g) $$ \\to $$ 3CO<sub... | ["A"] | null | (a) 4Fe(s) + 3O<sub>2</sub>(g) $$ \to $$ 2Fe<sub>2</sub>O<sub>3</sub>(s)
<br><br>$${{Moles\,of\,{O_2}} \over 3} = {{Moles\,of\,{Fe}} \over 4}$$
<br><br>$$ \Rightarrow $$ Moles of O<sub>2</sub> = $${3 \over 4}$$ $$ \times $$ $${1 \over {56}}$$ moles = $${1 \over {24.8}}$$ moles = $${3 \over {224}} \times 32$$ g of O<sub... | mcq | jee-main-2019-online-10th-april-evening-slot | 3,791 |
YaKfwPtJcpsSzoQMmN3rsa0w2w9jx55sska | chemistry | some-basic-concepts-of-chemistry | mole-concept | 5 moles of AB<sub>2</sub> weigh 125 × 10<sup>–3</sup>
kg and 10 moles of A<sub>2</sub>B<sub>2</sub> weigh 300 × 10<sup>–3</sup>
kg. The molar mass of A (M<sub>A</sub>)
and molar mass of B (M<sub>B</sub>) in kg mol are:
| [{"identifier": "A", "content": "M<sub>A</sub> = 10 \u00d7 10<sup>\u20133</sup>\n and M<sub>B</sub> = 5 \u00d7 10<sup>\u20133</sup>"}, {"identifier": "B", "content": "M<sub>A</sub> = 25 \u00d7 10<sup>\u20133</sup>\n and M<sub>B</sub> = 50 \u00d7 10<sup>\u20133</sup>"}, {"identifier": "C", "content": "M<sub>A</sub> = 5 ... | ["C"] | null | Molar weigh of 'A' = M<sub>A</sub>
<br><br>Molar weigh of 'B' = M<sub>B</sub>
<br><br>Molar mass of AB<sub>2</sub> = M<sub>A</sub> + 2M<sub>B</sub>
<br><br>Molar mass of A<sub>2</sub>B<sub>2</sub> = 2(M<sub>A</sub> + M<sub>B</sub>)
<br><br>5 mol AB<sub>2</sub> weighs 125 g
<br><br>$$ \therefore $$ AB<sub>2</sub> = 25 g... | mcq | jee-main-2019-online-12th-april-morning-slot | 3,792 |
7avp27bvbIPAKIwj1E3rsa0w2w9jx82nf67 | chemistry | some-basic-concepts-of-chemistry | mole-concept | 25 g of an unknown hydrocarbon upon burning produces 88 g of CO<sub>2</sub> and 9 g of H<sub>2</sub>O. This unknown
hydrocarbon contains : | [{"identifier": "A", "content": "18 g of carbon and 7 g of hydrogen"}, {"identifier": "B", "content": "20 g of carbon and 5 g of hydrogen"}, {"identifier": "C", "content": "22 g of carbon and 3 g of hydrogen"}, {"identifier": "D", "content": "24 g of carbon and 1 g of hydrogen"}] | ["D"] | null | C<sub>x</sub>H<sub>y</sub> + $$\left( {x + {y \over 4}} \right)$$O<sub>2</sub> $$ \to $$ xCO<sub>2</sub> + $${y \over 2}$$H<sub>2</sub>O
<br><br>No of moles of CO<sub>2</sub> produced = $${{88} \over {44}}$$ = 2 mol
<br>No of moles of H<sub>2</sub>O produced = $${9 \over {18}}$$ = 0.5 mol
<br><br>According to stoichiom... | mcq | jee-main-2019-online-12th-april-evening-slot | 3,793 |
7oDdEILc7npnNHTQRBjgy2xukf3noxwr | chemistry | some-basic-concepts-of-chemistry | mole-concept | The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ___________. | [] | null | 10 | H<sub>3</sub>PO<sub>2</sub> + NaOH $$ \to $$ NaH<sub>2</sub>PO<sub>2</sub> + H<sub>2</sub>O
<br><br>Using Stoichiometry
<br><br><span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;">Moles of H<sub>3</sub>PO<sub>2</sub> reacted</div>
<div s... | integer | jee-main-2020-online-3rd-september-evening-slot | 3,794 |
ytuTg6iZ987Ftn1rt6jgy2xukg3dv83p | chemistry | some-basic-concepts-of-chemistry | mole-concept | The average molar mass of chlorine is
35.5 g mol<sup>–1</sup>. The ratio of <sup>35</sup>Cl to <sup>37</sup>Cl in naturally
occuring chlorine is close to : | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "2 : 1"}, {"identifier": "C", "content": "3 : 1"}, {"identifier": "D", "content": "4 : 1"}] | ["C"] | null | Average molar mass = $${{{n_1}{M_1} + {n_2}{M_2}} \over {\left( {{n_1} + {n_2}} \right)}}$$
