MCQ
The minimum volume of water required to dissolve $0.1\,g$ lead $(II)$ chloride to get a saturated solution ($K_{SP}$ of $PbCl_2 = 3.2 \times 10^{-8}$; atomic mass of $Pb= 207\, u$) is......$L$
- A$1.798$
- B$0.36$
- C$17.95$
- ✓$0.18$
✓
Answer
Correct option: D.
$0.18$
d
${({K_{sp}})_{PbC{l_2}}} = 3.2 \times {10^{ - 8}} = 32 \times {10^{ - 9}}$
${({K_{sp}})_{PbC{l_2}}} = 3.2 \times {10^{ - 8}} = 32 \times {10^{ - 9}}$
$PbC{l_2} \leftrightarrow \mathop {P{b^{2 + }}}\limits_s + \mathop {2C{l^ - }}\limits_{2s} $
${K_{sp}} = [p{b^{2 + }}]{[C{l^ - }]^2}$
${K_{sp}} = 4{s^3} = 32 \times {10^{ - 9}}$
${s^3} = 8 \times {10^{ - 9}}$
$s = 2 \times {10^{ - 3}}\,M$
$\frac{w}{{M.W.}} \times \frac{1}{{{V_L}}} = 2 \times {10^{ - 3}}$
$\frac{{0.1}}{{278}} \times \frac{1}{{{V_L}}} = 2 \times {10^{ - 3}}$
${V_L} = \frac{{0.1 \times 1000}}{{278 \times 2}} = 0.18\,L$
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