MHT CET 2025 Chemistry Question Paper with Answer and Solution

(A) solution of $H_2$ gas in palladium is an example of a solid solution where a gas is the solute and a solid is the solvent.
In this system,the $H_2$ molecules are adsorbed onto the surface of the palladium metal and then diffuse into the metal lattice.
Therefore,the correct classification is gas as solute and solid as solvent.

(C) An ideal solution is one that obeys Raoult's law over the entire range of concentration.
Mixtures of liquids that have similar molecular structures and polarities form ideal solutions.
Benzene $(C_6H_6)$ and toluene $(C_6H_5CH_3)$ have similar structures and intermolecular forces,thus they form an ideal solution and obey Raoult's law.
Chloroform + acetone,carbon disulfide + acetone,and ethanol + acetone are examples of non-ideal solutions showing deviations from Raoult's law.

(D) According to Raoult's Law for a solution of two volatile liquids,$P_{total} = P_A^o x_A + P_B^o x_B$.
Given: $x_B = 0.4$,so $x_A = 1 - 0.4 = 0.6$.
Given: $P_A^o = 400 \ mm \ Hg$ and $P_{total} = 600 \ mm \ Hg$.
Substituting the values: $600 = (400 \times 0.6) + (P_B^o \times 0.4)$.
$600 = 240 + 0.4 P_B^o$.
$0.4 P_B^o = 600 - 240 = 360$.
$P_B^o = \frac{360}{0.4} = 900 \ mm \ Hg$.
Therefore,the correct option is $D$.

(A) mixture exhibits positive deviation from Raoult's law when the intermolecular forces between the components are weaker than the forces in the pure components.
In the mixture of $Ethanol$ and $Acetone$,the hydrogen bonding between $Ethanol$ molecules is disrupted by the addition of $Acetone$,leading to weaker interactions.
Therefore,the total vapor pressure of the solution is higher than that predicted by Raoult's law.
$Benzene$ and $Toluene$ form an ideal solution.
$Chloroform$ and $Acetone$ exhibit negative deviation due to strong hydrogen bonding between them.
$Phenol$ and $Aniline$ also exhibit negative deviation.

(D) An ideal solution is one that obeys Raoult's law over the entire range of concentration.
Mixtures of liquids that are structurally similar and have similar intermolecular forces form ideal solutions.
$Benzene$ and $toluene$ have similar structures and similar intermolecular forces,so they form an ideal solution and obey Raoult's law.
$Phenol$ and $aniline$ show negative deviation due to hydrogen bonding.
$Chloroform$ and $acetone$ show negative deviation due to hydrogen bonding.
$Ethanol$ and $acetone$ show positive deviation due to the disruption of hydrogen bonding.

(D) According to Raoult's Law for a binary solution of two volatile liquids $A$ and $B$:
$P_{total} = P_A + P_B = P_A^0 \chi_A + P_B^0 \chi_B$
Given:
$P_B^0 = 900 \ mm \ Hg$
$\chi_B = 0.4$
$P_{total} = 600 \ mm \ Hg$
Since $\chi_A + \chi_B = 1$,we have $\chi_A = 1 - 0.4 = 0.6$.
Substituting the values into the equation:
$600 = P_A^0(0.6) + (900)(0.4)$
$600 = P_A^0(0.6) + 360$
$P_A^0(0.6) = 600 - 360 = 240$
$P_A^0 = \frac{240}{0.6} = 400 \ mm \ Hg$
Thus,the vapour pressure of pure liquid $A$ is $400 \ mm \ Hg$.

(B) According to Raoult's Law,the total vapour pressure of a solution is given by $P_{total} = P_A^0 x_A + P_B^0 x_B$.
Given: $P_{total} = 500 \ mmHg$,$P_A^0 = 400 \ mmHg$,$P_B^0 = 575 \ mmHg$.
Since $x_A + x_B = 1$,we can write $x_A = 1 - x_B$.
Substituting these values into the equation: $500 = 400(1 - x_B) + 575x_B$.
$500 = 400 - 400x_B + 575x_B$.
$500 - 400 = 175x_B$.
$100 = 175x_B$.
$x_B = \frac{100}{175} = \frac{4}{7} \approx 0.57$.

