Question

An aqueous solution freezes at $-{2.55}^{o}C$. What is its boiling point?
$[K_b^{H_{2}O}=0.52K{m}^{-1};K_1^{H_{2}O}=1.86K{m}^{-1}]$

• A. ${107.0}^{o}C$
• B. ${100.6}^{o}C$
• C. ${100.1}^{o}C$
• D. ${100.7}^{o}C$

Answer 1

The correct option is D ${100.7}^{o}C$

$\Delta T_f=K_{f}m$

$m=\frac{\Delta T_f}{K_f}=\left(\frac{2.55}{1.86}\right)$

$\Delta T_b=K_{b}m$

$=0.52\left(\frac{02.55}{1.86}\right)$

$=0.71$

So $\Delta T_b=0.71{}^{o}C$

$T_{B.P}=100+0.71{}^{o}C$

$={100.7}^{o}C$

Hence, the correct option is $D$