Question

A solution of urea (mol. mass $56 \,g \,mol ^{-1}$ ) boils at $100.18^{\circ} C$ at the atmospheric pressure. If $k_{ f }$ and $k_{b}$ for water are $1.86$ and $0.512\, K\, kg\, mol ^{-1}$ respectively, the above solution will freeze at

• A. $-6.54^{\circ} C$
• B. $ 6.54^{\circ} C$
• C. $ 0.654^{\circ} C$
• D. $ -\,0.654^{\circ} C$

Answer 1

Answer: (D)

$\because \Delta T_{f}=k_{f} \times$ Molality of solution
$\Delta T_{b}=k_{b} \times$ Molality of solution
or$\frac{\Delta T_{f}}{\Delta T_{b}}=\frac{k_{f}}{k_{b}}$
Given that
$\Delta T_{b}=T_{2}-T_{1}=10018-100=0.18$
$k_{f}$ for water $=1.86\, K \,kg\, mol ^{-1}$
$k_{b}$ for water $=0.512 \,K\, kg\, mol ^{-1}$
$\therefore \frac{\Delta T_{f}}{018} =\frac{1.86}{0.512} $
$\Delta T_{f} =\frac{1.86 \times 018}{0.512} $
$=0.6539 \sim 0.654 $
$\Delta T_{f} =T_{1}-T_{2} $
$0.654 =0^{\circ} C -T_{2}$
$T_{2} =-0.654^{\circ} C$
$\left(T_{2} \rightarrow\right.$ Freezing point of aqueous urea solution)