(A) An ideal solution is one that obeys Raoult's law over the entire range of concentration and shows no change in enthalpy $(\Delta H_{mix} = 0)$ or volume $(\Delta V_{mix} = 0)$ upon mixing.
Benzene and toluene have similar molecular structures and polarities,resulting in similar intermolecular forces between the components.
Therefore,the mixture of benzene and toluene behaves nearly as an ideal solution.
Other options like phenol + aniline,chloroform + acetone,and ethanol + acetone show significant deviations from Raoult's law due to strong hydrogen bonding or dipole-dipole interactions.

(A) According to Henry's Law,the concentration of a dissolved gas $(C)$ is directly proportional to its partial pressure $(P)$ above the liquid surface.
The formula is given by: $C = K_H \times P$
Given:
$K_H = 0.15 \ mol \ dm^{-3} \ atm^{-1}$
$P = 0.15 \ atm$
Substituting the values:
$C = 0.15 \ mol \ dm^{-3} \ atm^{-1} \times 0.15 \ atm$
$C = 0.0225 \ mol \ dm^{-3}$
Since $1 \ mol \ dm^{-3} = 1 \ M$,the concentration is $0.0225 \ M$.

(C) According to Henry's law,the solubility $(S)$ of a gas in a liquid is given by the formula: $S = K_H \times P$
Where:
$K_H = 6.8 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}$
$P = 0.8 \ atm$
Substituting the values:
$S = (6.8 \times 10^{-4} \ mol \ dm^{-3} \ atm^{-1}) \times (0.8 \ atm)$
$S = 5.44 \times 10^{-4} \ mol \ dm^{-3}$
Since $1 \ mol \ dm^{-3} = 1 \ M$,the solubility is $5.44 \times 10^{-4} \ M$.

(C) The relative lowering of vapour pressure is given by the formula: $\frac{P^o - P}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (for dilute solutions).
Here,$n_2 = \frac{w_2}{M_2}$ and $n_1 = \frac{w_1}{M_1}$.
Given: $w_2 = 20 \ g$,$w_1 = 200 \ g$,$M_1 = 18 \ g \ mol^{-1}$ (for water),and $\frac{P^o - P}{P^o} = 0.02$.
Substituting the values: $0.02 = \frac{20 / M_2}{200 / 18}$.
$0.02 = \frac{20}{M_2} \times \frac{18}{200}$.
$0.02 = \frac{18}{10 M_2} = \frac{1.8}{M_2}$.
$M_2 = \frac{1.8}{0.02} = 90 \ g \ mol^{-1}$.
Thus,the correct option is $C$.

(B) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m$.
Given: $\Delta T_f = 0.046 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$m = 0.02 \ m$.
First,calculate the van't Hoff factor $(i)$:
$i = \frac{\Delta T_f}{K_f \times m} = \frac{0.046}{1.86 \times 0.02} = \frac{0.046}{0.0372} \approx 1.2366$.
For dissociation,the degree of dissociation $(\alpha)$ is given by $\alpha = \frac{i - 1}{n - 1}$.
Given $n = 2$,$\alpha = \frac{1.2366 - 1}{2 - 1} = 0.2366$.
Percent dissociation = $\alpha \times 100 = 0.2366 \times 100 = 23.66 \% \approx 23.6 \%$.
Therefore,the correct option is $B$.

(B) The osmotic pressure $(\pi)$ of a solution is given by the formula: $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
Given: $\pi = 12 \ atm$,$T = 300 \ K$,$R = 0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$.
Rearranging the formula for concentration: $C = \frac{\pi}{RT}$.
Substituting the values: $C = \frac{12}{0.0821 \times 300}$.
$C = \frac{12}{24.63} \approx 0.487 \ M$.
Therefore,the correct option is $B$.

