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1l58k28z7 | chemistry | solid-state | close-packing-in-crystals | <p>In a solid AB, A atoms are in ccp arrangement and B atoms occupy all the octahedral sites. If two atoms from the opposite faces are removed, then the resultant stoichiometry of the compound is A<sub>x</sub>B<sub>y</sub>. The value of x is ____________. [nearest integer]</p> | [] | null | 3 | $A$ atoms are in CCP contribution of $A$ is<br/><br/>
$$A=4$$<br/><br/>
If atoms from opposite faces are removed then<br/><br/>
$$
\begin{aligned}
& A=4-x \times \frac{1}{x} \\\\
& A=3
\end{aligned}
$$<br/><br/>
Value of $x=3$ | integer | jee-main-2022-online-26th-june-evening-shift | 3,473 |
1lh29b0ej | chemistry | solid-state | close-packing-in-crystals | <p>A compound is formed by two elements $$\mathrm{X}$$ and $$\mathrm{Y}$$. The element $$\mathrm{Y}$$ forms cubic close packed arrangement and those of element $$\mathrm{X}$$ occupy one third of the tetrahedral voids. What is the formula of the compound?</p> | [{"identifier": "A", "content": "$$\\mathrm{XY}_{3}$$"}, {"identifier": "B", "content": "$$\\mathrm{X_3Y}_{2}$$"}, {"identifier": "C", "content": "$$\\mathrm{X_3Y}$$"}, {"identifier": "D", "content": "$$\\mathrm{X_2Y}_{3}$$"}] | ["D"] | null | <p>A compound is formed by two elements $$\mathrm{X}$$ and $$\mathrm{Y}$$. The element $$\mathrm{Y}$$ forms a cubic close-packed (CCP) arrangement, and element $$\mathrm{X}$$ occupies one third of the tetrahedral voids.</p>
<p>In a CCP structure, there are 4 atoms of $$\mathrm{Y}$$ in a unit cell. This means there are ... | mcq | jee-main-2023-online-6th-april-morning-shift | 3,476 |
8Z5YSSag1cXd5PT9 | chemistry | solid-state | crystal-structure-of-solids | Na and Mg crystallize in BCC and FCC type crystals respectively, then the number of atoms of
Na and Mg present in the unit cell of their respective crystal is : | [{"identifier": "A", "content": "4 and 2"}, {"identifier": "B", "content": "9 and 14"}, {"identifier": "C", "content": "14 and 9"}, {"identifier": "D", "content": "2 and 4"}] | ["D"] | null | In bcc - points are at corners and one in the center of the unit cell.
<br><br>Number of atoms per unit cell $$ = 8 \times {1 \over 8} + 1 = 2.$$
<br><br>In fcc - points are at the corners and also centre of the six faces of each cell.
<br><br>Number of atoms per unit cell $$ = 8 \times {1 \over 8} + 6 \times {1 \ov... | mcq | aieee-2002 | 3,477 |
MnTPBx7eUUSPDHM9 | chemistry | solid-state | crystal-structure-of-solids | Total volume of atoms present in a face-centre cubic unit cell of a metal is (r is atomic radius) : | [{"identifier": "A", "content": "20/3 $$\\pi r^3$$"}, {"identifier": "B", "content": "24/3 $$\\pi r^3$$"}, {"identifier": "C", "content": "16/3 $$\\pi r^3$$"}, {"identifier": "D", "content": "12/3 $$\\pi r^3$$"}] | ["C"] | null | The face centered cubic unit cell contains $$4$$ atom
<br><br>$$\therefore$$ Total volume of atoms
<br><br>$$ = 4 \times {4 \over 3}\pi {r^3} = {{16} \over 3}\pi {r^3}$$ | mcq | aieee-2006 | 3,478 |
0uJERo1Cizf4UXUY | chemistry | solid-state | crystal-structure-of-solids | In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre
positions. If one atom of B is missing from one of the face centred points, the formula of the compound is : | [{"identifier": "A", "content": "AB<sub>2</sub>"}, {"identifier": "B", "content": "A<sub>2</sub>B<sub>3</sub>"}, {"identifier": "C", "content": "A<sub>2</sub>B<sub>5</sub>"}, {"identifier": "D", "content": "A<sub>2</sub>B"}] | ["C"] | null | No. of atoms in the corners
<br><br>$$\left( A \right) = 8 \times {1 \over 8} = 1$$
<br><br>No. of atoms at face centers
<br><br>$$\left( B \right) = 5 \times {1 \over 2} = 2.5$$
<br><br>$$\therefore$$ Formula is $$ = A{B_{2.5}}$$ or $${A_2}{B_5}$$ | mcq | aieee-2011 | 3,480 |
ddTbKpnahDDxHBZC | chemistry | solid-state | crystal-structure-of-solids | Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å . The radius of
sodium atom is approximately : | [{"identifier": "A", "content": "5.72 \u00c5"}, {"identifier": "B", "content": "0.93 \u00c5"}, {"identifier": "C", "content": "1.86 \u00c5"}, {"identifier": "D", "content": "3.22 \u00c5"}] | ["C"] | null | In $$bcc$$ the atoms touch along body diagonal
<br><br>$$\therefore$$ $$2r + 2r = \sqrt 3 a$$
<br><br>$$\therefore$$ $$r = {{\sqrt 3 a} \over 4} = {{\sqrt 3 \times 4.29} \over 4} = 1.857\mathop A\limits^ \circ $$ | mcq | jee-main-2015-offline | 3,482 |
DaSAsuPSM8VLjf3GVx95Q | chemistry | solid-state | crystal-structure-of-solids | All of the following share the same crystal structure except : | [{"identifier": "A", "content": "LiCl"}, {"identifier": "B", "content": "NaCl"}, {"identifier": "C", "content": "RbCl"}, {"identifier": "D", "content": "CsCl"}] | ["D"] | null | RbCl, LiCl, and NaCl have face centered cubic structure and CsCl body centered cubic structure. | mcq | jee-main-2018-online-15th-april-evening-slot | 3,484 |
9gLROTdBInaKJiqCLzlN7 | chemistry | solid-state | crystal-structure-of-solids | At 100<sup>o</sup>C, copper (Cu) has FCC unit cell structure with cell edge length of x $$\mathop A\limits^o $$. What is the approximate density of Cu (in g cm<sup>$$-$$3</sup>) at this temperature?
<br/>[Atomic Mass of Cu = 63.55 u] | [{"identifier": "A", "content": "$${{205} \\over {{x^3}}}$$"}, {"identifier": "B", "content": "$${{105} \\over {{x^3}}}$$"}, {"identifier": "C", "content": "$${{211} \\over {{x^3}}}$$"}, {"identifier": "D", "content": "$${{422} \\over {{x^3}}}$$"}] | ["D"] | null | FCC unit cell Z $$=$$ 4
<br><br>We know,
<br><br>d = $$Z \times {M \over {{N_A} \times {a^3}}}$$
<br><br>$$d = {{63.5 \times 4} \over {6 \times {{10}^{23}} \times {x^3} \times {{10}^{ - 24}}}}g/c{m^3}$$
<br><br>$$d = {{63.5 \times 4 \times 10} \over 6}g/c{m^3}$$
<br><br>$$d = {{423.33} \over {{x^3}}} \simeq \left( {{{4... | mcq | jee-main-2019-online-9th-january-evening-slot | 3,485 |
k5eMabjmavRIn34ZJosBg | chemistry | solid-state | crystal-structure-of-solids | Which premitive unit cell has unequal edge lengths (a $$ \ne $$ b $$ \ne $$ c) and all axial angles different from 90<sup>o</sup>? | [{"identifier": "A", "content": "Hexagonal "}, {"identifier": "B", "content": "Tetragonal "}, {"identifier": "C", "content": "Triclinic"}, {"identifier": "D", "content": "Monoclinic "}] | ["C"] | null | In Triclinic unit cell
<br><br>a $$ \ne $$ b $$ \ne $$ c & $$\alpha $$ $$ \ne $$ $$\beta $$ $$ \ne $$ $$\gamma $$ $$ \ne $$ 90<sup>o</sup> | mcq | jee-main-2019-online-10th-january-morning-slot | 3,486 |
z1RrQq5oYGmRs6tW3kQxj | chemistry | solid-state | crystal-structure-of-solids | A solid having density of 9$$ \times $$10<sup>3</sup> kg m<sup>–3</sup> forms face centred cubic crystals of edge length $$200\sqrt 2 $$ pm. What is the molar mass of the solid?
<br/>[Avogadro constant $$ \cong $$ 6 $$ \times $$ 10<sup>23</sup> mol<sup>–1</sup>
, $$\pi $$ $$ \cong $$ 3] | [{"identifier": "A", "content": "0.0305 kg mol<sup>\u20131</sup>"}, {"identifier": "B", "content": "0.4320 kg mol<sup>\u20131</sup>"}, {"identifier": "C", "content": "0.0216 kg mol<sup>\u20131</sup>"}, {"identifier": "D", "content": "0.0432 kg mol<sup>\u20131</sup>"}] | ["A"] | null | $$\rho $$ = $${{Z \times M} \over {{a^3} \times {N_A}}}$$
<br><br>$$ \Rightarrow $$ 9$$ \times $$10<sup>3</sup> = $${{4 \times M} \over {\left( {200\sqrt 2 \times {{10}^{ - 12}}} \right) \times 6 \times {{10}^{23}}}}$$
<br><br>$$ \Rightarrow $$ M = 0.0305 kg mol<sup>–1</sup> | mcq | jee-main-2019-online-11th-january-morning-slot | 3,487 |
OxpLwNaIXA2zsSLbSVUf3 | chemistry | solid-state | crystal-structure-of-solids | The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is : (Edge length is represented by 'a')
| [{"identifier": "A", "content": "0.134 a "}, {"identifier": "B", "content": "0.067 a"}, {"identifier": "C", "content": "0.047 a"}, {"identifier": "D", "content": "0.027 a "}] | ["B"] | null | a = 2(R + r)
<br><br>$$ \Rightarrow $$ $${a \over 2} = \left( {R + r} \right)$$ ......(1)
<br><br>For bcc, $$\sqrt 3 a$$ = 4R
<br><br>Using (i) and (ii)
<br><br>$${a \over 2} = \left( {{{a\sqrt 3 } \over 4} + r} \right)$$
<br><br>$$ \Rightarrow $$ r = $$a\left( {{{2 - \sqrt 3 } \over 4}} \right)$$
<br><br>$$ \Rightarro... | mcq | jee-main-2019-online-11th-january-evening-slot | 3,488 |
wyh7G1a54VgL5bnYASuQ2 | chemistry | solid-state | crystal-structure-of-solids | Element 'B' forms ccp structure and 'A' occupies half of the octahedral voids, while oxygen atoms
occupy all the tetrahedral voids, The structure of bimetallic oxide is : | [{"identifier": "A", "content": "AB<sub>2</sub>O<sub>4</sub>"}, {"identifier": "B", "content": "A<sub>2</sub>BO<sub>4</sub>"}, {"identifier": "C", "content": "A<sub>4</sub>B<sub>2</sub>O"}, {"identifier": "D", "content": "A<sub>2</sub>B<sub>2</sub>O"}] | ["A"] | null | We know, for cubic unit cell, only FCC has octahedral and tetrahedral voids.
<br><br>B forms ccp structure means B forms FCC structure.
<br><br>For FCC, z = 4
<br><br> We know for octahedral voids z = 4. In this lattice, A present in half of octahedral voids.
