JEE Main 2024 — Solutions Question with Solution

To solve the problem, we'll calculate the freezing point depression of the acetic acid solution in water.

1. Calculating Moles of Acetic Acid:

• Given volume of acetic acid: $5\text{mL}$


• Density of acetic acid: $1.2\text{g/mL}$


• Molar mass of acetic acid: $60\text{g/mol}$

$\text{Mass of acetic acid}=\text{Volume}\times \text{Density}=5\text{mL}\times 1.2\text{g/mL}=6\text{g}$

$\text{Moles of acetic acid}=\frac{\text{Mass}}{\text{Molar mass}}=\frac{6\text{g}}{60\text{g/mol}}=0.1\text{mol}$

1. Calculating Molality:

• Volume of water: $1\text{L}$


• Density of water: $1{\text{g/cm}}^3\Rightarrow 1\text{kg/L}$


• Hence, mass of water = $1\text{kg}$

$\text{Molality (m)}=\frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}=\frac{0.1\text{mol}}{1\text{kg}}=0.1\text{mol/kg}$

1. Degree of Dissociation ($\alpha$):

• Dissociation constant ($K_a$) of acetic acid: $6.25\times 10^{-5}$

$\alpha =\sqrt{\frac{K_a}{C}}=\sqrt{\frac{6.25\times 10^{-5}}{0.1}}=\sqrt{6.25\times 10^{-4}}=25\times 10^{-3}=0.025$

1. Van't Hoff factor (i):

• Acetic acid partially dissociates into 2 ions (CH${}_3$COO${}^{-}$ and H${}^{+}$)

$i=1+(\text{number of ions produced}-1)\times \alpha =1+(2-1)\times 0.025=1+0.025=1.025$

1. Freezing Point Depression ($\Delta T_f$):

• Freezing point depression constant ($K_f$) for water: $1.86\text{K kg/mol}$

$\Delta T_f=i\times K_f\times \text{Molality}=1.025\times 1.86\times 0.1=0.19065\text{K}$

1. Final Freezing Point:

• Pure water freezes at $0{}^{\circ }\text{C}$

$\text{Freezing point of solution}=0{}^{\circ }\text{C}-0.19065{}^{\circ }\text{C}=-0.19065{}^{\circ }\text{C}$

Here, $-x\times 10^{-2}{}^{\circ }\text{C}$ is given. Therefore, $x=19$ (nearest integer).

So, the final answer is:

$x=19$