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JEE MAIN - Chemistry (2024 - 5th April Evening Shift - No. 24)

Considering acetic acid dissociates in water, its dissociation constant is $$6.25 \times 10^{-5}$$. If $$5 \mathrm{~mL}$$ of acetic acid is dissolved in 1 litre water, the solution will freeze at $$-x \times 10^{-2}{ }^{\circ} \mathrm{C}$$, provided pure water freezes at $$0{ }^{\circ} \mathrm{C}$$.

$$x=$$ _________. (Nearest integer)

$$\begin{aligned} \text{Given :} \quad & \left(\mathrm{K}_{\mathrm{f}}\right)_{\text {water }}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}-1 \\ & \text { density of acetic acid is } 1.2 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { molar mass of water }=18 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { molar mass of acetic acid= } 60 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { density of water }=1 \mathrm{~g} \mathrm{~cm}^{-3} \end{aligned}$$

Acetic acid dissociates as $$\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{\oplus}$$

Answer
19

Explanation

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 $

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