1. Tardigrade
  2. Question
  3. Chemistry
  4. Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15° C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at -0.80 C. The solubility product of PbCl 2 formed is × 10-6 at 298 K. (Nearest integer) Given: K b =0.5 K kg mol -1 and K f =1.8 K kg mol -1. Assume molality to be equal to molarity in all cases.

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Q. Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at $100.15^{\circ} C$. When $0.2\, mol$ of $NaCl$ is added to the resulting solution, it was observed that the solution froze at $-0.8^0 C$. The solubility product of $PbCl _2$ formed is ___ $\times 10^{-6}$ at $298 \,K$. (Nearest integer)
Given : $K _{ b }=0.5\, K \,kg\, mol ^{-1}$ and $K _{ f }=1.8 \,K \,kg\, mol ^{-1}$. Assume molality to be equal to molarity in all cases.

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Solution:

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In final solution
$\Delta T _{ f }=0.8=1.8\left[\frac{0.3-3 x +0.2+0.2}{1}\right] $
$\Rightarrow x =\frac{2.3}{27}$
$ \Rightarrow K _{ sp }=\left(0.1-\frac{2.3}{27}\right)\left(0.2-\frac{4.6}{27}\right)^2=13 \times 10^{-6}$