JEE Chemistry 2023 Question with Solution

Given: $5 \, \text{g}$ of acetic acid is added to $500 \, \text{mL}$ of water, degree of dissociation $\alpha = 0.2$, and $K_f = 1.86 \, \text{K kg mol}^{-1}$.

Find: The depression in freezing point in the form $\times 10^{-3} \, ^\circ C$.

Molar mass of acetic acid $\left(\text{CH}_3\text{COOH}\right)$ is

$$2 \times 12 + 4 \times 1 + 2 \times 16 = 60 \, \text{g/mol}$$

Moles of acetic acid:

$$\frac{5 \, \text{g}}{60 \, \text{g/mol}} = \frac{1}{12} \, \text{mol}$$

Mass of water:

$$500 \, \text{mL} \times 1 \, \text{g/mL} = 500 \, \text{g} = 0.5 \, \text{kg}$$

Molality:

$$m = \frac{\frac{1}{12} \, \text{mol}}{0.5 \, \text{kg}} = \frac{1}{6} \, \text{mol/kg}$$

Acetic acid dissociates as

$$\text{CH}_3\text{COOH} \leftrightharpoons \text{CH}_3\text{COO}^- + \text{H}^+$$

For dissociation,

$$i = 1 + \alpha (n-1)$$

Here, $n = 2$, so

$$i = 1 + 0.2(2-1) = 1.2$$

Now use freezing point depression:

$$\Delta T_f = iK_f m$$

Substituting values,

$$\Delta T_f = 1.2 \times 1.86 \times \frac{1}{6} = 0.372 \, \text{K}$$

Since temperature difference in Kelvin and Celsius is the same,

$$\Delta T_f = 0.372 ^\circ \text{C} = 372 \times 10^{-3} \, ^\circ \text{C}$$

Therefore, the depression in freezing point is $372 \times 10^{-3} \, ^\circ \text{C}$, so the answer is 372.

Given: Weak electrolyte acetic acid with $20\%$ dissociation.

Find: The numerical value asked in $\times 10^{-3} \, ^\circ C$.

Use

$$i = 1 + (n-1)\alpha$$

For acetic acid, $n=2$ and $\alpha=0.2$, hence

$$i = 1.2$$

Also,

$$\Delta T_f = iK_f m$$

With molality written directly from the given mass and volume of water,

$$m = \frac{5/60}{0.5} = \frac{1}{6}$$

Thus,

$$\Delta T_f = 1.2 \times 1.86 \times \frac{1}{6} = 0.372 ^\circ \text{C}$$

So,

$$0.372 ^\circ \text{C} = 372 \times 10^{-3} \, ^\circ \text{C}$$

Therefore, the required numerical value is 372.

The second provided approach shows $3.72$, but that contradicts the detailed working and is a calculation error. The correct value is $0.372 ^\circ \text{C}$.