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JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)

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......\(L\)

  1. A \(1.798\)
  2. B \(0.36\)
  3. C \(17.95\)
  4. D \(0.18\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(0.18\)

Step-by-step Solution

Detailed explanation

\({({K_{sp}})_{PbC{l_2}}} = 3.2 \times {10^{ - 8}} = 32 \times {10^{ - 9}}\) \(PbC{l_2} \leftrightarrow \mathop {P{b^{2 + }}}\limits_s + \mathop {2C{l^ - }}\limits_{2s} \) \({K_{sp}} = [p{b^{2 + }}]{[C{l^ - }]^2}\) \({K_{sp}} = 4{s^3} = 32 \times {10^{ - 9}}\)…
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