MCQ
If the solubility product of $PbS$ is $8 \times 10^{-28}$, then the solubility of $PbS$ in pure water at $298\; K$ is $x \times 10^{-16}\; mol\; L ^{-1}$. The value of $x$ is $\dots$.
[Given $\sqrt2 = 1.41$]
- A$281$
- ✓$282$
- C$283$
- D$284$
[Given $\sqrt2 = 1.41$]
$S =\sqrt{ K _{ sp }}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$
$=2.82 \times 10^{-14}$
$=282 \times 10^{-16}$
Ans. $=282$
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