Question

The solubility product of AgCl is $1.5625\times {10}^{-10}$ at ${25}^0$C. Its solubility in grams per litre will be :

• A. $143.5$
• B. $108$
• C. $1.57\times {10}^{-8}$
• D. $1.79\times {10}^{-3}$

Answer 1

Answer: (D)

$1.79\times {10}^{-3}$

$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
S S

$S^2=1.5625\times {10}^{-10}$
$S=1.25\times {10}^{-5}mol{L}^{-1}$
Solubility in $g{L}^{-1}$ = Molar mass $\times S=143.5\times 1.25\times {10}^{-5}=1.79\times {10}^{-3}g{L}^{-1}$