<br><br>$$ \therefore $$ 35.5 = $${{35x + 37y} \over {x + y}}$$
<br><br>$$ \Rightarrow $$ 1.5y = –0.5x
<br><br>$$ \Rightarrow $$ $${x \over y} = {3 \over 1}$$ | mcq | jee-main-2020-online-6th-september-evening-slot | 3,795 |
M0Pyz1tg2ah7bG59We1klut7hve | chemistry | some-basic-concepts-of-chemistry | mole-concept | The NaNO<sub>3</sub> weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na<sup>+</sup> per mL is _________ g. (Rounded off to the nearest integer)<br/><br/>[Given : Atomic weight in g mol<sup>$$-$$1</sup> - Na : 23; N : 14; O : 16] | [] | null | 13 | Na<sup>+</sup> = 70 mg/mL<br><br>W<sub>Na+</sub> in 50 mL solution<br><br> = 70 $$\times$$ 50 mg<br><br>= 3500 mg<br><br>= 3.5 gm<br><br>Moles of Na<sup>+</sup> in 50 mL solution = $${{3.5} \over {23}}$$<br><br>Moles of NaNO<sub>3</sub> = moles of Na<sup>+</sup><br><br>= $${{3.5} \over {23}}$$ mol<br><br>Mass of NaNO<s... | integer | jee-main-2021-online-26th-february-evening-slot | 3,796 |
1krq71exh | chemistry | some-basic-concepts-of-chemistry | mole-concept | An average person needs about 10000 kJ energy per day. The amount of glucose (molar mass = 180.0 g mol<sup>$$-$$1</sup>) needed to meet this energy requirement is ____________ g.<br/><br/>(Use : $$\Delta$$<sub>C</sub>H(glucose) = $$-$$2700 kJ mol<sup>$$-$$1</sup>) | [] | null | 667 | 1 mole glucose give 2700 kJ energy, <br/><br/>so mole of glucose needed for 10<sup>4</sup> kJ energy<br/><br/>= $${{10000} \over {2700}} = 3.703$$ moles<br/><br/>Weight of glucose = 3.703 $$\times$$ 180 g/moles<br/><br/>= 666.666<br/><br/>$$\approx$$ 667 g<br/><br/>Hence, amount of glucose required is 667 g. | integer | jee-main-2021-online-20th-july-morning-shift | 3,798 |
1krrl3nas | chemistry | some-basic-concepts-of-chemistry | mole-concept | 4g equimolar mixture of NaOH and Na<sub>2</sub>CO<sub>3</sub> contains x g of NaOH and y g of Na<sub>2</sub>CO<sub>3</sub>. The value of x is ____________ g. (Nearest integer) | [] | null | 1 | Mass of NaOH = x<br><br>Moles of NaOH = $${x \over {40}}$$<br><br>Mass of Na<sub>2</sub>CO<sub>3</sub> = y<br><br>Moles of Na<sub>2</sub>CO<sub>3</sub> = $${y \over {106}}$$<br><br>$${x \over {40}} = {y \over {106}}$$<br><br>x + y = 4<br><br>x = 1.1, y = 2.9<br><br>x = 1.1 $$ \approx $$ 1 (nearest integer) | integer | jee-main-2021-online-20th-july-evening-shift | 3,799 |
1ktn2jgp0 | chemistry | some-basic-concepts-of-chemistry | mole-concept | The number of atoms in 8g of sodium is x $$\times$$ 10<sup>23</sup>. The value of x is ____________. (Nearest integer)<br/><br/>[Given : N<sub>A</sub> = 6.02 $$\times$$ 10<sup>23</sup> mol<sup>$$-$$1</sup><br/><br/>Atomic mass of Na = 23.0 u] | [] | null | 2 | Given, mass of Na = 8g<br/><br/>Molar mass of Na = 23 gmol<sup>$$-$$1</sup><br/><br/>$${{Weight\,of\,sodium\,atom} \over {Molecular\,mass\,of\,sodium\,atom}} = {{Number\,of\,atoms} \over {Avogadro's\,number}}$$<br/><br/>$${{8g} \over {23g}} = {{Number\,of\,atoms} \over {6.022 \times {{10}^{23}}}}$$<br/><br/>Number of a... | integer | jee-main-2021-online-1st-september-evening-shift | 3,800 |
