Q27 / 120 · VITEEE 2016 · Chemistry
At $298 K$, the conductivity of a saturated solution of AgCl in water is $2.6 \times 10^{-6} \mathrm{Scm}^{-1}$. Its solubility product at $298 K$.
Given : $\lambda^{\infty}\left(\mathrm{Ag}^{+}\right)=63.0 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$,
$\lambda^{\infty}\left(\mathrm{Cl}^{-}\right)=67.0 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
- A$2.0 \times 10^{-5} M^{2}$
- Correct answerB$4.0 \times 10^{-10} M^{2}$
- C$4.0 \times 10^{-16} M^{2}$
- D$2 \times 10^{-8} M^{2}$
Solution
Solubility $S=\frac{1000 k}{\lambda_{\mathrm{AgCl}}^{\circ}}=\frac{1000 \times 2.6 \times 10^{-6}}{\lambda_{\mathrm{Ag}^{+}}^{\circ}+\lambda_{\mathrm{Cl}^{-}}^{\circ}}$
$$
=\frac{2.6 \times 10^{-3}}{63+67}=2 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} ; K_{\mathrm{sp}}=S^{2}
$$
Hence, the correct answer is option: 'B'.