MHT CET 2021 Chemistry Question Paper with Answer and Solution

(A) In an $hcp$ structure,for each atom,there are three voids present (one octahedral void and two tetrahedral voids).
Number of atoms in $1 \ mol$ of the compound $= 6.022 \times 10^{23}$.
Total number of voids $=$ (Number of atoms) $\times 3$.
Total number of voids $= 6.022 \times 10^{23} \times 3$.
Total number of voids $= 18.066 \times 10^{23} = 1.8066 \times 10^{24}$.

(D) For a $BCC$ structure, the number of atoms per unit cell is $Z = 2$.
The formula for density is $d = \frac{Z \cdot M}{N_A \cdot a^3}$.
Given: $d = 10 \ g \ cm^{-3}$, $a = 200 \ pm = 200 \times 10^{-10} \ cm = 2 \times 10^{-8} \ cm$, and $N_A \approx 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $10 = \frac{2 \times M}{6.022 \times 10^{23} \times (2 \times 10^{-8})^3}$.
$10 = \frac{2 \times M}{6.022 \times 10^{23} \times 8 \times 10^{-24}}$.
$10 = \frac{2 \times M}{4.8176}$.
$M = \frac{10 \times 4.8176}{2} = 24.088 \ g \ mol^{-1}$.
Thus, the molar mass is approximately $24.0 \ g \ mol^{-1}$.

(D) The packing efficiency for different crystal structures is as follows:
$FCC$ (Face-Centered Cubic) = $74 \%$
$BCC$ (Body-Centered Cubic) = $68 \%$
Hexagonal Close Packing $(HCP)$ = $74 \%$
Simple cubic = $52.36 \%$
Therefore,the correct option is $D$.

(D) The atomic mass of potassium $(K)$ is $39 \ g/mol$.
Number of moles of potassium $= \frac{3.9 \ g}{39 \ g/mol} = 0.1 \ mol$.
Number of atoms $= \text{moles} \times N_{A} = 0.1 \ N_{A}$.
In a $BCC$ unit cell,the number of atoms per unit cell $(n)$ is $2$.
Therefore,the number of unit cells $= \frac{\text{Total number of atoms}}{n} = \frac{0.1 \ N_{A}}{2} = \frac{N_{A}}{20}$.

(C) Given: Molar mass $M = 180 \ g \ mol^{-1}$,Density $\rho = 18 \ g \ cm^{-3}$,Avogadro constant $N_A \approx 6.022 \times 10^{23} \ mol^{-1}$.
For a $BCC$ crystal,the number of atoms per unit cell $z = 2$.
The formula for density is $\rho = \frac{M \times z}{a^3 \times N_A}$.
Rearranging for the edge length $a$: $a^3 = \frac{M \times z}{\rho \times N_A}$.
Substituting the values: $a^3 = \frac{180 \times 2}{18 \times 6.022 \times 10^{23}} \approx \frac{360}{108.396 \times 10^{23}} \approx 3.32 \times 10^{-23} \ cm^3$.
To match the format,we write $a^3 = 33.2 \times 10^{-24} \ cm^3$.
Therefore,$a = \sqrt[3]{33.2} \times 10^{-8} \ cm$.

(B) In a simple cubic structure,the atoms touch along the edge of the unit cell.
Therefore,the relationship between the edge length $a$ and the atomic radius $r$ is $a = 2r$.
Given $a = 3.86 \ \mathring{A}$.
$r = \frac{a}{2} = \frac{3.86 \ \mathring{A}}{2} = 1.93 \ \mathring{A}$.
Converting to centimeters: $1 \ \mathring{A} = 10^{-8} \ cm$.
So,$r = 1.93 \times 10^{-8} \ cm$.

(C) The density formula for a unit cell is given by: $\rho = \frac{Z \times M}{N_A \times a^3}$,where $a^3$ is the volume of the unit cell $(V)$.
For a $BCC$ structure,the number of atoms per unit cell $(Z)$ is $2$.
Given: Atomic mass $(M)$ = $25 \ g \ mol^{-1}$,Density $(\rho)$ = $3 \ g \ cm^{-3}$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Rearranging the formula for volume $(V)$: $V = \frac{Z \times M}{\rho \times N_A}$.
Substituting the values: $V = \frac{2 \times 25}{3 \times 6.022 \times 10^{23}}$.
$V = \frac{50}{18.066 \times 10^{23}} \approx 2.767 \times 10^{-23} \ cm^3$.