<br><br>$$ \therefore $$ For A, z = 2
<br><br>For tetrahedra... | mcq | jee-main-2019-online-8th-april-morning-slot | 3,489 |
MKcfd78ZBvcfhSe4eYcLR | chemistry | solid-state | crystal-structure-of-solids | 10 mL of 1mM surfactant solution forms a
monolayer covering 0.24 cm<sup>2</sup> on a polar
substrate. If the polar head is approximated as
cube, what is its edge length? | [{"identifier": "A", "content": "1.0 pm"}, {"identifier": "B", "content": "2.0 nm"}, {"identifier": "C", "content": "0.1 nm"}, {"identifier": "D", "content": "2.0 pm"}] | ["D"] | null | No of moles formed = 10<sup>-3</sup> $$ \times $$ $${{10} \over {1000}}$$ = 10<sup>-5</sup>
<br><br>$$ \therefore $$ No of molecules formed = 10<sup>-5</sup> $$ \times $$ N<sub>A</sub>
<br><br>In unimolecular layer formation each cube occupy an area = a<sup>2</sup>
<br><br>$$ \therefore $$ Total area occupied = 10<sup>... | mcq | jee-main-2019-online-9th-april-evening-slot | 3,491 |
ZARaC3QCRDROhbCrYq3rsa0w2w9jx91dizy | chemistry | solid-state | crystal-structure-of-solids | The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic
structure are, respectively : | [{"identifier": "A", "content": "8 : 1 : 6"}, {"identifier": "B", "content": "4 : 2 : 1"}, {"identifier": "C", "content": "1 : 2 : 4"}, {"identifier": "D", "content": "4 : 2 : 3"}] | ["C"] | null | Z<sub>SC</sub> = 1
<br>Z<sub>BCC</sub> = 2
<br>Z<sub>FCC</sub> = 4
<br><br>$$ \therefore $$ Ratio = Z<sub>SC</sub> : Z<sub>BCC</sub> : Z<sub>FCC</sub> = 1 : 2 : 4 | mcq | jee-main-2019-online-12th-april-evening-slot | 3,493 |
qOMDyIOHIAUqZRTApCjgy2xukfjfltlj | chemistry | solid-state | crystal-structure-of-solids | A diatomic molecule X<sub>2</sub>
has a body-centred cubic
(bcc) structure with a cell edge of 300 pm. The
density of the molecule is 6.17 g cm<sup>–3</sup>. The number
of molecules present in 200 g of X<sub>2</sub>
is :
<br/>(Avogadro constant (N
<sub>A</sub>) = 6 $$ \times $$ 10<sup>23</sup> mol<sup>–1</sup>
) | [{"identifier": "A", "content": "8 N<sub>A</sub>"}, {"identifier": "B", "content": "40 N<sub>A</sub>"}, {"identifier": "C", "content": "4 N<sub>A</sub>"}, {"identifier": "D", "content": "2 N<sub>A</sub>"}] | ["C"] | null | d = $${{Z \times M} \over {{a^3} \times {N_A}}}$$
<br><br>$$ \Rightarrow $$ 6.17 = $${{2 \times M} \over {{{\left( {3 \times {{10}^{ - 8}}} \right)}^3} \times 6 \times {{10}^{23}}}}$$ [For BCC Z = 2]
<br><br>$$ \Rightarrow $$ M = 50 g/mol
<br><br>Number of moles in 200 gm = $${{{200} \over {50}}}$$ = 4
<br><br>$$ \ther... | mcq | jee-main-2020-online-5th-september-morning-slot | 3,494 |
xLNXDzsY06Ip66dpBEjgy2xukf2culmi | chemistry | solid-state | crystal-structure-of-solids | An element with molar mass 2.7 $$ \times $$ 10<sup>-2</sup> kg mol<sup>-1</sup> forms a cubic unit cell with edge length 405 pm. If its
density is 2.7 $$ \times $$ 10<sup>3</sup>
kg m<sup>-3</sup>, the radius of the element is approximately ______ $$ \times $$ 10<sup>-12</sup> m (to the
nearest integer). | [] | null | 143 | Molar mass of an element (M) = 27 gm mol<sup>–1</sup>
<br><br>Edge length of a cubic unit cell (a) = 405 pm = 4.05 × 10<sup>–8</sup> cm
<br><br>density of the element (d) = 2.7 gm/cc
<br><br>d = $${{Z \times M} \over {{N_A} \times {{\left( a \right)}^3}}}$$
<br><br>$$ \Rightarrow $$ 2.7 = $${{Z \times 27} \over {6 \tim... | integer | jee-main-2020-online-3rd-september-morning-slot | 3,495 |
OoLQt3h96k97JwY82J1klrghhlt | chemistry | solid-state | crystal-structure-of-solids | The coordination number of an atom in a body-centered cubic structure is _______.<br/>
[Assume that the lattice is made up of atoms.] | [] | null | 8 | Coordination number is the number of nearest neighbours of a central atom in the structure.<br/><br/>bcc has a coordination number of 8 and contains 2 atoms per unit cell.<br/><br/>This is because each atom touches four atoms in the layer above it, four in the layer below it and none in its own layer. | integer | jee-main-2021-online-24th-february-morning-slot | 3,496 |
I2bt6fXEk7hYmKeppM1kltbx658 | chemistry | solid-state | crystal-structure-of-solids | The unit cell of copper corresponds to a face centered cube of edge length 3.596 $$\mathop A\limits^o $$ with one copper atom at each lattice point. The calculated density of copper in kg/m<sup>3</sup> is ___________. [Molar mass of Cu : 63.54 g; Avogadro Number = 6.022 $$\times$$ 10<sup>23</sup>] | [] | null | 9077 | <p>Density of copper, $$d = {{Z \times M} \over {{a^3} \times {N_A}}}$$</p>
<p>Given, Z = 4, for fcc lattice,</p>
<p>M = 63.54 g mol<sup>$$-$$1</sup></p>
<p>= 63.54 $$\times$$ 10<sup>$$-$$3</sup> kg mol<sup>$$-$$1</sup>,</p>
<p>a = 3.596 $$\mathop A\limits^o $$ = 3.596 $$\times$$ 10<sup>$$-$$10</sup> m,</p>
<p>N<sub>A<... | integer | jee-main-2021-online-25th-february-evening-slot | 3,497 |
koLvGxf1HP6nlHYgyv1kmhuyun3 | chemistry | solid-state | crystal-structure-of-solids | A certain element crystallises in a bcc lattice of unit cell edge length 27$$\mathop A\limits^o $$. If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cel in $$\mathop A\limits^o $$ will be ____________. (Round off to the Nearest Integer).<br/><br/>[Assume each la... | [] | null | 33 | For BCC unit cell, $$\sqrt 3 a = 4R$$<br><br>$$a = {{4R} \over {\sqrt 3 }} = 27$$<br><br>$$R = {{27\sqrt 3 } \over 4}$$<br><br>For FCC unit cell<br><br>$$\sqrt 2 a = 4R$$<br><br>$$ \Rightarrow $$ $$a = {4 \over {\sqrt 2 }}\left( {{{27\sqrt 3 } \over 4}} \right)$$<br><br>$$ \Rightarrow $$ $$a = 27\sqrt {{3 \over 2}} $$<... | integer | jee-main-2021-online-16th-march-morning-shift | 3,498 |
1krrlhbne | chemistry | solid-state | crystal-structure-of-solids | Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where 50% of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is ____________. | [] | null | 8 | Carbon atoms occupy FCC lattice points as well as half of the tetrahedral voids. <br><br>Therefore number of carbon atoms per unit cell = 8 | integer | jee-main-2021-online-20th-july-evening-shift | 3,499 |
1krt71elu | chemistry | solid-state | crystal-structure-of-solids | A copper complex crystallising in a CCP lattice with a cell edge of 0.4518 nm has been revealed by employing X-ray diffraction studies. The density of a copper complex is found to be 7.62 g cm<sup>$$-$$3</sup>. The molar mass of copper complex is ____________ g mol<sup>$$-$$1</sup>. (Nearest integer)<br/><br/>[Given : ... | [] | null | 106 | $$d\left( {{{gm} \over {cc}}} \right) = {{4 \times {M \over {{N_A}}}} \over {{{(a\,cm)}^3}}}$$<br><br>$$7.62 = {{4 \times M/6.022 \times {{10}^{23}}} \over {{{(0.4518 \times {{10}^{ - 7}}cm)}^3}}} $$<br><br>$$\Rightarrow $$ M = 105.8 g/mol | integer | jee-main-2021-online-22th-july-evening-shift | 3,500 |
1l5c70bnl | chemistry | solid-state | crystal-structure-of-solids | <p>Atoms of element X form hcp lattice and those of element Y occupy $${2 \over 3}$$ of its tetrahedral voids. The percentage of element X in the lattice is ____________. (Nearest integer)</p> | [] | null | 43 | Since $X$ occupies hop lattice,
Number of particles of type $X$ in a unit cell $=6$
<br/><br/>
Number of particles of type $Y=\frac{2}{3} \times 12=8$
<br/><br/>
$\therefore$ Percentage of element $X=\frac{6}{14} \times 100$
<br/><br/>
$$
\begin{aligned}
&=\frac{300}{7} \\\\
&=42.85 \\\\
&\simeq 43 \%
\end{aligned}
$$ | integer | jee-main-2022-online-24th-june-morning-shift | 3,502 |
1l5w5ekzw | chemistry | solid-state | crystal-structure-of-solids | <p>An element X has a body centred cubic (bcc) structure with a cell edge of 200 pm. The density of the element is 5 g cm<sup>$$-$$3</sup>. The number of atoms present in 300 g of the element X is _______________.</p>
<p>Given : Avogadro constant, N<sub>A</sub> = 6.0 $$\times$$ 10<sup>23</sup> mol<sup>$$-$$1</sup>.</p> | [{"identifier": "A", "content": "5 N<sub>A</sub>"}, {"identifier": "B", "content": "6 N<sub>A</sub>"}, {"identifier": "C", "content": "15 N<sub>A</sub>"}, {"identifier": "D", "content": "25 N<sub>A</sub>"}] | ["D"] | null | $\rho=\frac{Z \times M}{a^{3} \times N_{\mathrm{A}}}$
<br/><br/>
$$
Z=2 \text { for } b c c
$$
<br/><br/>
$$
\begin{aligned}
& 5 \mathrm{~g} / \mathrm{cm}^{3}=\frac{2 \times M}{\left(200 \times 10^{-10} \mathrm{~cm}\right)^{3} \times 6.0 \times 10^{23}} \Rightarrow M=12 \mathrm{~g}
\end{aligned}
$$
<br/><br/>
$12 \math... | mcq | jee-main-2022-online-30th-june-morning-shift | 3,503 |
1l6mdzba2 | chemistry | solid-state | crystal-structure-of-solids | <p>An element M crystallises in a body centred cubic unit cell with a cell edge of $$300 \,\mathrm{pm}$$. The density of the element is $$6.0 \mathrm{~g} \mathrm{~cm}^{-3}$$. The number of atoms present in $$180 \mathrm{~g}$$ of the element is ____________ $$\times 10^{23}$$. (Nearest integer)</p> | [] | null | 22 | $M$ is body certred cubic, $\therefore Z=2$<br/><br/>
Let mass of 1 atom of $M$ is $A$<br/><br/>
Edge length $=300 \,\mathrm{pm}$<br/><br/>