1l57t743a | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>Two elements A and B which form 0.15 moles of A<sub>2</sub>B and AB<sub>3</sub> type compounds. If both A<sub>2</sub>B and AB<sub>3</sub> weigh equally, then the atomic weight of A is _____________ times of atomic weight of B.</p> | [] | null | 2 | <p>Let atomic weight of A = x</p>
<p>and atomic weight of B = y</p>
<p>$$\therefore$$ Molar mass of A<sub>2</sub>B = 2x + y</p>
<p>and molar mass of AB<sub>3</sub> = x + 3y</p>
<p>Now, weight of 0.15 moles of A<sub>2</sub>B = (2x + y) 0.15</p>
<p>and weight of 0.15 moles of AB<sub>3</sub> = (x + 3y) 0.15</p>
<p>Given, ... | integer | jee-main-2022-online-27th-june-morning-shift | 3,801 |
1l5amk19i | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>1 L aqueous solution of H<sub>2</sub>SO<sub>4</sub> contains 0.02 m mol H<sub>2</sub>SO<sub>4</sub>. 50% of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), 0.01 m mol of H<sub>2</sub>SO<sub>4</sub> are added. Total m mols of H<sub>2</sub>SO<sub>4</sub> in the final solution i... | [] | null | 0 | $\mathrm{n}_{\mathrm{H}_2 \mathrm{SO}_4}$ in $\mathrm{Sol}^{\mathrm{n}} \mathrm{A}=50 \%$ of original solution<br/><br/> $=0.01 \mathrm{~m} \mathrm{~mol}$.<br/><br/>
$\mathrm{n}_{\mathrm{H}_2 \mathrm{SO}_4}$ in Final solution $=0.01+0.01$<br/><br/> $=0.02 \,\mathrm{m\,mol}$ <br/><br/>$=0.00002 \times 10^3 \mathrm{m\,mo... | integer | jee-main-2022-online-25th-june-morning-shift | 3,802 |
1l6f7ystm | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>A sample of 4.5 mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is __________ g/mol. [Nearest integer]</p> | [] | null | 33 | $$\mathrm{R}-\mathrm{OH}+\mathrm{CH}_{3} \mathrm{Mgl} \Rightarrow \mathrm{R}-\mathrm{OMgl}+\mathrm{CH}_{4}$$<br/><br/>
moles of alcohol $(\mathrm{ROH}) \equiv$ moles of $\mathrm{CH}_{4}$<br/><br/> At STP,
[Assuming STP]
<br/><br/>
1 mole corresponds to $$22.7 \mathrm{~L}$$
<br/><br/>
Hence... | integer | jee-main-2022-online-25th-july-evening-shift | 3,803 |
1l6i4ah7e | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>Hemoglobin contains $$0.34 \%$$ of iron by mass. The number of Fe atoms in $$3.3 \mathrm{~g}$$ of hemoglobin is</p>
<p>(Given: Atomic mass of Fe is $$56 \,\mathrm{u}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$$.)</p> | [{"identifier": "A", "content": "$$1.21 \\times 10^{5}$$"}, {"identifier": "B", "content": "$$12.0 \\times 10^{16}$$"}, {"identifier": "C", "content": "$$1.21 \\times 10^{20}$$"}, {"identifier": "D", "content": "$$3.4 \\times 10^{22}$$"}] | ["C"] | null | According to the question,
<br/><br/>
$$100 \mathrm{~g}$$ of hemoglobin contains $$0.34 \mathrm{~g}$$ of iron<br/><br/> $$3.3 \mathrm{~g}$$ of hemoglobin contains $$\frac{0.34}{100} \times 3.3 \mathrm{~g}$$ of iron <br/><br/>moles of $$\mathrm{Fe}=\frac{0.34 \times 3.3}{100 \times 56}=\frac{\mathrm{N}}{\mathrm{N}_{\mat... | mcq | jee-main-2022-online-26th-july-evening-shift | 3,804 |
1l6i6o5k6 | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>A $$100 \mathrm{~mL}$$ solution of $$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}$$ on treatment with methanol produces $$2.24 \mathrm{~mL}$$ of a gas at STP. The weight of gas produced is _____________ mg. [nearest integer]</p> | [] | null | 3 | $$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{3}+\operatorname{MgBr}\left(\mathrm{OCH}_{3}\right)$$
<br/><br/>
As $$2.24 \,\mathrm{ml}$$ is formed at STP.