(D) Packing efficiency is defined as the fraction of the total volume of the unit cell occupied by the atoms.
For a $BCC$ (Body-Centered Cubic) structure:
$1$. The number of atoms per unit cell $(z)$ is $2$.
$2$. The relationship between the edge length $(a)$ and the atomic radius $(r)$ is $r = \frac{\sqrt{3}a}{4}$,or $a = \frac{4r}{\sqrt{3}}$.
$3$. The volume of the unit cell $(V)$ is $a^3 = (\frac{4r}{\sqrt{3}})^3 = \frac{64r^3}{3\sqrt{3}}$.
$4$. The volume occupied by $2$ atoms is $2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3$.
$5$. Packing efficiency = $\frac{\text{Volume of atoms}}{\text{Volume of unit cell}} \times 100 = \frac{\frac{8}{3} \pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100 = \frac{\sqrt{3} \pi}{8} \times 100 \approx 0.68 \times 100 = 68 \%$.

(C) For a $BCC$ unit cell,the number of atoms per unit cell,$Z = 2$.
The edge length $a = 400 \ pm = 400 \times 10^{-10} \ cm = 4 \times 10^{-8} \ cm$.
The molar mass $M = 100 \ g \ mol^{-1}$.
The density $d$ is given by the formula: $d = \frac{Z \times M}{N_A \times a^3}$.
Substituting the values: $d = \frac{2 \times 100}{6.022 \times 10^{23} \times (4 \times 10^{-8})^3}$.
$d = \frac{200}{6.022 \times 10^{23} \times 64 \times 10^{-24}}$.
$d = \frac{200}{6.022 \times 64 \times 10^{-1}} = \frac{200}{38.54} \approx 5.18 \ g \ cm^{-3}$.

(C) In one $BCC$ crystal unit cell,the number of atoms $(Z) = 2$.
Total number of atoms $= Z \times \text{Number of unit cells}$.
Total number of atoms $= 2 \times 1.8 \times 10^{20} = 3.6 \times 10^{20}$.

(A) Packing efficiency $\text{P.E.} = \frac{z \times \frac{4}{3} \pi r^3}{V} = \frac{2 \times \frac{4}{3} \times \pi (\frac{\sqrt{3} a}{4})^3}{a^3} = \frac{\sqrt{3} \pi}{8} \approx 0.68$ or $68 \%$.
Since $z = 2$,$r = \frac{\sqrt{3} a}{4}$,and $V = a^3$ for a $BCC$ unit cell.
In $BCC$ structure,$68 \%$ of the total volume is occupied by atoms.
Therefore,the percentage of unoccupied volume (void space) is $100 \% - 68 \% = 32 \%$.

(C) In a face-centred cubic $(FCC)$ unit cell,atoms are present at each corner and at each face centre.
Number of lattice points $=$ Number of corners $+$ Number of face centres
$= 8 + 6$
$= 14$

(D) In an $hcp$ structure,for each atom,there is $1$ octahedral void and $2$ tetrahedral voids.
Number of atoms $= 0.25 \ mol \times 6.022 \times 10^{23} \ atoms/mol = 1.5055 \times 10^{23} \ atoms$.
Number of octahedral voids $= \text{Number of atoms} = 1.5055 \times 10^{23} \approx 1.50 \times 10^{23}$.
Number of tetrahedral voids $= 2 \times \text{Number of atoms} = 2 \times 1.5055 \times 10^{23} = 3.011 \times 10^{23}$.