Density $=6 \mathrm{~g} / \mathrm{cm}^3$<br/><br/>
$$
\therefore 6 \mathrm{~g} / \mathrm{cm}^3=\frac{\mathrm{Z} \times \mathrm{A}}{\left(300 \times 10^{-10}\right)^3}=\frac{2 \time... | integer | jee-main-2022-online-28th-july-morning-shift | 3,504 |
1l6nxh1tk | chemistry | solid-state | crystal-structure-of-solids | <p>Metal $$\mathrm{M}$$ crystallizes into a fcc lattice with the edge length of $$4.0 \times 10^{-8} \mathrm{~cm}$$. The atomic mass of the metal is ____________ $$\mathrm{g} / \mathrm{mol}$$. (Nearest integer)</p>
<p>$$\left(\right.$$ Use : $$\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$$, density of... | [] | null | 87 | $\rho=\frac{Z M}{N_{A} a^{3}} \Rightarrow M=\frac{9.03 \times 6.02 \times 10^{23} \times\left(4 \times 10^{-8}\right)^{3}}{4}$
<br/><br/>
$$
\begin{aligned}
&=\frac{9.03 \times 6.02 \times 64 \times 10^{-1}}{4} \\\\
&=86.9 \mathrm{~g} \mathrm{~mol}^{-1} \\\\
&\approx 87 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$ | integer | jee-main-2022-online-28th-july-evening-shift | 3,505 |
ldqy6jqm | chemistry | solid-state | crystal-structure-of-solids | Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the $\mathrm{FeO}$ in the crystal is $4.0 \mathrm{~g} \mathrm{~cm}^{-3}$, then the number of $\mathrm{FeO}$ units present per unit cell is __________. (Nearest integer)
<br/><br/>
Given: Molar mass of $\mathrm{Fe}$ and ... | [] | null | 4 | <p>$$\mathrm{d=\frac{z\times m}{a^3}}$$</p>
<p>$$\mathrm{4=\frac{z\times72}{6\times10^{23}(5\times10^{-8})^3}}$$</p>
<p>$$\mathrm{4=\frac{z\times72}{6\times125\times10^{-1}}}$$</p>
<p>$$=\mathrm{z \approx4}$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 3,507 |
1ldulfzjs | chemistry | solid-state | crystal-structure-of-solids | <p>A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at $$\frac{1}{3}^{\mathrm{rd}}$$ of the total faces. The empirical formula of the compound is :</p> | [{"identifier": "A", "content": "$$\\mathrm{{X_2}{Y_{1.5}}}$$"}, {"identifier": "B", "content": "$$\\mathrm{{X_{3}}{Y_2}}$$"}, {"identifier": "C", "content": "$$\\mathrm{X{Y_{2.5}}}$$"}, {"identifier": "D", "content": "$$\\mathrm{{X_{2.5}}Y}$$"}] | ["B"] | null | $\begin{aligned} & \text { Number of } X \text { particles }=4 \times \frac{1}{8}+1=1.5 \\\\ & \text { Number of } Y \text { particles }=6 \times \frac{1}{3} \times \frac{1}{2}=1 \\\\ & \therefore \text { Empirical formula }=X_{1.5} Y_1=X_3 Y_2\end{aligned}$ | mcq | jee-main-2023-online-25th-january-morning-shift | 3,508 |
lgnzx8zu | chemistry | solid-state | crystal-structure-of-solids | Which of the following expressions is correct in case of a $\mathrm{CsCl}$ unit cell (edge length 'a')? | [{"identifier": "A", "content": "$\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}^{-}}=\\frac{\\sqrt{3}}{2} \\mathrm{a}$"}, {"identifier": "B", "content": "$\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}^{-}}=\\frac{\\mathrm{a}}{\\sqrt{2}}$"}, {"identifier": "C", "content": "$\\mathrm{r}_{\\mathrm{Cs... | ["A"] | null | $\mathrm{CsCl}$ has body centered type structure in which $\mathrm{Cs}^{+}$ occupies at corner of a cube and $\mathrm{Cl}^{-}$occupies the centre of the cube.<br/><br/>
$2 \mathrm{r}_{\mathrm{Cs}^{+}}+2 \mathrm{r}_{\mathrm{Cl}^{-}}=\sqrt{3} \mathrm{a}$ (where a is the edge length of the cube)<br/><br/>
$$
\mathrm{r}_{\... | mcq | jee-main-2023-online-15th-april-morning-shift | 3,509 |
1lgp3vvke | chemistry | solid-state | crystal-structure-of-solids | <p>Sodium metal crystallizes in a body centred cubic lattice with unit cell edge length of $$4~\mathop A\limits^o $$. The radius of sodium atom is __________ $$\times ~10^{-1}$$ $$\mathop A\limits^o $$ (Nearest integer)</p> | [] | null | 17 | In a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by :
<br/><br/>
$$\sqrt{3}a = 4r$$
<br/><br/>
Given the unit cell edge length (a) of sodium metal as 4 Å :
<br/><br/>
$$a = 4 ~\mathop A\limits^o$$
<br/><br/>
We can now solve for the radius (r) of th... | integer | jee-main-2023-online-13th-april-evening-shift | 3,510 |
1lgv0wa8m | chemistry | solid-state | crystal-structure-of-solids | <p>An atomic substance A of molar mass $$12 \mathrm{~g} \mathrm{~mol}^{-1}$$ has a cubic crystal structure with edge length of $$300 ~\mathrm{pm}$$. The no. of atoms present in one unit cell of $$\mathrm{A}$$ is ____________. (Nearest integer)</p>
<p>Given the density of $$\mathrm{A}$$ is $$3.0 \mathrm{~g} \mathrm{~mL}... | [] | null | 4 | <p>Given:</p>
<ul>
<li>Atomic substance A with molar mass $M = 12 \, \text{g mol}^{-1}$</li><br/>
<li>Cubic crystal structure with edge length $a = 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m}$</li><br/>
<li>Density $\rho = 3.0 \, \text{g mL}^{-1}$</li><br/>
<li>Avogadro's number $N_A = 6.02 \times 10^{23} \,... | integer | jee-main-2023-online-11th-april-morning-shift | 3,511 |
1lh33546b | chemistry | solid-state | crystal-structure-of-solids | <p>Number of crystal systems from the following where body centred unit cell can be found, is ____________.</p>
<p>Cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic</p> | [] | null | 3 | <p>Body-centered unit cells can be found in the following crystal systems among those listed:</p>
<ol>
<li>Cubic: Body-centered cubic (BCC) is one of the lattice structures that cubic systems can have.</li><br/>
<li>Tetragonal: A body-centered tetragonal system is also a possibility.</li><br/>
<li>Orthorhombic: The ort... | integer | jee-main-2023-online-6th-april-evening-shift | 3,513 |
N8vlbnr0i7Y4b8GC | chemistry | solid-state | defects-in-crystal | What type of crystal defect is indicated in the diagram below?
<br/><br/><img src="data:image/png;base64,UklGRjgPAABXRUJQVlA4ICwPAACwjQCdASoAAwgBP4G61mU2LjgnIfL54wAwCWlu4W3BG/Pb8y/4fwdfyH+28UdXXf/Z2f3v/C8O/mOn9uEcGPCTu+OOcPJdnWKJVAkTD8uoEiYfl1AkTD8uoEiYfl1AkS2vdchO65Cd1yE7rkSl5jD8uoEiYfl1AkTDFB+8uQnbzfQ7vGSYnc1/fnQUYH... | [{"identifier": "A", "content": "Interstitial defect"}, {"identifier": "B", "content": "Schottky defect"}, {"identifier": "C", "content": "Frenkel defect "}, {"identifier": "D", "content": "Frenkel and Schottky defects "}] | ["B"] | null | When equal number of cations and anions are missing from their regular lattice positions, we have schottky defect. This type of defects are more common in ionic compounds with high co-ordination number and where the size of positions have negative ions are almost equal e.g. $$NaCl$$ $$KCl$$ etc. | mcq | aieee-2004 | 3,514 |
Da5zXaLzbA2k3L9X | chemistry | solid-state | defects-in-crystal | Which type of ‘defect’ has the presence of cations in the interstitial sites? | [{"identifier": "A", "content": "Metal deficiency defect"}, {"identifier": "B", "content": "Schottky defect"}, {"identifier": "C", "content": "Vacancy defect"}, {"identifier": "D", "content": "Frenkel defect "}] | ["D"] | null | In Frenkel defect in a molecule an atom or ion (normally the cation) leave their original site and places itself in the interstitial site (area between all other cations and anions). Which is shown below
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267288/exam_images/... | mcq | jee-main-2018-offline | 3,515 |
WKzMyDo8fXgUgCRMa71kmkjgqtx | chemistry | solid-state | defects-in-crystal | KBr is doped with 10<sup>$$-$$5</sup> mole percent of SrBr<sub>2</sub>. The number of cationic vacancies in 1g of KBr crystal is ____________ 10<sup>14</sup>. (Round off to the Nearest Integer).<br/><br/>[Atomic Mass : K : 39.1 u, Br : 79.9 u N<sub>A</sub> = 6.023 $$\times$$ 10<sup>23</sup>] | [] | null | 5 | For every Sr<sup>+2</sup> ion, 1 cationic vacancy is created. Hence, no. of Sr<sup>+2</sup> ion = Number of cationic vacancies<br><br>Since mole percentage of SrBr<sub>2</sub> dropped is 10<sup>$$-$$5</sup> to that of total moles of KBr.<br><br>Hence,<br><br>No. of cationic vacancy $$ = {{{{10}^{-5}}} \over {100}} \tim... | integer | jee-main-2021-online-17th-march-evening-shift | 3,517 |
1ktb5km5o | chemistry | solid-state | defects-in-crystal | Given below are two statements.<br/><br/>Statement I : Frenkel defects are vacancy as well as interstitial defects.<br/><br/>Statement II : Frenkel defect leads to colour in ionic solids due to presence of F-centres.<br/><br/>Choose the most appropriate answer for the statements from the options given below : | [{"identifier": "A", "content": "Statement I is false but Statement II is true"}, {"identifier": "B", "content": "Both Statement I and Statement II are true"}, {"identifier": "C", "content": "Statement I is true but Statement II is false"}, {"identifier": "D", "content": "Both Statement I and Statement II are false"}] | ["C"] | null | <p>To answer this question, let's analyze each statement separately.</p>
<p><strong>Statement I: Frenkel defects are vacancy as well as interstitial defects.</strong></p>