<br/><br/>
Number of moles of ethane gas produced
<br/><br/>
$$
=\frac{2.24 X}{22.4}
$$
<br/><... | integer | jee-main-2022-online-26th-july-evening-shift | 3,805 |
1l6jm83xp | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>In Carius method of estimation of halogen, $$0.45 \mathrm{~g}$$ of an organic compound gave $$0.36 \mathrm{~g}$$ of $$\mathrm{AgBr}$$. Find out the percentage of bromine in the compound.</p>
<p>(Molar masses : $$\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$$)</p> | [{"identifier": "A", "content": "34.04%"}, {"identifier": "B", "content": "40.04%"}, {"identifier": "C", "content": "36.03%"}, {"identifier": "D", "content": "38.04%"}] | ["A"] | null | Mass of organic compound $=0.45 \,\mathrm{gm}$<br/><br/>
Mass of $\mathrm{AgBr}$ obtained $=0.36 \,\mathrm{gm}$<br/><br/>
$\therefore$ Moles of $\mathrm{AgBr}=\frac{0.36}{188}$<br/><br/>
$\therefore$ Mass of Bromine $=\frac{0.36}{188} \times 80=0.1532 \,\mathrm{gm}$<br/><br/>
$\therefore \%\, \mathrm{Br}$ in compound $... | mcq | jee-main-2022-online-27th-july-morning-shift | 3,806 |
1l6rj54no | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>Consider the reaction</p>
<p>$$4 \mathrm{HNO}_{3}(1)+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})$$</p>
<p>The amount of $$\mathrm{HNO}_{3}$$ required to produce $$110.0 \mathrm{~g}$$ of $$\mathrm... | [{"identifier": "A", "content": "32.2 g"}, {"identifier": "B", "content": "69.4 g"}, {"identifier": "C", "content": "91.5 g"}, {"identifier": "D", "content": "162.5 g"}] | ["C"] | null | $$
\begin{array}{r}
4 \mathrm{HNO}_{3}(\mathrm{l})+3 \mathrm{KCl}(\mathrm{s}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+ \\
2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})
\end{array}
$$
<br/><br/>$\because 110 \mathrm{~g}$ of $\mathrm{KNO}_{3} \Rightarrow$ moles of... | mcq | jee-main-2022-online-29th-july-evening-shift | 3,807 |
1l6rlznle | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344 mL of $$\mathrm{H}_{2}$$ gas at STP. The number of alcoholic hydrogens present in compound 'X' is ________.</p> | [] | null | 6 | <p>STP conditions define the volume of 1 mole of any gas as $22.4$ liters or $22400$ mL. The volume of hydrogen gas evolved here is $1.344$ mL, so we can calculate the number of moles of hydrogen gas ($H_2$) using the formula:</p>
<p>$n = \frac{V}{V_m}$ </p>
<p>where:</p>
<ul>
<li>$n$ is the number of moles,</li>
<li>$... | integer | jee-main-2022-online-29th-july-evening-shift | 3,809 |
ldoa0apq | chemistry | some-basic-concepts-of-chemistry | mole-concept | A sample of a metal oxide has formula $\mathrm{M}_{0.83} \mathrm{O}_{1.00}$. The metal $\mathrm{M}$ can exist in two oxidation states $+2$ and $+3$.<br/><br/> In the sample of $\mathrm{M}_{0.83} \mathrm{O}_{1.00}$, the percentage of metal ions existing in $+2$ oxidation state is __________ $\%$. (nearest integer) | [] | null | 59 | Let, metal $\mathrm{M}$ present x and (0.83 - x) in oxidation states $+2$ and $+3$ respectively.
<br/><br/>$$
\begin{aligned}
& 2 x+3(0.83-x)=2 \\\\
& x=0.49 \\\\
& \% M^{2+}=\frac{0.49}{0.83} \times 100 \\\\
& =59 \%
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-evening-shift | 3,810 |
1ldpq1921 | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>On complete combustion, $$0.492 \mathrm{~g}$$ of an organic compound gave $$0.792 \mathrm{~g}$$ of $$\mathrm{CO}_{2}$$. The % of carbon in the organic compound is ___________ (Nearest integer)</p> | [] | null | 44 | Weight of $\mathrm{C}$ in $0.792 \mathrm{gm} $ $\mathrm{CO}_2$
<br/><br/>$$
\begin{aligned}
& =\frac{12}{44} \times 0.792=0.216 \\\\
& \% \text { of } C \text { in compound }=\frac{0.216}{0.492} \times 100 \\\\
& =43.90 \%
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-morning-shift | 3,812 |
