Solubility product of AgCl is 1.5625×10^{-10} at 25,^oC . Its solubility in grams per litre will be

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6-2.Equilibrium-II (Ionic Equilibrium)

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The solubility product of $AgCl$ is $1.5625\times10^{-10}$ at $25\,^oC$. Its solubility in grams per litre will be

A

$143.5$

B

$108$

C

$1.57 \times {10^{ - 8}}$

D

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

Solution

${\text{S}} = \sqrt {{{\text{K}}_{{\text{SP}}}}} = \sqrt {1.5625 \times {{10}^{ – 10}}} $

$ = 1.25 \times {10^{ – 5}}\,{\text{M}}$

$ = 1.25 \times {10^{ – 5}} \times 143.5$

$ = 1.79 \times {10^{ – 3}}\,{\text{g}}/{\text{L}}$

Standard 11

Chemistry

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