(C) The elevation in boiling point is given by the formula: $\Delta T_b = T_b - T_b^\circ$
Given,$T_b = 319.8 \ K$ and $T_b^\circ = 319.5 \ K$.
So,$\Delta T_b = 319.8 \ K - 319.5 \ K = 0.3 \ K$.
The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is the molality.
Given $m = 0.12 \ m$.
Therefore,$K_b = \frac{\Delta T_b}{m} = \frac{0.3 \ K}{0.12 \ mol \ kg^{-1}} = 2.5 \ K \ kg \ mol^{-1}$.
Thus,the correct option is $C$.

(A) The formula for osmotic pressure $(\pi)$ is given by $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
First,calculate the molar concentration $(C)$:
$C = \frac{n}{V} = \frac{0.03 \ mol}{0.1 \ dm^3} = 0.3 \ mol \ dm^{-3}$.
Now,substitute the values into the formula:
$\pi = 0.3 \ mol \ dm^{-3} \times 0.0821 \ dm^3 \ atm \ mol^{-1} \ K^{-1} \times 300 \ K$.
$\pi = 0.3 \times 0.0821 \times 300 \ atm$.
$\pi = 7.389 \ atm \approx 7.4 \ atm$.

(C) The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$,where $i$ is the van't Hoff factor and $m$ is the molality.
For $0.1 \ m \ NaCl$,$i = 2$,so $\Delta T_f = 2 \times 0.1 = 0.2$.
For $0.05 \ m \ MgSO_4$,$i = 2$,so $\Delta T_f = 2 \times 0.05 = 0.1$.
For $1 \ m \ AlPO_4$,$i = 2$,so $\Delta T_f = 2 \times 1 = 2.0$.
For $0.05 \ m \ Al_2(SO_4)_3$,$i = 5$,so $\Delta T_f = 5 \times 0.05 = 0.25$.
Comparing the values,$1 \ m \ AlPO_4$ has the highest value of $i \times m$,therefore it exhibits the highest freezing point depression.

(A) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$.
Given: $\Delta T_f = T_f^{\circ} - T_f = 0^{\circ}C - (-0.95^{\circ}C) = 0.95 \ K$.
$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.95 = 1.86 \times m$.
$m = \frac{0.95}{1.86} \approx 0.51 \ mol \ kg^{-1}$.

(D) The elevation in boiling point is given by the formula $\Delta T_{b} = i \times K_{b} \times m$.
For a nonelectrolyte,the van't Hoff factor $i = 1$. Thus,for the nonelectrolyte solution,$\Delta T_{b} = 1 \times K_{b} \times m = x \ K$.
For $AlCl_3$,the dissociation is $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$.
The van't Hoff factor $i$ for $AlCl_3$ is $1 + 3 = 4$.
Therefore,for $1 \ m \ AlCl_3$ solution,$\Delta T_{b} = 4 \times K_{b} \times m = 4 \times (K_{b} \times m) = 4x \ K$.

(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$,where $\Delta T_{b}$ is the elevation in boiling point,$K_{b}$ is the ebullioscopic constant,and $m$ is the molality of the solution.
Given: $\Delta T_{b} = 0.2 \ K$ and $K_{b} = 0.52 \ K \ kg \ mol^{-1}$.
Rearranging the formula to solve for molality $(m)$: $m = \frac{\Delta T_{b}}{K_{b}}$.
Substituting the values: $m = \frac{0.2}{0.52} \approx 0.3846 \ mol \ kg^{-1}$.
Rounding to three decimal places,we get $m \approx 0.385 \ mol \ kg^{-1}$.
Therefore,the correct option is $C$.