1ldsdgdg7 | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>The volume of HCl, containing 73 g L$$^{-1}$$, required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water, is __________ mL. (Nearest Integer)</p>
<p>(Given : molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g mol$$^{-1}$$ respectively.)</p> | [] | null | 15 | <p>$$\mathrm{\mathop {Na + {H_2}O}\limits_{0.69\,g} \to \mathop {NaOH + {1 \over 2}{H_2}}\limits_{0.03\,moles}} $$</p>
<p>= 0.03 moles</p>
<p>$$\therefore 0.03=2\times\mathrm{V}$$</p>
<p>$$\mathrm{V=\frac{0.03}{2}L}$$</p>
<p>= 15 mL</p> | integer | jee-main-2023-online-29th-january-evening-shift | 3,813 |
1ldsdvyho | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of CO$$_2$$ was produced. The molar mass of compound is _____________ g mol$$^{-1}$$ (Nearest integer).</p> | [] | null | 200 | <p>Number of moles of $$C{O_2} = {{4.4} \over {44}} = 0.1$$</p>
<p>$$\therefore$$ Number of moles of C in 1 mole of compound = 10</p>
<p>$$\therefore$$ $$120 = {{60} \over {100}} \times (x)$$ [where x is molar mass of OC]</p>
<p>Molar mass = 200 g mol$$^{-1}$$</p> | integer | jee-main-2023-online-29th-january-evening-shift | 3,814 |
1ldu3z97f | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>Number of hydrogen atoms per molecule of a hydrocarbon A having 85.8% carbon is __________<br/><br/> (Given : Molar mass of A = 84 g mol$$^{-1}$$)</p> | [] | null | 12 | Molar mass of a hydrocarbon $(A)=84 \mathrm{~g} / \mathrm{mol}$
<br/><br/>
Mass of carbon in $1 \mathrm{~mol}$ of $(A)=\frac{85.8}{100} \times 84$
<br/><br/>
$$
=72 ~\mathrm{gm}
$$
<br/><br/>
Mass of hydrogen in $1 \mathrm{~mol}$ of $(A)=12 ~\mathrm{gm}$
<br/><br/>
$\therefore$ Number of $\mathrm{H}$-atoms in a molecul... | integer | jee-main-2023-online-25th-january-evening-shift | 3,815 |
1lgp3jml4 | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>$$1 \mathrm{~g}$$ of a carbonate $$\left(\mathrm{M}_{2} \mathrm{CO}_{3}\right)$$ on treatment with excess $$\mathrm{HCl}$$ produces $$0.01 \mathrm{~mol}$$ of $$\mathrm{CO}_{2}$$. The molar mass of $$\mathrm{M}_{2} \mathrm{CO}_{3}$$ is __________ $$\mathrm{g} ~\mathrm{mol}^{-1}$$. (Nearest integer)</p> | [] | null | 100 | When the carbonate reacts with HCl, it produces CO<sub>2</sub>, as shown in the following balanced equation:
<br/><br/>
$$\mathrm{M}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{MCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$$
<br/><br/>
From the problem, we know that 1 g of the carbonate produces 0.01 mol of CO<... | integer | jee-main-2023-online-13th-april-evening-shift | 3,818 |
1lgrkmdoz | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>A metal chloride contains $$55.0 \%$$ of chlorine by weight . $$100 \mathrm{~mL}$$ vapours of the metal chloride at STP weigh $$0.57 \mathrm{~g}$$. The molecular formula of the metal chloride is</p>
<p>(Given: Atomic mass of chlorine is $$35.5 \mathrm{u}$$)</p> | [{"identifier": "A", "content": "$$\\mathrm{MCl}_{2}$$"}, {"identifier": "B", "content": "$$\\mathrm{MCl}_{4}$$"}, {"identifier": "C", "content": "$$\\mathrm{MCl}_{3}$$"}, {"identifier": "D", "content": "$$\\mathrm{MCl}$$"}] | ["A"] | null | <p>The weight percent of chlorine in the compound is given as 55.0%. This implies that the weight percent of the metal is 45.0%. </p>
<p>Now, we need to find the molar mass of the compound. At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L. We know that 100 mL (or 0.1 L) of the metal chlorid... | mcq | jee-main-2023-online-12th-april-morning-shift | 3,819 |
1lgvv3573 | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>Match List I with List II</p>
<p><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;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-II, B-III, C-IV, D-I"}, {"identifier": "B", "content": "A-II, B-IV, C-I, D-III"}, {"identifier": "C", "content": "A-I, B-III, C-II, D-IV"}, {"identifier": "D", "content": "A-II, B-IV, C-III, D-I"}] | ["B"] | null | <p>A. 16 g of $\mathrm{CH_4~(g)}$</p>