(B) The packing efficiency of $68 \%$ corresponds to a Body-Centered Cubic $(BCC)$ unit cell.
For a $BCC$ unit cell,the number of atoms per unit cell is $Z = 2$.
The edge length $a = 3.3 \ \mathring{A} = 3.3 \times 10^{-8} \ cm$.
The density formula is given by $d = \frac{Z \cdot M}{N_A \cdot a^3}$,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Rearranging for molar mass $M$: $M = \frac{d \cdot N_A \cdot a^3}{Z}$.
Substituting the values: $M = \frac{8.57 \times 6.022 \times 10^{23} \times (3.3 \times 10^{-8})^3}{2}$.
$M = \frac{8.57 \times 6.022 \times 10^{23} \times 35.937 \times 10^{-24}}{2}$.
$M = \frac{185.48}{2} \approx 92.74 \ g \ mol^{-1}$.
Rounding to the nearest integer,$M = 93 \ g \ mol^{-1}$.

(D) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z \times M}{N_A \times a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N_A$ is Avogadro's number,and $a^3$ is the volume of the unit cell $(V)$.
Alternatively,$d = \frac{Z \times \text{mass of one atom}}{V}$.
For a $bcc$ structure,the number of atoms per unit cell is $Z = 2$.
Given: $d = 9 \ g \ cm^{-3}$,$V = 2.7 \times 10^{-23} \ cm^3$,and total mass $= 2.43 \ g$.
Let $N$ be the total number of atoms in $2.43 \ g$.
The mass of one atom is $\frac{2.43}{N}$.
Substituting the values into the density formula: $9 = \frac{2 \times (2.43 / N)}{2.7 \times 10^{-23}}$.
Rearranging for $N$: $N = \frac{2 \times 2.43}{9 \times 2.7 \times 10^{-23}}$.
$N = \frac{4.86}{24.3 \times 10^{-23}} = 0.2 \times 10^{23} = 2.0 \times 10^{22}$ atoms.

(A) In a face-centred cubic $(FCC)$ unit cell,particles are present at each of the $8$ corners and at the centre of each of the $6$ faces of the cube.

(A) In an $hcp$ structure,for each atom present,there is one octahedral void and two tetrahedral voids.
Number of atoms $= 0.4 \ mol \times N_A = 0.4 \times 6.022 \times 10^{23} \text{ atoms}$.
Number of tetrahedral voids $= 2 \times \text{Number of atoms}$.
Number of tetrahedral voids $= 2 \times 0.4 \times 6.022 \times 10^{23} = 4.8176 \times 10^{23} \approx 4.8 \times 10^{23}$.

(C) In an $FCC$ unit cell,the number of atoms at the corners is $8 \times \frac{1}{8} = 1$.
Since ions of $A$ are at the corners,the number of $A$ atoms per unit cell is $1$.
In an $FCC$ unit cell,the number of atoms at the face centres is $6 \times \frac{1}{2} = 3$.
Since ions of $B$ are at the face centres,the number of $B$ atoms per unit cell is $3$.
Therefore,the ratio of $A:B$ is $1:3$.
The formula of the compound is $AB_3$.

(A) In a $BCC$ unit cell,the number of particles $(Z) = 2$.
The relation between edge length '$a$' and radius '$r$' is $\sqrt{3} a = 4 r$,which gives $r = \frac{\sqrt{3} a}{4}$.
The volume occupied by particles is calculated as $Z \times \text{Volume of one sphere}$.
$\text{Volume} = 2 \times \frac{4}{3} \pi r^3$.
Substituting $r$: $\text{Volume} = 2 \times \frac{4}{3} \pi (\frac{\sqrt{3} a}{4})^3$.
$\text{Volume} = \frac{8}{3} \pi \times \frac{3 \sqrt{3} a^3}{64}$.
$\text{Volume} = \frac{\sqrt{3} \pi a^3}{8}$.

(D) The formula for density is $d = \frac{Z \cdot M}{N_A \cdot a^3}$.
Given:
$Z = 2$ (for $BCC$ structure)
$a = 400 \ pm = 400 \times 10^{-10} \ cm = 4 \times 10^{-8} \ cm$
$d = 4 \ g \ cm^{-3}$
$N_A = 6.022 \times 10^{23} \ mol^{-1}$
Rearranging for molar mass $M$:
$M = \frac{d \cdot N_A \cdot a^3}{Z}$
Substituting the values:
$M = \frac{4 \cdot 6.022 \times 10^{23} \cdot (4 \times 10^{-8})^3}{2}$
$M = \frac{4 \cdot 6.022 \times 10^{23} \cdot 64 \times 10^{-24}}{2}$
$M = 2 \cdot 6.022 \cdot 64 \cdot 10^{-1}$
$M = 77.08 \ g \ mol^{-1} \approx 77 \ g \ mol^{-1}$.