<p>This statement is true. A Frenkel defect, also known as a dislocation defect, occurs in a crystalline solid when an atom or ion leaves its norma... | mcq | jee-main-2021-online-26th-august-morning-shift | 3,518 |
1l56ai7q7 | chemistry | solid-state | defects-in-crystal | <p>The incorrect statement about the imperfections in solids is :</p> | [{"identifier": "A", "content": "Schottky defect decreases the density of the substance."}, {"identifier": "B", "content": "Interstitial defect increases the density of the substance."}, {"identifier": "C", "content": "Frenkel defect does not alter the density of the substance."}, {"identifier": "D", "content": "Vacanc... | ["D"] | null | The vacancy defect increases the density of substance.<br/><br/>
It does not change the density of the crystal. It only creates cationic vacancies. Frenkel
Defect causes vacancy defect at its original site and an interstitial defect at its new location. Therefore, it
does not change the density of the solid. | mcq | jee-main-2022-online-28th-june-morning-shift | 3,519 |
1l57stziz | chemistry | solid-state | defects-in-crystal | <p>Metal deficiency defect is shown by Fe<sub>0.93</sub>O. In the crystal, some Fe<sup>2+</sup> cations are missing and loss of positive charge is compensated by the presence of Fe<sup>3+</sup> ions. The percentage of Fe<sup>2+</sup> ions in the Fe<sub>0.93</sub>O crystals is __________. (Nearest integer)</p> | [] | null | 85 | $\mathrm{Fe}_{0.93} \mathrm{O}$
<br/><br/>
Let the number of $\mathrm{O}^{-2}$ ions be 100 and the number of $\mathrm{Fe}^{+2}$ ions be $\mathrm{X}$ The number of $\mathrm{Fe}^{+3}$ ions be $(93-\mathrm{X})$
<br/><br/>
$$
\begin{aligned}
&\therefore \,X(2)+(93-X) 3=200 \\\\
&279-X=200 \\\\
&X=79 \\\\
&\therefore \quad ... | integer | jee-main-2022-online-27th-june-morning-shift | 3,520 |
EECxyLQjIqV67Nd49Bjgy2xukfp2ui06 | chemistry | solid-state | interestitial-voids | An element crystallises in a face-centred cubic (fcc) unit cell with cell edge $$a$$. The distance
between the centres of two nearest octahedral voids in the crystal lattice is : | [{"identifier": "A", "content": "$$a$$"}, {"identifier": "B", "content": "$${a \\over 2}$$"}, {"identifier": "C", "content": "$${a \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${\\sqrt 2 a}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263655/exam_images/q1lnoip3a57h44bxdiqa.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Chemistry - Solid State Question 37 English Explanation">
<br><br>Dista... | mcq | jee-main-2020-online-5th-september-evening-slot | 3,522 |
ixtskjVoNE7LIOsnPD1klusttpb | chemistry | solid-state | interestitial-voids | The number of octahedral voids per lattice site in a lattice is _________. (Rounded off to the nearest integer) | [] | null | 1 | <p>Let us assume, the crystal has fcc or ccp lattice which has octahedral voids.</p>
<p>Number of lattice sites occupied = 8 corner + 6 face centres = 14</p>
<p>Number of octahedral voids = 12 edge centres + 1 body centre = 13</p>
<p>Number of octahedral void(s) per lattice site</p>
<p>$$ = {{13} \over {14}} = 0.928 \s... | integer | jee-main-2021-online-26th-february-evening-slot | 3,523 |
3vtRG37cz0uIYRi9 | chemistry | solid-state | structure-of-ionic-compounds | An ionic compound has a unit cell consisting of A ions at the corners of a cube and B
ions on the centres of the faces of the cube. The empirical formula for this compound
would be : | [{"identifier": "A", "content": "AB"}, {"identifier": "B", "content": "A<sub>2</sub>B "}, {"identifier": "C", "content": "AB<sub>3 </sub>"}, {"identifier": "D", "content": "A<sub>3</sub>B"}] | ["C"] | null | Number of A ions in the unit cell. $$ = {1 \over 8} \times 8 = 1$$
<br><br>Number of $$B$$ ions in the unit cell $$ = {1 \over 2} \times 6 = 3$$
<br><br>Hence empirical formula of the compound $$ = A{B_3}$$ | mcq | aieee-2005 | 3,525 |
gb6ztaMssnxMD7Av | chemistry | solid-state | structure-of-ionic-compounds | The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the
cation is 110 pm, the radius of the anion is : | [{"identifier": "A", "content": "288 pm"}, {"identifier": "B", "content": "398 pm"}, {"identifier": "C", "content": "618 pm"}, {"identifier": "D", "content": "144 pm"}] | ["D"] | null | For an Fcc crystal
<br><br>$${r_{cation}} + {r_{anion}} = {{edge\,\,length} \over 2};$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,110 + {r_{anion}}\,\, = {{508} \over 2}$$
<br><br>$${r_{anion}} = 254 - 110 = 144pm$$ | mcq | aieee-2010 | 3,526 |
dvTgtXlUwp2Xb8Am | chemistry | solid-state | structure-of-ionic-compounds | CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following
expressions is correct? | [{"identifier": "A", "content": "$${r_{C{s^ + }}} + {r_{C{l^ - }}} = {{\\sqrt 3 } \\over 2}a$$"}, {"identifier": "B", "content": "$${r_{C{s^ + }}} + {r_{C{l^ - }}} = {3 \\over 2}a$$ "}, {"identifier": "C", "content": "$${r_{C{s^ + }}} + {r_{C{l^ - }}} = \\sqrt 3 a$$"}, {"identifier": "D", "content": "$${r_{C{s^ + }}} +... | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266570/exam_images/iuqkwiyvbaofgz2gykqn.webp" loading="lazy" alt="JEE Main 2014 (Offline) Chemistry - Solid State Question 58 English Explanation">
<br><br>Relation between radius of cation, anion and edge length of the cube
<br>... | mcq | jee-main-2014-offline | 3,527 |
TswPb2b7XZp6vKRqdxjgy2xukg3e6aw3 | chemistry | solid-state | structure-of-ionic-compounds | A crystal is made up of metal ions 'M<sub>1</sub>' and 'M<sub>2</sub>'
and oxide ions. Oxide ions form a ccp lattice
structure. The cation 'M<sub>1</sub>' occupies 50% of
octahedral voids and the cation 'M<sub>2</sub>' occupies
12.5% of tetrahedral voids of oxide lattice. The
oxidation numbers of 'M<sub>1</sub>' and 'M... | [{"identifier": "A", "content": "+ 3, + 1"}, {"identifier": "B", "content": "+ 4, + 2"}, {"identifier": "C", "content": "+ 1, + 3"}, {"identifier": "D", "content": "+ 2, + 4"}] | ["D"] | null | O<sup>–2</sup> ions form ccp $$ \Rightarrow $$ O<sub>4</sub>.
<br><br>M<sub>1</sub> = 50% octahedral void = $${{50} \over {100}} \times 4$$ = 2
<br><br>M<sub>2</sub> = 12.5% tetrahedral void = $${{12.5} \over {100}} \times 8$$ = 1
<br><br>So formula is : (M<sub>1</sub>)<sub>2</sub>(M<sub>2</sub>)<sub>1</sub>O<sub>4</su... | mcq | jee-main-2020-online-6th-september-evening-slot | 3,528 |
1l5amdyv1 | chemistry | solid-state | structure-of-ionic-compounds | <p>The distance between Na<sup>+</sup> and Cl<sup>$$-$$</sup> ions in solid NaCl of density 43.1 g cm<sup>$$-$$3</sup> is _______________ $$\times$$ 10<sup>$$-$$10</sup> m. (Nearest Integer)</p>
<p>(Given : N<sub>A</sub> = 6.02 $$\times$$ 10<sup>23</sup> mol<sup>$$-$$1</sup>)</p> | [] | null | 1 | $\rho=\frac{Z \times M}{a^{3} \times N_{A}}$
<br/><br/>
$$
43.1=\frac{4 \times 58.5}{a^{3} \times 6.02 \times 10^{23}}
$$
<br/><br/>
$$
\begin{aligned}
& a^{3}=0.9 \times 10^{-23} \\\\
& =9 \times 10^{-24}
\end{aligned}
$$
<br/><br/>
$a=2.08 \times 10^{-8} \mathrm{~cm}$
<br/><br/>
$=2.08 \times 10^{-10} \mathrm{~m}$
<b... | integer | jee-main-2022-online-25th-june-morning-shift | 3,529 |
Q9p2CUSccdOnmyB2 | chemistry | solid-state | type-of-solids-and-their-properties | Which of the following exists as covalent crystals in the solid state? | [{"identifier": "A", "content": "Silicon "}, {"identifier": "B", "content": "Sulphur "}, {"identifier": "C", "content": "Phosphorous "}, {"identifier": "D", "content": "Iodine "}] | ["A"] | null | Among the given crystals only silicon is as a covalent solid. | mcq | jee-main-2013-offline | 3,530 |
1yVqUv6KGmf0LqnHFoj29 | chemistry | solid-state | type-of-solids-and-their-properties | Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance ? | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7u7rqd4/ff3d5d03-0de2-4325-ae09-692f448a4daf/08359b80-3019-11ed-ab58-4f43f7f8c851/file-1l7u7rqd5.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l7u7rqd4/ff3d5d03-0de2-4325-ae09-692f448a4daf/083... | ["C"] | null | In antiferro magnetic substance, magnetic dipoles are in opposite direction and cancel out eachother's magnetic moment. | mcq | jee-main-2018-online-15th-april-morning-slot | 3,531 |
IBuMPKx3rWORpIDf8hCvp | chemistry | solid-state | type-of-solids-and-their-properties | 0.27 g of a long chain fatty acid was dissolved
in 100 cm<sup>3</sup> of hexane. 10 mL of this solution
was added dropwise to the surface of water in
a round watch glass. Hexane evaporates and a
monolayer is formed. The distance from edge
to centre of the watch glass is 10 cm. What is
the height of the monolayer?<br/>
... | [{"identifier": "A", "content": "10<sup>\u20138</sup> m"}, {"identifier": "B", "content": "10<sup>\u20132</sup> m"}, {"identifier": "C", "content": "10<sup>\u20134</sup> m"}, {"identifier": "D", "content": "10<sup>\u20136</sup> m"}] | ["D"] | null | In 100 ml of hexane solution contains 0.27 g of fatty acid.
<br><br>$$ \therefore $$ In 10 ml of hexane solution contains 0.027 g of fatty acid.