(D) The formula for osmotic pressure $(\pi)$ is given by $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
Given values: $\pi = 1.5 \ atm$,$C = 0.05 \ M$,$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
Rearranging the formula to solve for $T$: $T = \frac{\pi}{CR}$.
Substituting the values: $T = \frac{1.5}{0.05 \times 0.0821} = \frac{1.5}{0.004105} \approx 365.4 \ K$.
Therefore,the correct option is $D$.

(B) The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$, where $i$ is the van't Hoff factor and $m$ is the molality. Assuming complete dissociation, $i$ equals the number of ions produced per formula unit.
For $A$: $0.1 \, m \, NaClO_4 \rightarrow i = 2$, so $\Delta T_f \propto 0.1 \times 2 = 0.2$.
For $B$: $0.05 \, m \, MgSO_4 \rightarrow i = 2$, so $\Delta T_f \propto 0.05 \times 2 = 0.1$.
For $C$: $0.08 \, m \, AlPO_4 \rightarrow i = 2$, so $\Delta T_f \propto 0.08 \times 2 = 0.16$.
For $D$: $0.06 \, m \, Al_2(SO_4)_3 \rightarrow i = 5$, so $\Delta T_f \propto 0.06 \times 5 = 0.3$.
Comparing the values, $0.1$ is the lowest, which corresponds to option $B$.

(B) The relative lowering of vapour pressure is given by Raoult's Law as: $\frac{P^o - P_s}{P^o} = \chi_{solute} = \frac{n_2}{n_1 + n_2}$.
Given: Mass of solute $(w_2)$ $= 0.56 \ g$,Molar mass of solute $(M_2)$ $= 60 \ g \ mol^{-1}$,Mass of solvent $(w_1)$ $= 100 \ g$,Molar mass of water $(M_1)$ $= 18 \ g \ mol^{-1}$.
Moles of solute $(n_2)$ $= \frac{0.56}{60} \approx 0.00933 \ mol$.
Moles of solvent $(n_1)$ $= \frac{100}{18} \approx 5.556 \ mol$.
Since $n_2$ is very small compared to $n_1$,we can approximate $\frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$.
Relative lowering of vapour pressure $= \frac{0.00933}{5.556} \approx 0.00168 \approx 0.0017$.

(A) Osmotic pressure $(\pi)$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Since $R$ and $T$ are constant,$\pi \propto i \times C$.
For complete dissociation:
$A. 0.2 \ m \ KCl: i = 2, C = 0.2, \pi \propto 2 \times 0.2 = 0.4$
$B. 0.3 \ m \ MgSO_4: i = 2, C = 0.3, \pi \propto 2 \times 0.3 = 0.6$
$C. 0.1 \ m \ BaCl_2: i = 3, C = 0.1, \pi \propto 3 \times 0.1 = 0.3$
$D. 0.5 \ m \ Al_2(SO_4)_3: i = 5, C = 0.5, \pi \propto 5 \times 0.5 = 2.5$
Comparing the values: $2.5 (D) > 0.6 (B) > 0.4 (A) > 0.3 (C)$.
Thus,the decreasing order is $D > B > A > C$.

(A) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality of the solution.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
Moles of urea = $\frac{\text{given mass}}{\text{molar mass}} = \frac{15 \ g}{60 \ g \ mol^{-1}} = 0.25 \ mol$.
Mass of solvent (water) = $1000 \ g = 1 \ kg$.
Therefore,$m = \frac{0.25 \ mol}{1 \ kg} = 0.25 \ mol \ kg^{-1}$.
Now,$\Delta T_{b} = 0.52 \ K \ kg \ mol^{-1} \times 0.25 \ mol \ kg^{-1} = 0.13 \ K$.