(A) Given: Edge length $a = 500 \ pm = 500 \times 10^{-10} \ cm = 5 \times 10^{-8} \ cm$.
Density $d = 4 \ g \ cm^{-3}$.
For $BCC$ structure, the number of atoms per unit cell $Z = 2$.
The formula for density is $d = \frac{Z \times M}{N_A \times a^3}$.
Rearranging for molar mass $M$: $M = \frac{d \times N_A \times a^3}{Z}$.
Substituting the values: $M = \frac{4 \times 6.022 \times 10^{23} \times (5 \times 10^{-8})^3}{2}$.
$M = \frac{4 \times 6.022 \times 10^{23} \times 125 \times 10^{-24}}{2}$.
$M = 2 \times 6.022 \times 125 \times 10^{-1} = 150.55 \ g \ mol^{-1} \approx 150 \ g \ mol^{-1}$.

(D) In $BCC$ structure,atoms are present at each corner and at the body center of the unit cell.
$No. \text{ of atoms per unit cell } (Z) = 8 \times \frac{1}{8} + 1 \times 1$
$= 1 + 1 = 2$
Therefore,there are $2$ particles per unit cell in a $BCC$ structure.

(B) For a simple cubic structure, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $a = 2r$.
Given that the edge length $a = 336 \ pm$.
Substituting the value into the formula: $336 \ pm = 2r$.
Therefore, $r = \frac{336 \ pm}{2} = 168 \ pm$.

(C) For a $BCC$ unit cell, the relationship between the edge length $a$ and the atomic radius $r$ is given by $4r = \sqrt{3}a$.
Given $a = 352 \ pm$.
Substituting the value: $r = \frac{\sqrt{3} \times 352}{4}$.
$r = \frac{1.732 \times 352}{4} = 152.4 \ pm$.

(A) The density formula for a unit cell is $d = \frac{Z \cdot M}{N_A \cdot a^3}$.
For a $BCC$ structure,the number of atoms per unit cell is $Z = 2$.
Given: $d = 8.55 \ g \ cm^{-3}$,$M = 93 \ g \ mol^{-1}$,$N_A \approx 6.022 \times 10^{23} \ mol^{-1}$.
Rearranging for $a^3$: $a^3 = \frac{Z \cdot M}{d \cdot N_A}$.
Substituting the values: $a^3 = \frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}} \approx 3.61 \times 10^{-23} \ cm^3$.
Therefore,the edge length $a = (3.61 \times 10^{-23})^{1/3} \ cm$.

(A) In a simple cubic unit cell,the atoms are present only at the corners of the cube.
These atoms touch each other along the edge of the cube.
Let '$a$' be the edge length of the unit cell and '$r$' be the radius of the atom.
Since the atoms touch along the edge,the edge length '$a$' is equal to the sum of the radii of the two touching atoms.
Therefore,$a = r + r = 2r$.

(A) The formula for density is $d = \frac{Z \cdot M}{N_{A} \cdot a^3}$.
Rearranging to solve for $Z$: $Z = \frac{d \cdot N_{A} \cdot a^3}{M}$.
Given: $d = 7 \ g \ cm^{-3}$, $a = 300 \ pm = 300 \times 10^{-10} \ cm = 3 \times 10^{-8} \ cm$, $M = 52 \ g \ mol^{-1}$, and $N_{A} = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $Z = \frac{7 \times 6.022 \times 10^{23} \times (3 \times 10^{-8})^3}{52}$.
$Z = \frac{7 \times 6.022 \times 10^{23} \times 27 \times 10^{-24}}{52} \approx \frac{1138.158 \times 10^{-1}}{52} \approx \frac{113.8}{52} \approx 2.18 \approx 2$.
Since $Z = 2$, the unit cell is Body centred cubic $(BCC)$.