<br><br>Volume of fatty acid present on the round glass = $${{0.027} \over {0.9}}$$
<br><br>As here Area of fatty acid layer = Area of round plate = $$\pi {r^2}$$
<br><br>$$ \... | mcq | jee-main-2019-online-8th-april-evening-slot | 3,532 |
dqmI96SC2iDG6849831kmm0k74t | chemistry | solid-state | type-of-solids-and-their-properties | A hard substance melts at high temperature and is an insulator in both solid and in molten state. This solid is most likely to be a/an : | [{"identifier": "A", "content": "Covalent solid"}, {"identifier": "B", "content": "Molecular solid"}, {"identifier": "C", "content": "Ionic solid"}, {"identifier": "D", "content": "Metallic solid"}] | ["A"] | null | Covalent or network solid are insulator (except graphite) and have very high melting point. | mcq | jee-main-2021-online-18th-march-evening-shift | 3,533 |
1krq5ildk | chemistry | solid-state | type-of-solids-and-their-properties | Given below are two statements. One is labelled as Assertion A nd the other is labelled as Reason R.<br/><br/>Assertion A : Sharp glass edge becomes smooth on heating it upto its melting point.<br/><br/>Reason R : The viscosity of glass decreases on melting.<br/><br/>Choose the most appropriate answer from the options ... | [{"identifier": "A", "content": "A is true but R is false"}, {"identifier": "B", "content": "Both A and R are true but R is NOT the correct explanation of A."}, {"identifier": "C", "content": "A is false but R is true."}, {"identifier": "D", "content": "Both A and R are true and R is the correct explanation of A."}] | ["B"] | null | <p>On heating the glass, it melts and takes up rounded shape at the
edges, which has minimum surface area. This is due to the property
of surface tension of liquids and not due to decrease in viscosity.</p>
<p>Viscosity generally decreases as the temperature increases.</p>
<p>Hence, both A and R are true but R is not t... | mcq | jee-main-2021-online-20th-july-morning-shift | 3,534 |
1ks1ihsr8 | chemistry | solid-state | type-of-solids-and-their-properties | Select the correct statements<br/><br/>(A) Crystalline solids have long range order.<br/><br/>(B) Crystalline solids are isotropic<br/><br/>(C) Amorphous solid are sometimes called pseudo solids.<br/><br/>(D) Amorphous solids soften over a range of temperatures.<br/><br/>(E) Amorphous solids have a definite heat of fus... | [{"identifier": "A", "content": "(A), (B), (E) only"}, {"identifier": "B", "content": "(B), (D) only"}, {"identifier": "C", "content": "(C), (D) only"}, {"identifier": "D", "content": "(A), (C), (D) only"}] | ["D"] | null | (A) Crystalline solids have definite arrangement of constituent particles and have long range order.<br><br>(C), (D) Different constituent particles of an amorphous solid have different bond strengths and soften over a range of temperatures. | mcq | jee-main-2021-online-27th-july-evening-shift | 3,535 |
s5aaCHQtyDoQrezp | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking k<sub>f</sub> for water as 1.85, the freezing point of the solution will be nearest to | [{"identifier": "A", "content": "-0.360<sup>o</sup>C"}, {"identifier": "B", "content": "-0.260<sup>o</sup>C"}, {"identifier": "C", "content": "+0.480<sup>o</sup>C"}, {"identifier": "D", "content": "-0.480<sup>o</sup>C"}] | ["D"] | null | $$\Delta {T_f} = {K_f} \times m \times i;$$
<br><br>$$\Delta {T_f} = 1.855 \times 0.2 \times 1.3 = {0.480^ \circ }C$$
<br><br>$$\therefore$$ $$\,\,\,{T_f} = 0 - {0.480^ \circ }C = - {0.480^ \circ }C$$
<br><br>$$\mathop {\left( {HX} \right.}\limits_{1 - 0.3} \,\,\rightleftharpoons\,\,\mathop {{H^ + }}\limits_{0.3} +... | mcq | aieee-2003 | 3,536 |
PoXBfP93pWIo259f | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | If liquids A and B form an ideal solution | [{"identifier": "A", "content": "the entropy of mixing is zero"}, {"identifier": "B", "content": "the free energy of mixing is zero"}, {"identifier": "C", "content": "the free energy as well as the entropy of mixing are each zero"}, {"identifier": "D", "content": "the enthalpy of mixing is zero"}] | ["D"] | null | When $$A$$ and $$B$$ from an ideal solution, $$\Delta {H_{mix}} = 0$$ | mcq | aieee-2003 | 3,537 |
vfaGK6GR608F07PE | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | Which one of the following statements is false? | [{"identifier": "A", "content": "Raoult\u2019s law states that the vapour pressure of a components over a solution is\nproportional to its mole fraction "}, {"identifier": "B", "content": "Two sucrose solutions of same molality prepared in different solvents will have the same\nfreezing point depression "}, {"identifi... | ["B"] | null | $$\Delta {T_f} = {K_f} \times m \times i.$$ Since $${K_f}$$ has different values for different solvents, hence even if the $$m$$ is the same $$\Delta {T_f}$$ will be different | mcq | aieee-2004 | 3,538 |
RI2Bv42Km2tBeFSQ | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | If $$\alpha$$ is the degree of dissociation of Na<sub>2</sub>SO<sub>4</sub>, the vant Hoff’s factor (i) used for
calculating the molecular mass is : | [{"identifier": "A", "content": "1 + $$\\alpha$$"}, {"identifier": "B", "content": "1 + $$2\\alpha$$"}, {"identifier": "C", "content": "1 - $$\\alpha$$"}, {"identifier": "D", "content": "1 - $$2\\alpha$$"}] | ["B"] | null | $$N{a_2}\,S{O_4}\,\rightleftharpoons\mathop {2N{a^ + }}\limits_{2\alpha } \,\, + \,\,\mathop {SO_4^{ - - }}\limits_\alpha $$
<br><br>Vant. Hoff's factor $$\,\,\,i = {{1 - \alpha + 2\alpha + \alpha } \over 1} = 1 + 2\alpha $$ | mcq | aieee-2005 | 3,540 |
gT9UW738T4wk2zvr | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | Among the following mixtures, dipole-dipole as the major interaction, is present in | [{"identifier": "A", "content": "benzene and ethanol"}, {"identifier": "B", "content": "acetonitrile and acetone "}, {"identifier": "C", "content": "KCl and water"}, {"identifier": "D", "content": "benzene and carbon tetrachloride"}] | ["B"] | null | <b>Acetonitrile $$\mathop {\left( {C{H_3}} \right.}\limits^{\delta + } \,$$ $$\,\, - \,\,\,C \equiv \,\,$$ $$\mathop {\left. N \right)}\limits^{\delta - } $$ and acetone</b>
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263290/exam_images/rroiix1g0nd5wlzmhukh.webp" lo... | mcq | aieee-2006 | 3,541 |
FNZbeT7PKHNHVm0r | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous
solution, the change in freezing point of water (∆T<sub>f</sub>), when 0.01 mol of sodium sulphate is dissolved
in 1 kg of water, is (K<sub>f</sub> = 1.86 K kg mol<sup>–1</sup>) | [{"identifier": "A", "content": "0.0372 K"}, {"identifier": "B", "content": "0.0558 K"}, {"identifier": "C", "content": "0.0744 K"}, {"identifier": "D", "content": "0.0186 K"}] | ["B"] | null | Sodium sulphate dissociates as
<br><br>$$N{a_2}S{O_4}\left( s \right) \to 2N{a^ + } + SO_4^{ - - }$$
<br><br>hence van't hoff factor $$i=3$$
<br><br>Now $$\Delta {T_f} = i\,{k_f}.m = 3 \times 1.86 \times 0.01$$
<br><br>$$ = 0.0558\,K$$ | mcq | aieee-2010 | 3,542 |
UxOAOLB0n6FCdGiW | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | The degree of dissociation ($$\alpha$$ ) of a weak electrolyte, A<sub>x</sub>B<sub>y</sub> is related to van’t Hoff factor (i) by the expression : | [{"identifier": "A", "content": "$$\\alpha = {{i - 1} \\over {x + y + 1}}$$"}, {"identifier": "B", "content": "$$\\alpha = {{x + y - 1} \\over {i - 1}}$$"}, {"identifier": "C", "content": "$$\\alpha = {{x + y + 1} \\over {i - 1}}$$"}, {"identifier": "D", "content": "$$\\alpha = {{i - 1} \\over {(x + y - 1)}}$$"}] | ["D"] | null | $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{A_x}\,{B_y}\,\,\rightleftharpoons\,\,{}_x{A^{y + }} + {}_y{B^{x - }}$$
<br><br>$$t = 0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$$
<br><br>$${t_{eq}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,x\alpha \,\,... | mcq | aieee-2011 | 3,543 |
kaDUZp1Vne9f6oX2 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | Consider separate solutions of 0.500 M C<sub>2</sub>H<sub>5</sub>OH(aq), 0.100 M Mg<sub>3</sub>(PO<sub>4</sub>)<sub>2</sub>(aq), 0.250 M KBr(aq) and 0.125
M Na<sub>3</sub>PO<sub>4</sub>(aq) at 25<sup>o</sup>C. Which statement is true about these solutions, assuming all salts to be strong electrolytes? | [{"identifier": "A", "content": "0.125 M Na<sub>3</sub>PO<sub>4</sub>(aq) has the highest osmotic pressure."}, {"identifier": "B", "content": "0.500 M C<sub>2</sub>H<sub>5</sub>OH(aq) has the highest osmotic pressure"}, {"identifier": "C", "content": "They all have the same osmotic pressure"}, {"identifier": "D", "cont... | ["C"] | null | $$\pi = i\,CRT$$
<br><br>$$^\pi {C_2}{H_3}OH$$
<br><br>$$ = 1 \times 0.500 \times R \times T = 0.5{\mkern 1mu} RT$$
<br><br>$${}^\pi M{g_3}{\left( {P{O_4}} \right)_2}$$
<br><br>$$ = 5 \times 0.100 \times R \times T$$ $$ = 0.5RT$$
<br><br>$${}^\pi KBr = 2 \times 0.250 \times R \times T = 0.5\,RT$$
<br><br>$${}^\pi N{a_... | mcq | jee-main-2014-offline | 3,545 |
mRK2n4zCCX5hpeq9yRRaF | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | An aqueous solution of a salt MX<sub>2</sub> at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is : | [{"identifier": "A", "content": "0.33"}, {"identifier": "B", "content": "0.50"}, {"identifier": "C", "content": "0.67"}, {"identifier": "D", "content": "0.80"}] | ["B"] | null | <p>Let us assume that degree of dissociation is $$\alpha$$.</p>
<p>$$\matrix{
{M{X_2}} & { \to {M^{2 + }} + } & {2X + } \cr
{(1 - \alpha )} & \alpha & {2\alpha } \cr
} $$</p>
<p>Thus, after dissociation total number of moles formed (n) = 3.</p>
<p>Now, we know degree of dissociation is</p>
<p>$$\alpha = {... | mcq | jee-main-2016-online-10th-april-morning-slot | 3,546 |
0NrKLjtuON0zWg9M | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | The freezing point of benzene decreases by 0.45<sup>0</sup>C when 0.2 g of acetic acid is added to 20g of benzene. If
acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be:
(K<sub>f</sub> for benzene = 5.12 K kg mol<sup>–1</sup>) | [{"identifier": "A", "content": "80.4 %"}, {"identifier": "B", "content": "74.6 %"}, {"identifier": "C", "content": "94.6 %"}, {"identifier": "D", "content": "64.6 %"}] | ["C"] | null | $$\Delta $$T<sub>f</sub> = i $$ \times $$ K<sub>f</sub> $$ \times $$ m
<br><br>$$ \Rightarrow $$ 0.45 = i $$ \times $$ 5.12 $$ \times $$ $${{0.2 \times 1000} \over {60 \times 20}}$$
<br><br>$$ \Rightarrow $$ i = 0.527
<br><br>2CH<sub>3</sub>COOH ⇌ (CH<sub>3</sub>COOH)<sub>2</sub>
<br> &nbs... | mcq | jee-main-2017-offline | 3,547 |
H9JnV1FMWZHsTQX4Irz4X | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | 5 g of Na<sub>2</sub>SO<sub>4</sub> was dissolved in x g of H<sub>2</sub>O. The change in freezing point was found to be
3.82<sup>o</sup>C. If Na<sub>2</sub>SO<sub>4</sub> is 81.5% ionised,
the value of x
<br/><br/>(K<sub>f</sub> for water=1.86<sup>o</sup>C kg mol<sup>−1</sup>) is approximately :
<br/><br/>(molar ma... | [{"identifier": "A", "content": "15 g"}, {"identifier": "B", "content": "25 g"}, {"identifier": "C", "content": "45 g"}, {"identifier": "D", "content": "65 g"}] | ["C"] | null | <table class="tg">
<tbody><tr>
<th class="tg-13k7"></th>
<th class="tg-13k7">Na<sub>2</sub>SO<sub>4</sub></th>
<th class="tg-13k7">$$\rightleftharpoons$$ </th>
<th class="tg-13k7">2Na<sup>+</sup></th>
<th class="tg-13k7">+</th>
<th class="tg-60hs">SO<sub>4</sub><sup>-2</sup></th>
</tr>
<tr... | mcq | jee-main-2017-online-8th-april-morning-slot | 3,548 |
HCjxAsmCGrVE9z34BSAOB | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | Molecules of benzoic acid (C<sub>6</sub>H<sub>5</sub>COOH) dimerise in benzene. 'w' g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimmer in the solution is 80, then w is – (Its given that K<sub>f</sub> = 5 K kg mol<sup>–1</... | [{"identifier": "A", "content": "1.5 g"}, {"identifier": "B", "content": "1.8 g"}, {"identifier": "C", "content": "1.0 g"}, {"identifier": "D", "content": "2.4 g"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266616/exam_images/jdjuoyplzv6a2jvihjeu.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Chemistry - Solutions Question 107 English Explanation">
<br>We know, ... | mcq | jee-main-2019-online-12th-january-evening-slot | 3,550 |
vYAPREOHuQqFPnH4gftdv | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | Molal depression constant for a solvent is
4.0 kg mol<sup>–1</sup>. The depression in the freezing
point of the solvent for 0.03 mol kg<sup>–1</sup> solution
of K<sub>2</sub>SO<sub>4</sub> is :<br/>
(Assume complete dissociation of the
electrolyte) | [{"identifier": "A", "content": "0.18 K"}, {"identifier": "B", "content": "0.24 K"}, {"identifier": "C", "content": "0.36 K"}, {"identifier": "D", "content": "0.12 K"}] | ["C"] | null | K<sub>2</sub>SO<sub>4</sub> $$ \to $$ 2K<sup>+</sup> + SO<sub>4</sub><sup>2-</sup>
<br><br> Van’t Hoff Factor (i) = 3
<br><br>$$ \therefore $$ $$\Delta $$T<sub>f</sub> = ik<sub>f</sub>m
<br><br>= 3 $$ \times $$ 4 $$ \times $$ 0.03 = 0.36 K | mcq | jee-main-2019-online-9th-april-evening-slot | 3,551 |
b0PA5RHbPrV9ubW6zma8B | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | K<sub>2</sub>Hgl<sub>4</sub> is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is: | [{"identifier": "A", "content": "1.6"}, {"identifier": "B", "content": "2.2"}, {"identifier": "C", "content": "2.0"}, {"identifier": "D", "content": "1.8"}] | ["D"] | null | K<sub>2</sub>Hgl<sub>4</sub> is 40% ionised.