(A) The elevation in boiling point is given by the formula: $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{n_{solute}}{W_{solvent} \text{ (in kg)}}$.
Substituting the values: $0.3 \ K = 1.8 \ K \ kg \ mol^{-1} \times \frac{n_{solute}}{0.3 \ kg}$.
$n_{solute} = \frac{0.3 \times 0.3}{1.8} \ mol$.
$n_{solute} = \frac{0.09}{1.8} \ mol = 0.05 \ mol$.
Rounding to the nearest option,the correct value is $0.051 \ mol$.

(B) The formula for osmotic pressure $(\pi)$ is given by $\pi = iCRT$.
Since the solute is non-volatile and non-electrolyte,the van't Hoff factor $(i)$ is $1$.
Given:
Concentration $(C)$ = $0.5 \ M$
Temperature $(T)$ = $300 \ K$
Gas constant $(R)$ = $0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$
Substituting the values:
$\pi = 1 \times 0.5 \ mol \ dm^{-3} \times 0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1} \times 300 \ K$
$\pi = 0.5 \times 0.0821 \times 300 \ atm$
$\pi = 12.315 \ atm \approx 12.32 \ atm$.
Therefore,the correct option is $B$.

(C) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.
Given: $\Delta T_{b} = 0.6 \ K$,$K_{b} = 2 \ K \ kg \ mol^{-1}$,and moles of solute $= 0.01 \ mol$.
Substituting the values: $0.6 = 2 \times \frac{0.01}{W_{solvent}}$.
$W_{solvent} = \frac{2 \times 0.01}{0.6} = \frac{0.02}{0.6} = \frac{1}{30} \ kg$.
$W_{solvent} \approx 0.0333 \ kg$.
Thus,the correct option is $C$.

(D) The boiling point elevation is given by the formula $\Delta T_b = i \times K_b \times m$,where $i$ is the van't Hoff factor and $m$ is the molality. Since $K_b$ is constant,$\Delta T_b$ is proportional to $i \times m$.
For complete dissociation,$i$ equals the number of ions produced per formula unit.
$A$: $KCl \rightarrow K^+ + Cl^-$,$i = 2$. $\Delta T_b \propto 2 \times 0.2 = 0.4$.
$B$: $NaCl \rightarrow Na^+ + Cl^-$,$i = 2$. $\Delta T_b \propto 2 \times 0.1 = 0.2$.
$C$: $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$,$i = 4$. $\Delta T_b \propto 4 \times 1 = 4.0$.
$D$: $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$,$i = 3$. $\Delta T_b \propto 3 \times 0.05 = 0.15$.
Comparing the values,$0.15 < 0.2 < 0.4 < 4.0$. Thus,$0.05 \ m \ MgCl_2$ exhibits the minimum boiling point elevation.

(A) The relative lowering of vapour pressure is given by the mole fraction of the solute,which is $\frac{n_2}{n_1 + n_2}$.
Given:
Mass of urea $(w_2)$ $= 3 \ g$
Molar mass of urea $(M_2)$ $= 60 \ g \ mol^{-1}$
Moles of urea $(n_2)$ $= \frac{3}{60} = 0.05 \ mol$
Mass of water $(w_1)$ $= 50 \ g$
Molar mass of water $(M_1)$ $= 18 \ g \ mol^{-1}$
Moles of water $(n_1)$ $= \frac{50}{18} \approx 2.778 \ mol$
Relative lowering of vapour pressure $= \frac{n_2}{n_1 + n_2} = \frac{0.05}{2.778 + 0.05} = \frac{0.05}{2.828} \approx 0.01768 \approx 0.018$.

(C) The formula for boiling point elevation is given by $\Delta T_b = K_b \times m$,where $\Delta T_b$ is the boiling point elevation,$K_b$ is the molal boiling point elevation constant,and $m$ is the molality of the solution.
Given: $\Delta T_b = 1.75 \ K$ and $K_b = 3 \ K \ kg \ mol^{-1}$.
Substituting the values into the formula: $1.75 = 3 \times m$.
Solving for $m$: $m = \frac{1.75}{3} \approx 0.5833 \ m$.
Rounding to two decimal places,we get $0.58 \ m$.