(A) For a $BCC$ (Body-Centered Cubic) structure, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $\sqrt{3} a = 4 r$.
Given that the edge length $a = 600 \ pm$.
Substituting the value of $a$ into the formula: $r = \frac{\sqrt{3} a}{4} = \frac{\sqrt{3} \times 600}{4}$.
Calculating the value: $r = \sqrt{3} \times 150 \ pm$.

(A) Substances like iron $(Fe)$,cobalt $(Co)$,nickel $(Ni)$,gadolinium $(Gd)$,and chromium dioxide $(CrO_2)$ are attracted very strongly by an external magnetic field.
Such substances are called ferromagnetic substances.
These materials can be permanently magnetized even after the removal of the external magnetic field.
Among the given options,$Ni$ is a ferromagnetic element.

(A) Substances like iron $(Fe)$,cobalt $(Co)$,nickel $(Ni)$,gadolinium $(Gd)$,and $CrO_2$ are attracted very strongly by a magnetic field. Such substances are called ferromagnetic substances. Besides strong attraction,these substances can be permanently magnetized.

(A) In an amalgam of mercury with sodium,$Hg$ (mercury) is the solute and $Na$ (sodium) is the solvent.
Since mercury is a liquid and sodium is a solid,this forms a solution of a liquid in a solid.
Therefore,the correct classification is liquid in solid.

(A) Given: Vapour pressure of pure solvent,$P_1^0 = 240 \ mm Hg$.
Vapour pressure of solution,$P_1 = 216 \ mm Hg$.
According to Raoult's Law,the relative lowering of vapour pressure is equal to the mole fraction of the solute $(x_2)$:
$\frac{P_1^0 - P_1}{P_1^0} = x_2$
$\frac{240 - 216}{240} = x_2$
$x_2 = \frac{24}{240} = 0.1$
The mole fraction of the solvent $(x_1)$ is given by $x_1 = 1 - x_2$.
$x_1 = 1 - 0.1 = 0.9$.

(A) In a positive deviation from Raoult's law,the solute-solvent intermolecular forces are weaker than the solute-solute and solvent-solvent interactions.
$Ethanol + Acetone$ shows positive deviation because the hydrogen bonding in pure ethanol is disrupted by the addition of acetone.
$Chloroform + Acetone$ shows negative deviation due to the formation of strong hydrogen bonds between them.
$Benzene + Toluene$ forms an ideal solution.
$Phenol + Aniline$ shows negative deviation due to strong hydrogen bonding between the two components.

(B) According to Raoult's law for non-volatile solutes: $\frac{P^{\circ} - P_s}{P^{\circ}} = X_{\text{solute}}$
Here,$P^{\circ} = 32 \ mm \ Hg$,$n_{\text{solute}} = 0.1 \ mol$.
The number of moles of solvent (water) is $n_{\text{solvent}} = \frac{16.2 \ g}{18 \ g/mol} = 0.9 \ mol$.
The mole fraction of the solute is $X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} = \frac{0.1}{0.1 + 0.9} = \frac{0.1}{1.0} = 0.1$.
Substituting the values: $\frac{32 - P_s}{32} = 0.1$.
$32 - P_s = 32 \times 0.1 = 3.2$.
$P_s = 32 - 3.2 = 28.8 \ mm \ Hg$.

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P_1^0 - P_s}{P_1^0} = X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$
Mass of water $= 1.8 \times 10^{-2} \ kg = 18 \ g$.
Moles of water $(n_{\text{solvent}})$ $= \frac{18 \ g}{18 \ g/mol} = 1 \ mol$.
Given $n_{\text{solute}} = 0.1 \ mol$ and $P_1^0 = 24 \ mm \ Hg$.
Substituting the values: $\frac{24 - P_s}{24} = \frac{0.1}{0.1 + 1} = \frac{0.1}{1.1} = \frac{1}{11}$.
$24 - P_s = \frac{24}{11} \approx 2.18 \ mm \ Hg$.
$P_s = 24 - 2.18 = 21.82 \ mm \ Hg \approx 21.84 \ mm \ Hg$.