<br><br>$$ \therefore $$ $$\alpha $$ = $${{40} \over {100}}$$ = 0.4
<br><br>K<sub>2</sub>[Hgl<sub>4</sub>] $$ \to $$ 2K<sup>+</sup> + [Hgl<sub>4</sub>]<sup>2+</sup>
<br><br>N = $${{2 + 1} \over 1}$$ = 3
<br><br>i = 1 + (N - 1)$$\alpha $$
<br><br>= 1 + (3 - 1)0.4
<br><br>= 1 ... | mcq | jee-main-2019-online-11th-january-evening-slot | 3,552 |
YK46Ml11TsVXbVtuDEFy2 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | The freezing point of a diluted milk sample is found to be –0.2<sup>o</sup>C, while it should have been –0.5<sup>o</sup>C for pure milk. How much water has been added to pure milk to make the diluted sample? | [{"identifier": "A", "content": "1 cup of water to 2 cups of pure milk "}, {"identifier": "B", "content": "2 cups of water to 3 cups of pure milk"}, {"identifier": "C", "content": "3 cups of water to 2 cups of pure milk"}, {"identifier": "D", "content": "1 cup of water to 3 cups of pure milk"}] | ["C"] | null | We know,
<br><br>$$\Delta $$T<sub>f</sub> = i $$ \times $$ k<sub>f</sub> $$ \times $$ m
<br><br>$$ \therefore $$ $$\Delta $$T<sub>f<sub>Dil. Milk</sub></sub> = (1) $$ \times $$ k<sub>f</sub> $$ \times $$ m<sub>dil</sub> = 0.2 ...(1)
<br><br>$$\Delta $$T<sub>f<sub>Pure Milk</sub></sub> = (1) $$ \times $$ k<sub>f</sub> ... | mcq | jee-main-2019-online-11th-january-morning-slot | 3,553 |
aOVkl6GdbBZ74td1d57k9k2k5idupa6 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | How much amount of NaCl should be added
to 600 g of water ($$\rho $$ = 1.00 g/mL) to decrease
the freezing point of water to – 0.2 °C ?
______. <br/>
(The freezing point depression constant for water = 2K kg mol<sup>–1</sup>) | [] | null | 1.74to1.76 | $$\Delta $$T<sub>f</sub> = 0.2<sup>o</sup> C
<br><br>$$ \therefore $$ $$\Delta $$T<sub>f</sub> = ik<sub>f</sub>.m
<br><br>i = 2 for NaCl
<br><br>$$ \therefore $$ 0.2 = 2$$ \times $$2$$ \times $$$${{{W_{NaCl}} \times 1000} \over {58.5 \times 600}}$$
<br><br>$$ \Rightarrow $$ W<sub>NaCl</sub> = $${{58.5 \times 600 \times... | integer | jee-main-2020-online-9th-january-morning-slot | 3,554 |
ZzQIFmCNxSL59jA5cSjgy2xukfuqu70t | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | The elevation of boiling point of 0.10 m
aqueous CrCl<sub>3</sub>.xNH<sub>3</sub> solution is two times that
of 0.05 m aqueous CaCl<sub>2</sub> solution. The value of
x is ______.
<br/><br/>[Assume 100% ionisation of the complex and
CaCl<sub>2</sub>, coordination number of Cr as 6, and that
all NH<sub>3</sub> molecules... | [] | null | 5 | Molality of CaCl<sub>2</sub> solution = 0.05 m
<br><br>$$\Delta $$T<sub>b</sub> = i K<sub>b</sub> m = 3 × K<sub>b</sub> × 0.05 = 0.15 K<sub>b</sub>
<br><br>Molality of CrCl<sub>3</sub>.xNH<sub>3</sub> = 0.10 m
<br><br>$$\Delta $$T<sub>b</sub><sup>'</sup> = i K<sub>b</sub> $$ \times $$ 0.10
<br><br>Given, $$\Delta $$T<s... | integer | jee-main-2020-online-6th-september-morning-slot | 3,555 |
uW74yMHbP02VSEhYty1klrh0sh6 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | When 9.45 g of CICH<sub>2</sub>COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of CICH<sub>2</sub>COOH is x $$ \times $$ 10<sup>-3</sup>. <br/>The value of x is ________. (Rounded
off to the nearest integer)<br/>
[K<sub>f(H<sub>2</sub>0)</sub> = 1.86 K kg mol<sup>-1</sup>] | [] | null | 34.4 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxfqatp2/3d8efe2b-8583-464f-a1f5-0d7506d18635/93f38760-6227-11ec-95de-59bb36457aaa/file-1kxfqatp3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxfqatp2/3d8efe2b-8583-464f-a1f5-0d7506d18635/93f38760-6227-11ec-95de-59bb36457aaa/fi... | integer | jee-main-2021-online-24th-february-morning-slot | 3,556 |
cseUSE7QYEyLMbilXb1klrvhq0b | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | C<sub>6</sub>H<sub>6</sub> freezes at 5.5$$^\circ$$C. The temperature at which a solution of 10g of C<sub>4</sub>H<sub>10</sub> in 200g of C<sub>6</sub>H<sub>6</sub> freeze is __________ $$^\circ$$C. (The molal freezing point depression constant of C<sub>6</sub>H<sub>6</sub> is 5.12$$^\circ$$C/m.) | [] | null | 1 | Pure solvent : C<sub>6</sub>H<sub>6</sub>(l)<br/><br/>Given, <br/><br/>$$T_f^o = 5.5^\circ C$$<br/><br/>$${K_f} = 5.12^\circ C/m \Rightarrow m = 200g$$<br/><br/>$${m_{solute}} = 10g$$<br/><br/>Molar mass of solute $${C_4}{H_{10}} = 12 \times 4 + 10 = 58$$<br/><br/>Solute (C<sub>4</sub>H<sub>10</sub>) is non-dissociativ... | integer | jee-main-2021-online-24th-february-evening-slot | 3,557 |
ErkMDtDt9VNWviRbL81kluefvo5 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | 224 mL of SO<sub>2(g)</sub> at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36g of water. The lowering of vapour pressure of solution (assuming the solution in dilute) (P$$_{({H_2}O)}^o$$ $$-$$ 24 mm of Hg) is x $$\times$$ 10<sup>$$-$$2</sup> mm of Hg... | [] | null | 18TO24 | moles of SO<sub>2</sub> = $${{224} \over {22400}}$$ = 0.01
<br><br>moles of NaOH = molarity × volume (in litre)
<br>= 0.1 × 0.1
<br>= 0.01 moles
<br><br>The balanced equation is
<br><br>SO<sub>2</sub> + 2NaOH $$ \to $$ Na<sub>2</sub>SO<sub>3</sub> + H<sub>2</sub>O
<br><br>$$ \therefore $$ Here NaOH is limiting Reagent.... | integer | jee-main-2021-online-26th-february-morning-slot | 3,560 |
5zCB4afeCDViAlb4Mo1klusqnm8 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be $$-$$0.93$$^\circ$$C (K<sub>f</sub>(H<sub>2</sub>O) = 1.86 K kg mol<sup>$$-$$1</sup>). The number (n) of benzoic acid molecules associated (assuming 100% association) is ___________. | [] | null | 2 | $\underset{\text{Benzoic acid}}{n \mathrm{PhCOOH}} \stackrel{\text { Association }}{\longrightarrow}(\mathrm{PhCOOH})_n$<br/><br/>
Assuming $100 \%$ association ( $\alpha=1$ ),<br/><br/>
$$
\Rightarrow i=1-\alpha\left(1-\frac{1}{n}\right)=\frac{1}{n}[\because \alpha+1]
$$
<br/><br/>
Now, $\Delta T_f=K_f \times m \times... | integer | jee-main-2021-online-26th-february-evening-slot | 3,561 |
ZuLYXy4UjO0ejqLCfW1kmhv2pib | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | AB<sub>2</sub> is 10% dissociated in water to A<sup>2+</sup> and B<sup>$$-$$</sup>. The boiling point of a 10.0 molal aqueous solution of AB<sub>2</sub> is __________$$^\circ$$C. (Round off to the Nearest Integer).<br/><br/>[Given : Molal elevation constant of water K<sub>b</sub> = 0.5 K kg mol<sup>$$-$$1</sup> boiling... | [] | null | 106 | AB<sub>2</sub> $$ \leftrightharpoons $$ A<sup>+</sup> + 2B<sup>$$-$$</sup><br><br>$$ \therefore $$ For AB<sub>2</sub>, n = 3<br><br>i = 1 + (n $$-$$ 1)$$\alpha$$<br><br>= 1 + (3 $$-$$ 1) $$\times$$ 0.1<br><br>= 1.2<br><br>Now, $$\Delta$$T<sub>b</sub> = K<sub>b</sub> (im)<br><br>$$ \Rightarrow $$ T<sub>b</sub> $$-$$ T$$... | integer | jee-main-2021-online-16th-march-morning-shift | 3,562 |
QufovOamD6kZVi4S1k1kmlo037h | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | 2 molal solution of a weak acid HA has a freezing point of 3.885$$^\circ$$C. The degree of dissociation of this acid is ___________ $$\times$$ 10<sup>$$-$$3</sup>. (Round off to the Nearest Integer).<br/><br/>[Given : Molal depression constant of water = 1.85 K kg mol<sup>$$-$$1</sup> Freezing point of pure water = 0$$... | [] | null | 50 | $$\Delta$$T<sub>f</sub> = K<sub>f</sub> (im)<br><br>$$ \Rightarrow $$ 3.885 = i $$\times$$ 1.85 $$\times$$ 2<br><br>$$ \Rightarrow $$ i = 1.05<br><br>Also, we know,<br><br>i = 1 + (n $$-$$ 1) $$\alpha$$<br><br>here n = number of particle obtained upon the dissociation of one particle.<br><br>$$HA\rightleftharpoons H^{+... | integer | jee-main-2021-online-18th-march-morning-shift | 3,563 |
te3cQFjnps1SW9JLRX1kmm2a100 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52$$^\circ$$C. The percentage association of A is __________. (Round off to the Nearest Integer).<br/><br/>[Use : K<sub>b</sub> for water = 0.52 K kg mol<sup>$$-$$1</sup> Boiling point of water = 100$$^\circ$$C] | [] | null | 100 | $$\Delta$$T<sub>b</sub> = Boiling point of the solution $$-$$ Boiling point of the pure solvent<br><br>= 100.52 $$-$$ 100<br><br>= 0.52<br><br>$$ \therefore $$ $$\Delta$$T<sub>b</sub> = K<sub>b</sub> (iM)<br><br>$$ \Rightarrow $$ 0.52 = i $$\times$$ 0.52 $$\times$$ 2<br><br>$$ \Rightarrow $$ i = $${1 \over 2}$$<br><br>... | integer | jee-main-2021-online-18th-march-evening-shift | 3,564 |
1ks1k0poa | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is __________ $$\times$$ 10<sup>$$-$$2</sup>.<br/><br/>(Round off to the nearest integer) | [] | null | 125 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266901/exam_images/sxv0sdssoxgbqzuowyuk.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267257/exam_images/qgdabwpcxnspfeff7tpz.webp"><img src="https://res.c... | integer | jee-main-2021-online-27th-july-evening-shift | 3,566 |
1ktjyfpus | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | 1.22 g of an organic acid is separately dissolved in 100 g of benzene (K<sub>b</sub> = 2.6 K kg mol<sup>$$-$$1</sup>) and 100 g of acetone (K<sub>b</sub> = 1.7 K kg mol<sup>$$-$$1</sup>). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases ... | [] | null | 13 | With benzene as solvent<br><br>$$\Delta$$T<sub>b</sub> = i K<sub>b</sub> m<br><br>$$\Delta$$T<sub>b</sub> = $${1 \over 2}$$ $$\times$$ 2.6 $$\times$$ $${{1.22/{M_w}} \over {100/1000}}$$ .... (1)<br><br>With Acetone as solvent<br><br>$$\Delta$$T<sub>b</sub> = i K<sub>b</sub> m<br><br>0.17 = 1 $$\times$$ 1.7 $$\times$$ $... | integer | jee-main-2021-online-31st-august-evening-shift | 3,568 |
1l54a08ot | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198$$^\circ$$C. The percentage of dissociation of the acid is ___________. (Nearest integer)</p>
<p>[Given : Density of acetic acid is 1.02 g mL<sup>$$-$$1</sup>, Molar mas... | [] | null | 5 | $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m}$
<br/><br/>
Moles of solute $($ acetic acid $)=\frac{1.2 \times 1.02}{60}$
<br/><br/>
As moles of solute are very less.
