For a saturated solution of AgCl at $25^{\circ} C, k =3.4 \times 10^{-6} ohm ^{-1} cm ^{-1}$ and that of $H _{2} O (\ell)$ used is $2.02 \times 10^{-6} ohm ^{-1} cm ^{-1}$.
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For a saturated solution of $AgCl$ at $25^{\circ} C , k =3.4 \times 10^{-6} ohm ^{-1} cm ^{-1}$ and that of $H _{2} O (\ell)$ used is $2.02 \times 10^{-6} ohm ^{-1} cm ^{-1} \cdot \lambda_{m}^{0}$ for $AgCl$ is $138 ohm ^{-1} cm ^{2} mol ^{-1}$, then the solubility of $AgCl$ in moles per liter will be:

(A) $10^{-5}$

(B) $10^{-10}$

(C) $10^{-14}$

(D) $10^{-16}$

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The solubility of AgCl in moles per liter will be 10–5. So, The correct option of this question is (A).

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