(A) An ideal solution is one that obeys Raoult's law over the entire range of concentration and shows no change in enthalpy or volume upon mixing.
Benzene and toluene have similar molecular structures and intermolecular forces,making their mixture behave nearly as an ideal solution.
Chloroform and acetone show negative deviation due to hydrogen bonding.
Phenol and aniline show negative deviation due to strong hydrogen bonding.
Ethanol and acetone show positive deviation due to the disruption of hydrogen bonds.

(B) According to Raoult's law for non-volatile solutes: $\frac{P_1^0 - P_s}{P_1^0} = X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}}$.
Number of moles of water $(n_{\text{water}})$ = $\frac{36 \ g}{18 \ g/mol} = 2 \ mol$.
Given $n_{\text{solute}} = 1 \ mol$ and $P_1^0 = 400 \ mm \ Hg$.
Substituting the values: $\frac{400 - P_s}{400} = \frac{1}{1 + 2} = \frac{1}{3}$.
$400 - P_s = \frac{400}{3} = 133.33 \ mm \ Hg$.
$P_s = 400 - 133.33 = 266.67 \ mm \ Hg \approx 267 \ mm \ Hg$.

(C) According to Henry's Law,the solubility of a gas is given by the formula: $\text{Solubility} = K_H \times P_{gas}$.
Given,$K_H = 0.16 \ mol \ L^{-1} \ bar^{-1}$ and $\text{Solubility} = 0.08 \ mol \ L^{-1}$.
Rearranging the formula to solve for pressure: $P_{gas} = \frac{\text{Solubility}}{K_H}$.
Substituting the values: $P_{gas} = \frac{0.08 \ mol \ L^{-1}}{0.16 \ mol \ L^{-1} \ bar^{-1}} = 0.5 \ bar$.

(B) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to the partial pressure $(p)$ of the gas: $S = K_H \times p$.
Given:
Solubility $(S)$ = $7 \times 10^{-4} \ mol \ L^{-1}$
Pressure $(p)$ = $1 \ bar$
Substituting the values into the formula:
$7 \times 10^{-4} \ mol \ L^{-1} = K_H \times 1 \ bar$
Therefore,$K_H = \frac{7 \times 10^{-4} \ mol \ L^{-1}}{1 \ bar} = 7 \times 10^{-4} \ mol \ L^{-1} \ bar^{-1}$.

(C) According to Henry's law,$p = K_H \times x$,where $p$ is the partial pressure of the gas,$K_H$ is the Henry's law constant,and $x$ is the mole fraction of the gas in the solution.
$1$. $K_H$ depends on the nature of the gas and increases with an increase in temperature.
$2$. From the relation $p = K_H \times x$,we can write $x = p / K_H$. This shows that for a given pressure $p$,the solubility $(x)$ is inversely proportional to $K_H$.
$3$. Therefore,a higher value of $K_H$ implies lower solubility of the gas in the liquid.
$4$. Thus,the statement 'Higher the value of $K_H$ at a given pressure,higher is the solubility of the gas in the liquids' is incorrect.

(A) The formula for depression in freezing point is $\Delta T_f = K_f \cdot m$.
First,calculate the molality $(m)$ of the solution:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{5 \ g / 180 \ g \ mol^{-1}}{0.1 \ kg} = \frac{5}{18} \ mol \ kg^{-1} \approx 0.2778 \ mol \ kg^{-1}$.
Given $\Delta T_f = 2.15 \ K$.
Substituting the values into the formula: $2.15 = K_f \cdot (5/18)$.
$K_f = \frac{2.15 \times 18}{5} = 7.74 \ K \ kg \ mol^{-1}$.

(C) The boiling point of a liquid is defined as the temperature at which its vapour pressure becomes equal to the applied or external pressure.

(D) For two solutions to have no net flow across a semipermeable membrane,they must be isotonic,meaning their molar concentrations must be equal.
$M_{urea} = \frac{12 \ g / 60 \ g \ mol^{-1}}{1 \ dm^3} = 0.2 \ M$
$M_{glucose} = \frac{36 \ g / 180 \ g \ mol^{-1}}{1 \ dm^3} = 0.2 \ M$
Since the molar concentrations of urea and glucose in option $D$ are equal $(0.2 \ M)$,the osmotic pressures are equal $(\pi_1 = \pi_2)$.
Therefore,no net flow of solvent occurs between these two solutions.

