Question

The solubility product of $AgCl$ is $1.8 \times 10^{-10}$, the minimum volume $($ in $L )$ of water required to dissolve $1.9 \,mg$ of $AgCl$ is approximately.( molecular weight of $AgCl =143.5$ )

• A. 10
• B. 2
• C. 1
• D. 20

Answer 1

Answer: (C)

Solubility $=\sqrt{1.8} \times 10^{-5} M $
$=1.34 \times 10^{-5} M $
$=1.34 \times 10^{-5} \times 143.5 \,gm / litre $
$=192.29 \times 10^{-5} gm / litre$
$=1.92 \,mg / litre$
so required volume $=1 L $