MCQ
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$108$
- ✓$1.57 \times {10^{ - 8}}$
- D$1.79 \times {10^{ - 3}}$
✓
Answer
Correct option: C.
$1.57 \times {10^{ - 8}}$
c
${\text{S}} = \sqrt {{{\text{K}}_{{\text{SP}}}}} = \sqrt {1.5625 \times {{10}^{ - 10}}} $
${\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}}$
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