(A) Given: $W_2 = 6 \ g$ (mass of solute),$W_1 = 100 \ g$ (mass of solvent),$K_b = 0.52 \ K \ kg \ mol^{-1}$,$T_b = 100.52^{\circ} C$.
Elevation in boiling point $\Delta T_b = T_b - T_b^{\circ} = 100.52^{\circ} C - 100^{\circ} C = 0.52 \ K$.
Using the formula for molar mass of solute: $M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$.
Substituting the values: $M_2 = \frac{1000 \times 0.52 \times 6}{0.52 \times 100}$.
$M_2 = \frac{1000 \times 6}{100} = 60 \ g \ mol^{-1}$.

(D) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$.
Given that the solution is decimolal,the molality $m = 0.1 \ mol \ kg^{-1}$.
Given $K_b = 0.52 \ ^{\circ}C \ kg \ mol^{-1}$.
Calculating the elevation in boiling point: $\Delta T_b = 0.52 \times 0.1 = 0.052 \ ^{\circ}C$.
The boiling point of pure water $T_b^{\circ}$ is $100 \ ^{\circ}C$.
The boiling point of the solution $T_b$ is given by $T_b = T_b^{\circ} + \Delta T_b$.
$T_b = 100 \ ^{\circ}C + 0.052 \ ^{\circ}C = 100.052 \ ^{\circ}C$.

(B) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m$.
For the glucose solution: $T_{b} - T_{b}^{\circ} = K_{b} \cdot m$.
$100.16^{\circ} C - 100^{\circ} C = K_{b} \times 0.1 \ m$.
$0.16 = K_{b} \times 0.1$,so $K_{b} = 1.6 \ ^{\circ} C \ kg \ mol^{-1}$.
Now,for the sucrose solution: $\Delta T_{b} = K_{b} \times m = 1.6 \times 0.5 = 0.80^{\circ} C$.
The boiling point of the sucrose solution is $T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100^{\circ} C + 0.80^{\circ} C = 100.80^{\circ} C$.

(A) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is equal to the mole fraction of the solute:
$\frac{P^{\circ} - P_s}{P^{\circ}} = X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$
Given: $n_{solute} = 2 \ mol$,$n_{solvent} = 20 \ mol$,$P^{\circ} = 32 \ mm \ Hg$
Substituting the values:
$\frac{32 - P_s}{32} = \frac{2}{2 + 20} = \frac{2}{22} = \frac{1}{11}$
$32 - P_s = 32 \times \frac{1}{11} \approx 2.91 \ mm \ Hg$
$P_s = 32 - 2.91 = 29.09 \ mm \ Hg \approx 29.1 \ mm \ Hg$

(A) The formula for freezing point depression is $\Delta T_{f} = K_{f} \cdot m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{6 / M}{100 / 1000} = \frac{6}{M} \times 10 = \frac{60}{M} \ mol \ kg^{-1}$.
Given $\Delta T_{f} = 0.93 \ K$ and $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.93 = 1.86 \times \frac{60}{M}$.
$M = \frac{1.86 \times 60}{0.93} = 2 \times 60 = 120 \ g \ mol^{-1}$.

(C) The molal depression constant $(K_f)$ is defined as the depression in freezing point for a $1 \ molal$ solution.
Since $K_f = \frac{\Delta T_f}{m}$,the unit involves a temperature difference.
$A$ temperature difference of $1^{\circ} C$ is equivalent to a temperature difference of $1 \ K$.
Therefore,the numerical value of the molal depression constant remains the same when expressed in $K \ kg \ mol^{-1}$.
Thus,$2.77^{\circ} C \ kg \ mol^{-1} = 2.77 \ K \ kg \ mol^{-1}$.

(B) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^{\circ} - P_{S}}{P^{\circ}} = X_{\text{solute}}$
Given:
Lowering in vapour pressure,$P^{\circ} - P_{S} = 2.5 \ mm \ Hg$
Vapour pressure of pure solvent,$P^{\circ} = 250 \ mm \ Hg$
Substituting the values:
$X_{\text{solute}} = \frac{2.5}{250}$
$X_{\text{solute}} = 0.01$