Q20 / 75 · JEE Main 2025 (8 Apr Shift 2) · Chemistry

Figure

  • Correct answer
    A
    $1.38 \times 10^{-3}$
  • B
    $1.1 \times 10^{-2}$
  • C
    $1.90 \times 10^{-3}$
  • D
    $1.89 \times 10^{-1}$

Solution

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f}} \mathrm{m} \\ & 0.2=\mathrm{i} \times 1.8 \times 0.1 \\ & \mathrm{i}=\frac{20}{18}=\frac{10}{9} \\ & \text { For } \mathrm{HA}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-} \\ & \mathrm{t}=0 \quad 1 \\ & \mathrm{t}=\mathrm{t}_{\mathrm{eq}} 1-\alpha \quad \alpha \\ & \mathrm{i}=1+\alpha \\ & \frac{10}{9}=1+\alpha \\ & \alpha=\frac{1}{9} \\ & \mathrm{~K}_{\mathrm{eq}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\ & 0.1\left(\frac{1}{9}\right)^2 \\ & =\frac{1}{1-\frac{1}{9}}=\frac{1}{720} \\ & \mathrm{~K}_{\mathrm{eq}}=1.38 \times 10^{-3}\end{aligned}$

Hence, the correct answer is option: 'A'.

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