Question

What is freezing point of solution containing $8.1\, g$ of $HBr$ in $100\, g$ of water, assuming the acid to be $90 \%$ ionised. $\left(K_{f}\right.$ for water $=1.86\, kg\, mol ^{-1}$ and molar mass of $HBr =81$ )

• A. $-0.35^{\circ} C$
• B. $-1.35^{\circ} C$
• C. $-2.35^{\circ} C$
• D. $-3.53^{\circ} C$

Answer 1

Answer: (D)

Dissociation of HBr occurs as follows

$i_{\text {total }}=1+0.9 \Rightarrow 1.9$
$\because \Delta T_{f}=k_{f} \times i \times m$
$\Rightarrow k_{f} \times i \times \frac{w}{m} \times \frac{1000}{100}$
$\Delta T_{f}=1.86 \times 1.9 \times \frac{8.1}{81} \times \frac{1000}{100}$
$=3.53^{\circ} C$
$\therefore T_{f}=T-\Delta T_{f}=0-3.53=-3.53^{\circ} C$