Question

What is freezing point of solution containing $8.1g$ of $HBr$ in $100g$ of water, assuming the acid to be $90\%$ ionised. $(K_f$ for water $=1.86kgmo{l}^{-1}$ 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

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