Question

A one litre solution is prepared by dissolving some solid lead-nitrate in water. The solution was found to boil at ${100.15}^{\circ }C$. To the resulting solution $0.2moleNaCl$ was added. The resulting solution was found to freeze at $-{0.83}^{\circ }C$. Determine solubility product of $PbC{l}_2$. Given $K_b=0.5$ and $K_f=1.86$. Assume molality to be equal to molarity in all case.

• A. $3\times {10}^{-5}$
• B. $9.0\times {10}^{-5}$
• C. $23\times {10}^{-5}$
• D. $1.5\times {10}^{-5}$

Answer 1

Answer: (D)

$1.5\times {10}^{-5}$

$\Delta T_b=i{K}_{b}m$

$0.15=3\times 0.5\times m$

Therefore, $m=0.1$

Since these solutions contains two salts

$\Delta T_f=k_{f}m$

$0.83=1.86\times [20.2+3s]$

$s=1.54\times {10}^{-3}$

$K_{sp}=4s^3=1.5\times {10}^{-5}$