<p>The molar mass of $\mathrm{CH_4}$ (Methane) is 16 g/mol. Therefore, 16 g of $\mathrm{CH_4}$ is equivalent to 1 mole of $\mathrm{CH_4}$. Furthermore, each molecule of $\mathrm{CH_4}$ has 10 electrons (6 from Carbon and 4 from Hydrogen). As a result, 1 mole of $\mathrm{CH_4}$ (or ... | mcq | jee-main-2023-online-10th-april-evening-shift | 3,820 |
1lgyg3ikm | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>The number of molecules and moles in 2.8375 litres of O$$_2$$ at STP are respectively</p> | [{"identifier": "A", "content": "1.505 $$\\times$$ 10$$^{23}$$ and 0.250 mol"}, {"identifier": "B", "content": "7.527 $$\\times$$ 10$$^{22}$$ and 0.250 mol"}, {"identifier": "C", "content": "7.527 $$\\times$$ 10$$^{23}$$ and 0.125 mol"}, {"identifier": "D", "content": "7.527 $$\\times$$ 10$$^{22}$$ and 0.125 mol"}] | ["D"] | null | <p>At STP, one mole of any gas occupies 22.4 liters. Therefore, 2.8375 liters of oxygen gas at STP is equal to 0.125 moles. The number of molecules of oxygen gas in 0.125 moles is calculated by multiplying the number of moles by Avogadro's number, which is $$6.022 × 10^{23} $$ molecules/mol. This gives us a total o... | mcq | jee-main-2023-online-10th-april-morning-shift | 3,821 |
1lh04nq5b | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>$$0.5 \mathrm{~g}$$ of an organic compound $$(\mathrm{X})$$ with $$60 \%$$ carbon will produce __________ $$\times 10^{-1} \mathrm{~g}$$ of $$\mathrm{CO}_{2}$$ on complete combustion.</p> | [] | null | 11 | Mass of Carbon $=12$<br/><br/>
Molar Mass of $\mathrm{CO}_2=12+(16 \times 2)=44$<br/><br/>
Mass of Compound $=0.5 \mathrm{~g}$<br/><br/>
$$
\begin{aligned}
&\% \text { of } \mathrm{C} =\frac{\text { Molar mass of } \mathrm{C} \times \text { Mass Of } \mathrm{CO}_2}{\text { Mass Of Compound } \times \text { Molar Mass }... | integer | jee-main-2023-online-8th-april-morning-shift | 3,822 |
1lh29u8d5 | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>If 5 moles of $$\mathrm{BaCl}_{2}$$ is mixed with 2 moles of $$\mathrm{Na}_{3} \mathrm{PO}_{4}$$, the maximum number of moles of $$\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}$$ formed is ___________ (Nearest integer)</p> | [] | null | 1 | <p>Given the balanced chemical equation:</p>
<p>$3 \mathrm{BaCl}_2 + 2 \mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3\mathrm{(PO}_4)_2 + 6 \mathrm{NaCl}$</p>
<p>We can see that 3 moles of $\mathrm{BaCl}_2$ react with 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$ to produce 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$.</p... | integer | jee-main-2023-online-6th-april-morning-shift | 3,823 |
jaoe38c1lscsbmyf | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>Volume of $$3 \mathrm{M} \mathrm{~NaOH}$$ (formula weight $$40 \mathrm{~g} \mathrm{~mol}^{-1}$$ ) which can be prepared from $$84 \mathrm{~g}$$ of $$\mathrm{NaOH}$$ is __________ $$\times 10^{-1} \mathrm{dm}^3$$.</p> | [] | null | 7 | <p>First, let's calculate the number of moles of NaOH that can be prepared from $$84\ \text{g}$$ of NaOH. The molar mass of NaOH is given as $$40\ \text{g/mol}$$.</p>
<p>The number of moles (n) is calculated using the formula:</p>
$$ n = \frac{mass}{molar\ mass} $$
<p>So for our case:</p>
$$ n = \frac{84\ \text{g}}... | integer | jee-main-2024-online-27th-january-evening-shift | 3,824 |
jaoe38c1lse8hulm | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>Molar mass of the salt from $$\mathrm{NaBr}, \mathrm{NaNO}_3, \mathrm{KI}$$ and $$\mathrm{CaF}_2$$ which does not evolve coloured vapours on heating with concentrated $$\mathrm{H}_2 \mathrm{SO}_4$$ is ________ $$\mathrm{g} \mathrm{~mol}{ }^{-1}$$.</p>