<br/><br/>
So, take molarity and molality the same.
<br/><br/>
$0.0198=\mathrm{i} \times 1.85 \times \frac{1... | integer | jee-main-2022-online-29th-june-morning-shift | 3,569 |
1l54z5wi5 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4 K. The ratio of molal elevation constant to molal depression constant (K<sub>b</sub>/K<sub>f</sub>) is _________.</p> | [] | null | 3 | $$\begin{aligned}
&\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{iK}_{\mathrm{b}} \mathrm{m} \\\\
&\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \mathrm{m} \\\\
&\frac{4}{4}=\frac{\mathrm{K}_{\mathrm{b}} 1.5}{\mathrm{~K}_{\mathrm{f}} 4.5} \\\\
&\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=3
\end{aligned}$... | integer | jee-main-2022-online-29th-june-evening-shift | 3,570 |
1l55nwodi | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>2.5 g of protein containing only glycine (C<sub>2</sub>H<sub>5</sub>NO<sub>2</sub>) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03 $$\times$$ 10<sup>$$-$$3</sup> bar. The total number of glycine units present in the protein is ____________.</p>
<p... | [] | null | 330 | Since,
<br/><br/>
$$
\pi=\mathrm{icR} \mathrm{T}
$$<br/><br/>
$5.03 \times 10^{-3}=\frac{2.5}{M} \times \frac{1000}{500} \times 0.083 \times 300$
<br/><br/>
Molar mass of protein $=24751.5 \mathrm{~g} / \mathrm{mol}$
<br/><br/>
Number of glycine units in protein $=\frac{24751.5}{75}$
<br/><br/>
$$
=330
$$ | integer | jee-main-2022-online-28th-june-evening-shift | 3,571 |
1l57sxmat | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>2 g of a non-volatile non-electrolyte solute is dissolved in
200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 8. The elevation in boiling points of A and B are in the ratio $${x \over y}$$ (x : y). The value of y is ______________. (Nearest integer)</p> | [] | null | 8 | $\Delta T b=k b m$
<br/><br/>
$$
\begin{aligned}
&\frac{\left(\Delta T_{b}\right)_{A}}{\left(\Delta T_{b}\right)_{B}}=\frac{\left(k_{b}\right)_{A}}{\left(k_{b}\right)_{B}} \\\\
&=\frac{1}{8}=\frac{x}{y} \\\\
&\therefore y=8
\end{aligned}
$$ | integer | jee-main-2022-online-27th-june-morning-shift | 3,572 |
1l58efxbx | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>A 0.5 percent solution of potassium chloride was found to freeze at $$-$$0.24$$^\circ$$C. The percentage dissociation of potassium chloride is ______________. (Nearest integer)</p>
<p>(Molal depression constant for water is 1.80 K kg mol<sup>$$-$$1</sup> and molar mass of KCl is 74.6 g mol<sup>$$-$$1</sup>)</p> | [] | null | 98 | $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{b}} \mathrm{m}$
<br/><br/>
$$
\begin{aligned}
&\mathrm{i}=\frac{0.24 \times 99.5 \times 74.6}{1.80 \times 0.5 \times 1000} \\\\
&=1.98 \\\\
&\alpha=\frac{\mathrm{i}-1}{\mathrm{n}-1}=\frac{0.98}{1}=0.98 = 98 \,\%
\end{aligned}
$$ | integer | jee-main-2022-online-26th-june-morning-shift | 3,573 |
1l59qkfik | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2$$^\circ$$C. The percentage association of solute A in water, is :</p>
<p>[Given : Molar mass of A = 93 g mol<sup>$$-$$1</sup>. Molal depression constant of water is 1.86 K kg mol<sup>$$-$$1</s... | [{"identifier": "A", "content": "50%"}, {"identifier": "B", "content": "60%"}, {"identifier": "C", "content": "70%"}, {"identifier": "D", "content": "80%"}] | ["D"] | null | Since, $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f} m}$
<br/><br/>
$$
\begin{aligned}
&m=\frac{0.7}{93} \times \frac{1000}{42} \\\\
&0.2=i \times 1.86 \times \frac{0.7 \times 1000}{93 \times 42} \\\\
&i=0.6 \\\\
&\alpha=\frac{i-1}{\frac{1}{n}-1}=\frac{0.6-1}{\frac{1}{2}-1}=0.8
\end{aligned}
$$
<br/><br/>
Hen... | mcq | jee-main-2022-online-25th-june-evening-shift | 3,575 |
1l6e0nxvl | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>The depression in freezing point observed for a formic acid solution of concentration $$0.5 \mathrm{~mL} \mathrm{~L}^{-1}$$ is $$0.0405^{\circ} \mathrm{C}$$. Density of formic acid is $$1.05 \mathrm{~g} \mathrm{~mL}^{-1}$$. The Van't Hoff factor of the formic acid solution is nearly : (Given for water $$\mathrm{k}_{... | [{"identifier": "A", "content": "0.8"}, {"identifier": "B", "content": "1.1"}, {"identifier": "C", "content": "1.9"}, {"identifier": "D", "content": "2.4"}] | ["C"] | null | $$\Delta \mathrm{T}_{\mathrm{f}}$$ of formic acid $$=0.0405^{\circ} \mathrm{C}$$
<br/><br/>
Concentration $$=0.5 \mathrm{~mL} / \mathrm{L}$$
<br/><br/>
and density $$=1.05 \mathrm{~g} / \mathrm{mL}$$
<br/><br/>
$$\therefore$$ Mass of formic acid in solution $$=1.05 \times 0.5 \mathrm{~g}$$
<br/><br/>
$$
=0.525 \mathrm... | mcq | jee-main-2022-online-25th-july-morning-shift | 3,576 |
1l6f64rq1 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is</p> | [{"identifier": "A", "content": "1 : 4"}, {"identifier": "B", "content": "1 : 0.25"}, {"identifier": "C", "content": "1 : 0.20"}, {"identifier": "D", "content": "1 : 5"}] | ["B"] | null | $$\Delta T_{f}=i k_{f} \times m$$
<br/><br/>
$$
\frac{\Delta T_{\mathrm{f}(\mathrm{A})}}{\Delta \mathrm{T}_{\mathrm{f}(\mathrm{B})}}=\frac{1}{4}
$$
<br/><br/>
$$\frac{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{A}}} \times 1}{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\m... | mcq | jee-main-2022-online-25th-july-evening-shift | 3,577 |
1ldo3sksy | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>$$20 \%$$ of acetic acid is dissociated when its $$5 \mathrm{~g}$$ is added to $$500 \mathrm{~mL}$$ of water. The depression in freezing point of such water is _________ $$\times 10^{-3}{ }^{\circ} \mathrm{C}$$.</p>
<p>Atomic mass of $$\mathrm{C}, \mathrm{H}$$ and $$\mathrm{O}$$ are 12,1 and 16 a.m.u. respectively.<... | [] | null | 372 | $\begin{aligned} & \mathrm{i}=1+(\mathrm{n}-1) \alpha \\\\ & \Rightarrow \mathrm{i}=1+0.2(2-1)=1.2 \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=1.2 \times 1.86 \times \frac{5 \times 1000}{60 \times 500} \\\\ & \Delta \mathrm{t}_{\ma... | integer | jee-main-2023-online-1st-february-evening-shift | 3,578 |
ldo9dqg6 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | Evaluate the following statements for their correctness.
<br/><br/>
A. The elevation in boiling point temperature of water will be same for $0.1 \mathrm{M} \, \mathrm{NaCl}$ and $0.1 \mathrm{M}$ urea.
<br/><br/>
B. Azeotropic mixtures boil without change in their composition.
<br/><br/>
C. Osmosis always takes place fr... | [{"identifier": "A", "content": "A, B and D only"}, {"identifier": "B", "content": "A and C only"}, {"identifier": "C", "content": "B and D only"}, {"identifier": "D", "content": "B, D and E only"}] | ["C"] | null | (A) Elevation in boiling point temperature of water will
be higher for 0.1 M NaCl as compared to 0.1 M
urea.
<br/><br/>(B) Azeotropic mixtures boil without change in their
composition
<br/><br/>(C) Osmosis always takes place from hypotonic (low
concentration of solute) solution to hypertonic
(high concentration of solu... | mcq | jee-main-2023-online-31st-january-evening-shift | 3,579 |
1ldokmfw7 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M $$\mathrm{AgNO_3}$$ solution when titrated using $$\mathrm{K_2CrO_4}$$ as an indicator. What is the depression in freezing point of KCl solution of the given concentration? _________ (Nearest integer).</p>
<p>(Given : $$\mathrm{K_f=2.0~K~kg~mol^{-... | [] | null | 3 | $25 \times M=20 \times 1$
<br/><br/>$$
\begin{aligned}
M & =\frac{20}{25}=\frac{4}{5}=0.8 \\\\
\Delta T_{f} & =(\mathrm{i})\left(\mathrm{K}_{\mathrm{f}}\right)(\mathrm{m}) \\\\
& =(2)(2)\left(\frac{4}{5}\right)=\frac{16}{5}=3.2
\end{aligned}
$$ | integer | jee-main-2023-online-1st-february-morning-shift | 3,580 |
ldqy4uri | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | Lead storage battery contains $38 \%$ by weight solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$. The van't Hoff factor is $2.67$ at this concentration. The temperature in Kelvin at which the solution in the battery will freeze is ________. (Nearest integer).