(C) The formula for freezing point depression is $\Delta T_{f} = i \times K_{f} \times m$.
Given: $\Delta T_{f} = 0 - (-0.54) = 0.54 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.18 \ m$.
Substituting the values: $0.54 = i \times 1.86 \times 0.18$.
Calculating $i$: $i = \frac{0.54}{1.86 \times 0.18} = \frac{0.54}{0.3348} \approx 1.612$.

(C) The depression in freezing point is given by the formula $\Delta T_{f} = i \times K_{f} \times m$.
For a $1 \ m$ urea solution,urea is a non-electrolyte,so the van't Hoff factor $i = 1$. Thus,$\Delta T_{f} = 1 \times K_{f} \times 1 = K_{f} = x \ K$.
For a $1 \ m$ $CaCl_{2}$ solution,$CaCl_{2}$ dissociates as $CaCl_{2} \rightarrow Ca^{2+} + 2Cl^{-}$,so the van't Hoff factor $i = 3$.
Thus,$\Delta T_{f} = 3 \times K_{f} \times 1 = 3 \times x \ K = 3x \ K$.

(B) The formula for osmotic pressure $(\pi)$ is given by: $\pi = i \times C \times R \times T$
Where:
$i$ (van't Hoff factor) = $1.6$
$C$ (molarity) = $0.2 \ M$
$R$ (gas constant) = $0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$
$T$ (temperature) = $300 \ K$
Substituting the values:
$\pi = 1.6 \times 0.2 \times 0.0821 \times 300$
$\pi = 0.32 \times 24.63$
$\pi = 7.8816 \ atm$
Therefore,the osmotic pressure is approximately $7.88 \ atm$.

(B) . The depression in freezing point $(\Delta T_f)$ is given by the formula: $\Delta T_f = K_f \times m$,where $K_f$ is the cryoscopic constant and $m$ is the molality of the solution.
Given: $\Delta T_f = 1.8 \ K$ and $m = 0.4 \ m$.
Rearranging the formula to solve for $K_f$: $K_f = \frac{\Delta T_f}{m}$.
Substituting the values: $K_f = \frac{1.8 \ K}{0.4 \ mol \ kg^{-1}} = 4.5 \ K \ kg \ mol^{-1}$.
Therefore,the correct option is $B$.

(C) The formula for boiling point elevation is given by: $\Delta T_{b} = K_{b} \times m$.
Here,$\Delta T_{b} = 0.39 \ K$ and $K_{b} = 0.52 \ K \ kg \ mol^{-1}$.
Rearranging the formula to solve for molality $(m)$: $m = \frac{\Delta T_{b}}{K_{b}}$.
Substituting the values: $m = \frac{0.39}{0.52} = 0.75 \ mol \ kg^{-1}$.
Therefore,the correct option is $C$.

(B) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{W_{solute} \times 1000}{M_{solute} \times W_{solvent(g)}}$.
Given: $\Delta T_{f} = 0.2 \ K$,$W_{solute} = 5 \ g$,$W_{solvent} = 50 \ g$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.2 = 1.86 \times \frac{5 \times 1000}{M_{solute} \times 50}$.
$0.2 = 1.86 \times \frac{100}{M_{solute}}$.
$M_{solute} = \frac{1.86 \times 100}{0.2} = \frac{186}{0.2} = 930 \ g \ mol^{-1}$.

