MCQ
$PbCl_2$ is dissolved in water to make its saturated solution. What will be the freezing point of this solution.

Given : $K_f (H_2O)$ = $2\ K\ kg\ mole^{-1}$, $K_{sp} (PbCl_2)$ = $4 × 10^{-6}$

(Assume molarity to be equal to molality) .....$^oC$

  • A
    $-0.04$
  • $-0.06$
  • C
    $-0.02$
  • D
    $-0.6$

Answer

Correct option: B.
$-0.06$
b
$PbC{l_2}(s) \leftrightarrow \mathop {P{b^{ + 2}}(aq.)}\limits_s  + \mathop {2C{l^ - }(aq)}\limits_{2s} $

$4 s^{3}=4 \times 10^{-6}$

$s=10^{-2}$

$\Delta \mathrm{T}_{\mathrm{f}}=3 \times 2 \times 10^{-2}=0.06$

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