<br/><br/>
Given $\mathrm{K}_{f}=1.8 \mathrm{~K} \mathrm{~kg} \m... | [] | null | 243 | <p>$$\mathrm{\Delta T_f=K_f~i~m}$$</p>
<p>$$ = 1.8 \times 2.67 \times {{{{38} \over {98}}} \over {0.062}} = 30$$</p>
<p>$$\therefore$$ It freeze at $$273-30=243$$ $$\mathrm{K}$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 3,581 |
1ldsd375e | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-III, B-I, C-IV, D-II"}, {"identifier": "B", "content": "A-III, B-I, C-II, D-IV"}, {"identifier": "C", "content": "A-III, B-II, C-I, D-IV"}, {"identifier": "D", "content": "A-I, B-III, C-II, D-IV"}] | ["B"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial,... | mcq | jee-main-2023-online-29th-january-evening-shift | 3,582 |
1ldste4zz | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15$$^\circ$$C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at $$-0.8^\circ$$ C. The solubility product of PbCl$$_2$$ formed is __________ $$\times$$ 10$$^{-6}$$ at 298 K. (N... | [] | null | 13 | $0.15=3 \times 0.5 \times \mathrm{M}$
<br/><br/>
$$
\begin{aligned}
& \mathrm{M}_{\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}}=0.1 \text { molar } \\\\
& \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \longrightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3} \\\\
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\... | integer | jee-main-2023-online-29th-january-morning-shift | 3,583 |
1ldu1q203 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>The number of pairs of the solutions having the same value of the osmotic pressure from the following is _________.</p>
<p>(Assume 100% ionization)</p>
<p>A. 0.500 $$\mathrm{M~C_2H_5OH~(aq)}$$ and 0.25 $$\mathrm{M~KBr~(aq)}$$</p>
<p>B. 0.100 $$\mathrm{M~K_4[Fe(CN)_6]~(aq)}$$ and 0.100 $$\mathrm{M~FeSO_4(NH_4)_2SO_4~... | [] | null | 4 | $\pi=i C R T$
<br/><br/>The following pairs of solutions have same value of osmotic pressure<br/><br/>
(A) $0.500 \mathrm{M} ~\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}) \mathrm{i}=1$ and $0.25 \mathrm{M} ~\mathrm{KBr}(\mathrm{aq})$ $i=2$<br/><br/>
$$
\begin{aligned}
& \pi_1=0.5 \times 1 \times R T =0.5 R T... | integer | jee-main-2023-online-25th-january-evening-shift | 3,584 |
1lgq57n5s | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>Solution of $$12 \mathrm{~g}$$ of non-electrolyte (A) prepared by dissolving it in $$1000 \mathrm{~mL}$$ of water exerts the same osmotic pressure as that of $$0.05 ~\mathrm{M}$$ glucose solution at the same temperature. The empirical formula of $$\mathrm{A}$$ is $$\mathrm{CH}_{2} \mathrm{O}$$. The molecular mass of... | [] | null | 240 | To solve this problem, we will first calculate the osmotic pressure of the 0.05 M glucose solution and then use that information to determine the molecular mass of compound A.
<br/><br/>
1. Osmotic pressure equation:
<br/><br/>
$$\Pi = iMRT$$
<br/><br/>
where
$$\Pi$$ is the osmotic pressure,
$$i$$ is the van't Hoff f... | integer | jee-main-2023-online-13th-april-morning-shift | 3,585 |
1lgyhw1k3 | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | <p>If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be ___________% higher than the expected/theoretical freezing point. (Nearest integer)</p> | [] | null | 30 | <p>The degree of dissociation, often represented as $\alpha$, is the fraction of a mole of a substance that has dissociated into ions in solution. For a weak monobasic acid, this degree of dissociation can increase the number of particles in solution, which can in turn affect colligative properties such as the freezing... | integer | jee-main-2023-online-10th-april-morning-shift | 3,587 |
lsbmfv5f | chemistry | solutions | abnormal-colligative-property-and-van't-hoff-factor | We have three aqueous solutions of $\mathrm{NaCl}$ labelled as ' $\mathrm{A}$ ', ' $\mathrm{B}$ ' and ' $\mathrm{C}$ ' with concentration $0.1 \mathrm{M}$, $0.01 \mathrm{M}$ and $0.001 \mathrm{M}$, respectively. The value of van 't Hoff factor(i) for these solutions will be in the order : | [{"identifier": "A", "content": "$\\mathrm{i}_{\\mathrm{A}}<\\mathrm{i}_{\\mathrm{C}}<\\mathrm{i}_{\\mathrm{B}}$"}, {"identifier": "B", "content": "$\\mathrm{i}_{\\mathrm{A}}<\\mathrm{i}_{\\mathrm{B}}<\\mathrm{i}_{\\mathrm{C}}$"}, {"identifier": "C", "content": "$\\mathrm{i}_{\\mathrm{A}}>\\mathrm{i}_{\\mathrm{B}}>\\ma... | ["B"] | null | <style>
table {
border-collapse: collapse;
width: 100%;
}
th, td {
border: 1px solid black;
padding: 8px;
text-align: center;
}
.header {
background-color: #0000FF;
color: white;
}
</style>
<table>
<tr>
<th class="header">Salt</th>
<th class="header" colspan="3">Values... | mcq | jee-main-2024-online-1st-february-morning-shift | 3,588 |
A5qdxMcm7s6dPyHC | chemistry | solutions | depression-in-freezing-point | Freezing point of an aqueous solution is (-0.186)<sup>o</sup>C. Elevation of boiling point of the same solution is K<sub>b</sub> = 0.512 <sup>o</sup>C, K<sub>f</sub> = 1.86 <sup>o</sup>C, find the increase in boiling point. | [{"identifier": "A", "content": "0.186 <sup>o</sup>C"}, {"identifier": "B", "content": "0.0512 <sup>o</sup>C"}, {"identifier": "C", "content": "0.092 <sup>o</sup>C"}, {"identifier": "D", "content": "0.2732 <sup>o</sup>C"}] | ["B"] | null | $$\Delta {T_b} = {K_b}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;$$
<br><br>$$\Delta {T_f} = {K_f}{{{W_B}} \over {{M_B} \times {W_A}}} \times 1000;$$
<br><br>$${{\Delta {T_b}} \over {\Delta {T_f}}} = {{{K_b}} \over {{K_f}}} = {{\Delta {T_b}} \over { - 0.186}}$$
<br><br>$$ = {{0.512} \over {1.86}}$$
<br><br>$$ =... | mcq | aieee-2002 | 3,589 |
Raghw5Pv8ekP1S85 | chemistry | solutions | depression-in-freezing-point | Equimolar solutions in the same solvent have | [{"identifier": "A", "content": "Same boiling point but different freezing point"}, {"identifier": "B", "content": "Same freezing point but different boiling point "}, {"identifier": "C", "content": "Same boiling and same freezing points "}, {"identifier": "D", "content": "Different boiling and different freezing point... | ["C"] | null | Equimolar solutions of normal solutes in the same solvent will have the same boiling point and same freezing point. | mcq | aieee-2005 | 3,590 |
Ov2iNZd93pMjbXe6 | chemistry | solutions | depression-in-freezing-point | Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be
added to 4 kg of water to prevent it from freezing at −6<sup>o</sup>C will be :
[K<sub>f</sub> for water = 1.86 K kg mol<sup>−1</sup> , and molar mass of ethylene glycol = 62 g mol<sup>−1</sup> ) | [{"identifier": "A", "content": "204.30 g"}, {"identifier": "B", "content": "400.00 g "}, {"identifier": "C", "content": "304.60 g"}, {"identifier": "D", "content": "804.32 g"}] | ["D"] | null | Given $${K_r} = 1.86\,K\,kg\,mo{l^{ - 1}}$$
<br><br>$$\Delta {T_f} = 0 - \left( { - 6} \right) = {6^ \circ }C$$
<br><br>As we know that
<br><br>$$\Delta {T_f} = {K_f} \times \,\,molality$$
<br><br>$$ = {{{K_f} \times 1000 \times mass\,\,of\,\,solute} \over {molar\,\,mass\,\,of\,\,solute\,\, \times \,\,mass\,\,of\,\,so... | mcq | aieee-2011 | 3,591 |
uPhJ9fJb3qbYEviqMUKWP | chemistry | solutions | depression-in-freezing-point | A solution containing 62 g ethylene glycol in 250 g water is cooled to $$-$$ 10<sup>o</sup>C. If K<sub>f</sub> for water is 1.86 K kg mol<sup>$$-$$1</sup> , the amount of water (in g) separated as ice is : | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "64"}, {"identifier": "D", "content": "16"}] | ["C"] | null | Here water is solvent and ethylene glycol is solute.
<br/><br/>We know, Depression of freezing point,
<br/><br/>$$\Delta $$T<sub>f</sub> = K<sub>f</sub> . m
<br/><br/>$$\Delta {T_f} = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{250} \over {1000}}} \right)}}$$
<br/><br/>= 7.44
<br/><br/>We know, f... | mcq | jee-main-2019-online-9th-january-evening-slot | 3,592 |
cLJziJMbqdaggnBOU6H3u | chemistry | solutions | depression-in-freezing-point | Elevation in the boiling point for 1 molar solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between K<sub>b</sub> and K<sub>f</sub> is | [{"identifier": "A", "content": "K<sub>b</sub> = K<sub>f</sub>"}, {"identifier": "B", "content": "K<sub>b</sub> = 0.5 K<sub>f</sub>"}, {"identifier": "C", "content": "K<sub>b</sub> = 1.5 K<sub>f</sub>"}, {"identifier": "D", "content": "K<sub>b</sub> = 2 K<sub>f</sub>"}] | ["D"] | null | $${{\Delta {T_b}} \over {\Delta {T_f}}} = {{i.m \times {k_b}} \over {i \times m \times {k_f}}}$$
<br><br>$${2 \over 2} = {{1 \times 1 \times {k_b}} \over {1 \times 2 \times {k_f}}}$$
<br><br>$${k_b} = 2{k_f}$$ | mcq | jee-main-2019-online-10th-january-evening-slot | 3,593 |
49J2TKThr2kEsCqY1QD6K | chemistry | solutions | depression-in-freezing-point | Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is - | [{"identifier": "A", "content": "4A"}, {"identifier": "B", "content": "2A"}, {"identifier": "C", "content": "3A"}, {"identifier": "D", "content": "A"}] | ["C"] | null | For same freezing point,
<br><br>($$\Delta $$T<sub>f</sub>)<sub>X</sub> = ($$\Delta $$T<sub>f</sub>)<sub>Y</sub>
<br><br>$$ \Rightarrow $$ k<sub>f</sub> m<sub>x</sub>
= k<sub>f</sub> m<sub>y</sub>
<br><br>$$ \Rightarrow $$ $${{4 \times 1000} \over {A \times 96}} = {{12 \times 1000} \over {M \times 88}}$$
<br><br>$$ \R... | mcq | jee-main-2019-online-12th-january-morning-slot | 3,594 |
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