(A) Osmotic pressure $(\pi)$ is given by the formula $\pi = CRT = \frac{n}{V}RT = \frac{w}{M \times V}RT$.
For the same temperature and volume,$\pi$ is proportional to $\frac{w}{M}$.
For urea $(M = 60 \ g \ mol^{-1})$:
$A: \frac{3}{60} = 0.05 \ mol \ L^{-1}$
$B: \frac{6}{60} = 0.1 \ mol \ L^{-1}$
For sucrose $(M = 342 \ g \ mol^{-1})$:
$17.1 \ g \ L^{-1}: \frac{17.1}{342} = 0.05 \ mol \ L^{-1}$
$34.2 \ g \ L^{-1}: \frac{34.2}{342} = 0.1 \ mol \ L^{-1}$
Comparing the values,$3 \ g \ L^{-1}$ urea $(0.05 \ mol \ L^{-1})$ and $17.1 \ g \ L^{-1}$ sucrose $(0.05 \ mol \ L^{-1})$ have the same molar concentration,hence the same osmotic pressure.
Thus,the correct option is $A$.

(C) Osmotic pressure $(\pi)$ is a colligative property given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Since the solutions are equimolar ($C$ is constant) and at the same temperature,$\pi$ is directly proportional to the van't Hoff factor $(i)$.
For complete ionization,$i$ equals the number of ions produced per formula unit:
$A$. $KCl \rightarrow K^+ + Cl^-$ $(i = 2)$
$B$. $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$ $(i = 3)$
$C$. $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$ $(i = 4)$
$D$. $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ $(i = 5)$
Comparing the $i$ values: $2 < 3 < 4 < 5$.
Therefore,the increasing order of osmotic pressure is $KCl < BaCl_2 < AlCl_3 < Al_2(SO_4)_3$.

(B) The formula for elevation in boiling point is given by: $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute $(n_{solute})$ divided by the mass of the solvent in kilograms $(W_{solvent(kg)})$.
So,$\Delta T_{b} = K_{b} \times \frac{n_{solute}}{W_{solvent(kg)}}$.
Given: $\Delta T_{b} = 0.8 \ K$,$K_{b} = 2 \ K \ kg \ mol^{-1}$,and $W_{solvent(kg)} = 0.5 \ kg$.
Substituting the values: $0.8 = 2 \times \frac{n_{solute}}{0.5}$.
$0.8 = 4 \times n_{solute}$.
$n_{solute} = \frac{0.8}{4} = 0.2 \ mol$.

(C) The relative lowering of vapour pressure is given by the formula: $\frac{P^\circ - P_s}{P^\circ} = 0.018$.
Given: $P^\circ = 18 \ mm \ Hg$.
Substituting the values: $\frac{18 - P_s}{18} = 0.018$.
$18 - P_s = 0.018 \times 18$.
$18 - P_s = 0.324$.
$P_s = 18 - 0.324 = 17.676 \ mm \ Hg$.
Rounding to two decimal places,we get $P_s \approx 17.68 \ mm \ Hg$.

(B) The depression in freezing point is given by the formula: $\Delta T_{f} = K_{f} \times m$.
Here,$\Delta T_{f} = T_{f}^{\circ} - T_{f} = 0 \ ^{\circ}C - (-0.36 \ ^{\circ}C) = 0.36 \ K$.
Given $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.36 = 1.86 \times m$.
Therefore,$m = \frac{0.36}{1.86} \approx 0.1935 \ mol \ kg^{-1}$.
Rounding to three decimal places,the molality is $0.193 \ mol \ kg^{-1}$.

(C) The formula for osmotic pressure $(\Pi)$ is given by $\Pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Since $R$ and $T$ are constant,$\Pi \propto C$.
For $0.5 \ M$ urea solution,$\Pi_1 = 0.5RT = x$.
For $1 \ M$ urea solution,$\Pi_2 = 1RT$.
Dividing the two equations: $\frac{\Pi_2}{x} = \frac{1RT}{0.5RT} = 2$.
Therefore,$\Pi_2 = 2x$.