<p>(Molar mass in $$\mathrm{g} \mathrm{~mol}^{-1}: \mathrm{Na}: 2... | [] | null | 78 | <p>$$\mathbf{C a F}_2$$ does not evolve any gas with concentrated $$\mathrm{H}_2 \mathrm{SO}_4$$.</p>
<p>$$\mathrm{NaBr} \rightarrow$$ evolve $$\mathrm{Br}_2$$</p>
<p>$$\mathrm{NaNO}_3 \rightarrow$$ evolve $$\mathrm{NO}_2$$</p>
<p>$$\mathrm{KI} \rightarrow$$ evolve $$\mathrm{I}_2$$</p> | integer | jee-main-2024-online-31st-january-morning-shift | 3,826 |
jaoe38c1lsfpilz3 | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>If $$50 \mathrm{~mL}$$ of $$0.5 \mathrm{M}$$ oxalic acid is required to neutralise $$25 \mathrm{~mL}$$ of $$\mathrm{NaOH}$$ solution, the amount of $$\mathrm{NaOH}$$ in $$50 \mathrm{~mL}$$ of given $$\mathrm{NaOH}$$ solution is ______ g.</p> | [] | null | 4 | <p>Equivalent of Oxalic acid $$=$$ Equivalents of $$\mathrm{NaOH}$$</p>
<p>$$\begin{aligned}
& 50 \times 0.5 \times 2=25 \times \mathrm{M} \times 1 \\
& \mathrm{M}_{\mathrm{NaOH}}=2 \mathrm{M} \\
& \mathrm{W}_{\mathrm{NaOH}} \text { in } 50 \mathrm{ml}=2 \times 50 \times 40 \times 10^{-3} \mathrm{~g}=4 \mathrm{g}
\end{... | integer | jee-main-2024-online-29th-january-evening-shift | 3,827 |
1lsgz0zjb | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>$$0.05 \mathrm{~cm}$$ thick coating of silver is deposited on a plate of $$0.05 \mathrm{~m}^2$$ area. The number of silver atoms deposited on plate are ________ $$\times 10^{23}$$. (At mass $$\mathrm{Ag}=108, \mathrm{~d}=7.9 \mathrm{~g} \mathrm{~cm}^{-3}$$)</p> | [] | null | 11 | <p>$$\begin{aligned}
&\begin{aligned}
& \text { Volume of silver coating }=0.05 \times 0.05 \times 10000 \\
& =25 \mathrm{~cm}^3 \\
& \text { Mass of silver deposited }=25 \times 7.9 \mathrm{~g} \\
& \text { Moles of silver atoms }=\frac{25 \times 7.9}{108} \\
& \text { Number of silver atoms }=\frac{25 \times 7.9}{108... | integer | jee-main-2024-online-30th-january-morning-shift | 3,828 |
lv0vys72 | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>$$\mathrm{Xg}$$ of ethylamine is subjected to reaction with $$\mathrm{NaNO}_2 / \mathrm{HCl}$$ followed by water; evolved dinitrogen gas which occupied $$2.24 \mathrm{~L}$$ volume at STP. X is _________ $$\times 10^{-1} \mathrm{~g}$$.</p> | [] | null | 45 | <p>Moles of $$\mathrm{N}_2=0.1$$</p>
<p>$$\begin{aligned}
\text { Mass of } \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 & =(0.1) \times 45 \\
& =4.5 \mathrm{~gm} \\
& =45 \times 10^{-1} \\
& =45
\end{aligned}
$$</p> | integer | jee-main-2024-online-4th-april-morning-shift | 3,829 |
lv2erz2h | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>From $$6.55 \mathrm{~g}$$ of aniline, the maximum amount of acetanilide that can be prepared will be ________ $$\times 10^{-1} \mathrm{~g}$$.</p> | [] | null | 95 | <p>To determine the maximum amount of acetanilide that can be prepared from 6.55 g of aniline, we need to use stoichiometry. Let's go through the process step by step.</p>
<p>1. <strong>Molecular weights calculation:</strong></p>
<p>The molecular weight of aniline (C<sub>6</sub>H<sub>5</sub>NH<sub>2</sub>) is calcula... | integer | jee-main-2024-online-4th-april-evening-shift | 3,830 |
lv7v49so | chemistry | some-basic-concepts-of-chemistry | mole-concept | <p>An organic compound has $$42.1 \%$$ carbon, $$6.4 \%$$ hydrogen and remainder is oxygen. If its molecular weight is 342 , then its molecular formula is :</p> | [{"identifier": "A", "content": "$$\\mathrm{C}_{14} \\mathrm{H}_{20} \\mathrm{O}_{10}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{C}_{12} \\mathrm{H}_{20} \\mathrm{O}_{12}$$\n"}, {"identifier": "D", "content": "$$\\mathrm{... | ["B"] | null | <p>To determine the molecular formula of the compound, we first calculate the empirical formula based on the given percentages of carbon, hydrogen, and oxygen. Then we use the molecular weight to find the molecular formula.</p>
