Solid Lead nitrate is dissolved in 1 ,litre of water. The solutio… - Vidyadip

MCQ

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 $-0.8^{\circ}\,C$. The solutbility product of $PbCl _2$ formed is $...........\times 10^{-6}$ at $298\,K$. (Nearest integer) Given : $K _{ b }=0.5\,K\,kg\,mol ^{-1}$ and $K _{ f }=1.8\,kg\,mol ^{-1}$. Assume molality to be equal to molarity in all cases.

✓
$13$
B
$12$
C
$11$
D
$10$

✓

Answer

Correct option: A.

$13$

a
Let a mole $Pb \left( NO _3\right)_2$ be added

$Pb \left( NO _3\right)_2 \rightarrow Pb ^{2+}+2 NO _3^{-}$

a $\quad\quad\quad\quad\quad\quad$ a $\quad\qquad 2 a$

$\Delta T _{ b }=0.15=0.5[3 a ] \Rightarrow a =0.1$

$Pb _{( aq )}^{2+}+2 Cl _{( aq )}^{-} \rightarrow PbCl _2( s )$

$\begin{array}{llc} t =0 & 0.1 & 0.2 \\ t =\infty & (0.1- x ) & (0.2-2 x )\end{array}$

In final solution

$\Delta T _{ f }=0.8=1.8\left[\frac{0.3-3 x +0.2+0.2}{1}\right]$

$\Rightarrow x =\frac{2.3}{27}$

$\Rightarrow K _{ sp }=\left(0.1-\frac{2.3}{27}\right)\left(0.2-\frac{4.6}{27}\right)^2=13 \times 10^{-6}$

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