The $\mathrm{K}_{\mathrm{sp}}$ of silver carbonate is $5.0 \times 10^{-13} \mathrm{M}^{3}$. Hence solubility…
- $13.80 \mathrm{ppm}$
- $21.9 \mathrm{ppm}$
- $6.40 \mathrm{ppm}$
- $10.95 \mathrm{ppm}$
Solution

$\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^{3}$
$\mathrm{~s}=\sqrt[3]{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=\sqrt[3]{\frac{5 \times 10^{-13}}{4}}=5 \times 10^{-5} \mathrm{M}$
$=10^{-5} \times 276=1.38 \times 10^{-2} \mathrm{~g} / \mathrm{L} 10^{3} \mathrm{~g}$ water (or solution) contains $1.38 \times 10^{-2} \mathrm{~g} \mathrm{Ag}_{2} \mathrm{CO}_{3}$
$\therefore 10^{6} \mathrm{~g}$ water (or solution) contains $13.8 \mathrm{~g}$ $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ Solubility $=13.8 \mathrm{ppm}$ ,
Asked in: JEE-TOPICTESTS-CHEMISTRY