(C) The elevation in boiling point is given by the formula $\Delta T_b = i \times K_b \times m$. Since the molality $(m)$ and the ebullioscopic constant $(K_b)$ are the same for all solutions,the elevation in boiling point depends directly on the van't Hoff factor $(i)$.
For complete dissociation:
$KCl \rightarrow K^+ + Cl^-$ $(i = 2)$
$NaCl \rightarrow Na^+ + Cl^-$ $(i = 2)$
$AlCl_3 \rightarrow Al^{3+} + 3Cl^-$ $(i = 4)$
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$ $(i = 3)$
Since $AlCl_3$ has the highest van't Hoff factor $(i = 4)$,it will exhibit the maximum boiling point elevation.

(C) The formula for depression in freezing point is given by: $\Delta T_f = K_f \times m$
Where:
$\Delta T_f$ is the depression in freezing point $(0.2 \ K)$
$K_f$ is the cryoscopic constant
$m$ is the molality of the solution $(0.18 \ m)$
Rearranging the formula to solve for $K_f$:
$K_f = \frac{\Delta T_f}{m} = \frac{0.2 \ K}{0.18 \ mol \ kg^{-1}} \approx 1.11 \ K \ kg \ mol^{-1}$
Therefore,the correct option is $C$.

(C) The formula for the depression in freezing point is $\Delta T_f = K_f \times m$.
Given:
Molality $(m)$ = $0.15 \ m$
Cryoscopic constant $(K_f)$ = $1.5 \ K \ kg \ mol^{-1}$
Calculation:
$\Delta T_f = 1.5 \ K \ kg \ mol^{-1} \times 0.15 \ mol \ kg^{-1} = 0.225 \ K$.
Therefore,the correct option is $C$.

(C) The formula for osmotic pressure $(\pi)$ is given by: $\pi = i \times C \times R \times T$
Where:
$i = 1.125$ (van't Hoff factor)
$C = 0.1 \ M$ (molar concentration)
$R = 0.0821 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$ (gas constant)
$T = 300 \ K$ (temperature)
Substituting the values:
$\pi = 1.125 \times 0.1 \times 0.0821 \times 300$
$\pi = 1.125 \times 0.1 \times 24.63$
$\pi = 1.125 \times 2.463$
$\pi = 2.770875 \ atm$
Rounding to two decimal places,we get $\pi \approx 2.77 \ atm$.

(A) Sorption is a phenomenon where both adsorption and absorption occur simultaneously.
$1$. When charcoal is added to a methylene blue solution,the dye molecules are adsorbed on the surface of the charcoal,while the solvent molecules are absorbed into the bulk of the charcoal. This is a classic example of sorption.
$2$. Dipping chalk in ink involves both adsorption on the surface and absorption into the porous structure of the chalk,which is also an example of sorption.
$3$. Hydrogen gas on platinum and oxygen on nickel are primarily examples of adsorption.
Since both $A$ and $B$ represent sorption,in the context of standard chemistry textbooks,the addition of charcoal to methylene blue is the most frequently cited example.

(A) Silver nanoparticles $(AgNPs)$ are well-known for their potent antimicrobial properties. They effectively disrupt the cell membrane and metabolic processes of bacteria like $E. coli$,making them highly efficient for water purification and disinfection.

(B) The extent of adsorption of a gas on a solid adsorbent depends on the ease of liquefaction of the gas.
Easily liquefiable gases have higher critical temperatures $(T_c)$ and stronger van der Waals forces of attraction,leading to greater adsorption.
The critical temperatures $(T_c)$ of the given gases are:
$SO_2$ $(430 \ K)$ > $O_2$ $(154 \ K)$ > $N_2$ $(126 \ K)$ > $H_2$ $(33 \ K)$.
Since $SO_2$ has the highest critical temperature,it is the most easily liquefiable and thus adsorbed to the greatest extent.

(A) The graph shows the variation of the extent of adsorption $(x/m)$ with pressure $(p)$ at different temperatures.
Physical adsorption is an exothermic process.
According to Le Chatelier's principle,for an exothermic process,the extent of adsorption decreases with an increase in temperature.
Therefore,the highest amount of gas is adsorbed at the lowest temperature.
From the given graph,the lowest temperature is $195 \ K$.