QJEEM24Chd51d: JEE Main 2024 Chemistry - Class 12 JEEM24Chd51d Solution: Considering acetic acid dissociates in water, its dissociation…

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:

$x = 19$