Question

An aqueous solution of NaCl shows the depression of freezing point of water equal to 0.372 K. The boiling point of BaCl${}_2$ solution of same molality will be:

[K${}_f$ (H${}_2$O) = 1.86 K kg mol${}^{-1}$; K${}_b$ (H${}_2$O) = 0.52 K mol${}^{-1}]$

• A. ${100.52}^{\circ }C$
• B. ${100.104}^{\circ }C$
• C. ${101.56}^{\circ }C$
• D. ${100.156}^{\circ }C$

Answer 1

Answer: (D)

${100.156}^{\circ }C$

$\frac{\Delta T_f}{\Delta T_b}=\frac{i_1{K}_{f}m}{i_2{K}_{b}m}=\frac{2\times 1.86}{3\times 0.52}=\frac{0.372}{\Delta T_b}$

$\Delta T_f=0.372$.

Hence, $\Delta T_b={0.156}^{\circ }C$

i.e., $T_b=100+0.156={100.156}^{\circ }C$

Hence, the correct option is $\text{D}$.