### Solutions Question 6

#### Question 6 - 29 January - Shift 1

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.

## Show Answer

#### Answer: (13)

#### Solution:

Let a mole $Pb(NO_3)_2$ be added

$Pb(NO_3)_2 \to Pb^{2+}+2 NO_3^{-}$

a mat $2 a$

$\Delta T_b=0.15=0.5[3 a] \Rightarrow a=0.1$

$ Pb _{(aq)}^{2+}+2 Cl _{(aq)}^{-} \to PbCl_2(s) $

$t=0$ 0.1 0.2

$t=\infty \quad(0.1-x) \quad(0.2-2 x)$

In final solution

$\Delta T_f=0.8=1.8[\frac{0.3-3 x+0.2+0.2}{1}]$

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

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