<p>Step 1: Calculating the moles of each element per 100 g of the compound</p>
<ul>
<li>F... | mcq | jee-main-2024-online-5th-april-morning-shift | 3,831 |
cIb7tW44TFvHbqdR | chemistry | some-basic-concepts-of-chemistry | quantitative-measures-in-chemical-equations | In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight . Molecular weight of the compound is 108 g mol<sup>-1</sup> . Molecular formula of compound is | [{"identifier": "A", "content": "C<sub>2</sub>H<sub>6</sub>N<sub>2</sub>"}, {"identifier": "B", "content": "C<sub>3</sub>H<sub>4</sub>N"}, {"identifier": "C", "content": "C<sub>6</sub>H<sub>8</sub>N<sub>2</sub>"}, {"identifier": "D", "content": "C<sub>9</sub>H<sub>12</sub>N<sub>3</sub>"}] | ["C"] | null | <table class="tg">
<tbody><tr>
<th class="tg-amwm">Element</th>
<th class="tg-amwm">Weight ratio</th>
<th class="tg-amwm">No of atoms of element</th>
<th class="tg-amwm">Simplest Ratio</th>
</tr>
<tr>
<td class="tg-amwm">C</td>
<td class="tg-baqh">9</td>
<td class="tg-baqh">9/12 = 3/4<... | mcq | aieee-2002 | 3,833 |
KAOBLLxcrPv5D22n | chemistry | some-basic-concepts-of-chemistry | quantitative-measures-in-chemical-equations | A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO<sub>2</sub>. The empirical formula of the hydrocarbon is | [{"identifier": "A", "content": "C<sub>2</sub>H<sub>4</sub> "}, {"identifier": "B", "content": "C<sub>3</sub>H<sub>4</sub> "}, {"identifier": "C", "content": "C<sub>6</sub>H<sub>5</sub> "}, {"identifier": "D", "content": "C<sub>7</sub>H<sub>8</sub>"}] | ["D"] | null | Required reaction,
<br><br>C<sub>$$x$$</sub>H<sub>$$y$$</sub> + $$\left( {x + {y \over 4}} \right)$$O<sub>2</sub> $$ \to $$$$x$$CO<sub>2</sub> + $${{y \over 2}}$$H<sub>2</sub>O
<br><br>0.72 gm of H<sub>2</sub>O = $${{{0.72} \over {18}}}$$ mole of H<sub>2</sub>O = 0.04 mole of H<sub>2</sub>O
<br><br>In one H<sub>2</sub>... | mcq | jee-main-2013-offline | 3,834 |
sEZTscq20HMkpB6y | chemistry | some-basic-concepts-of-chemistry | quantitative-measures-in-chemical-equations | Experimentally it was found that a metal oxide has formula M<sub>0.98</sub>O. Metal M, present as M<sup>2+</sup> and M<sup>3+</sup> in its oxide. Fraction of the metal which exists as M<sup>3+</sup> would be | [{"identifier": "A", "content": "7.01%"}, {"identifier": "B", "content": "4.08%"}, {"identifier": "C", "content": "6.05%"}, {"identifier": "D", "content": "5.08%"}] | ["B"] | null | Given metal oxide = M<sub>0.98</sub>O
<br>We know oxidation number of O = (-2)
<br>Now assume oxidaton no of M = $$x$$
<br><br>$$\therefore$$ 0.98$$x$$ + 1$$ \times $$ (-2) = 0
<br>$$ \Rightarrow x =$$ $${{200} \over {98}}$$
<br>This represent the charge in one atom of M. As you can see the charge of M is in the range ... | mcq | jee-main-2013-offline | 3,835 |
7PYAO24NRxqfwJqtl1cLV | chemistry | some-basic-concepts-of-chemistry | quantitative-measures-in-chemical-equations | An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are :
<br/>(Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023 $$ \times $$ 10<sup>23</sup> mol<sup>-1</sup>) | [{"identifier": "A", "content": "6.023 $$ \\times $$ 10<sup>20</sup>"}, {"identifier": "B", "content": "6.023 $$ \\times $$ 10<sup>9</sup>"}, {"identifier": "C", "content": "6.023 $$ \\times $$ 10<sup>21</sup>"}, {"identifier": "D", "content": "6.023 $$ \\times $$ 10<sup>23</sup>"}] | ["A"] | null | % of Cl = 3.55
<br><br>$$\therefore\,\,\,\,$$ In 100 g chlorohydrocarbon 3.55 gm Cl present.
<br><br>In 1 gm chlorohydrocarbon Cl present
<br><br>= $${{3.55} \over {100}}$$
<br><br>= 0.0355 gm
<br><br>$$\therefore\,\,\,\,$$ No of Moles of Cl = $${{0.0355} \over {35.5}}$$
<br><br>= ... | mcq | jee-main-2018-online-16th-april-morning-slot | 3,837 |
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