Question

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

• A. $0.85^{\circ} C$
• B. $-3.53^{\circ} C$
• C. $0^{\circ} C$
• D. $-0.35^{\circ} C$

Answer 1

Answer: (B)

$\Delta T _{ f }= i \times K _{ f } \times m$

$\underset{1-\alpha}{HBr} \rightarrow \underset{\alpha}{H ^{+}}+ \underset{\alpha}{Br^{-}}$

Total ions $=1-\alpha+\alpha+\alpha=1+\alpha$

$\therefore i =1+\alpha$

Given; $K _{ f }=1.86\, K\, mol ^{-1}$

mass of $HBr =8.1 \,g$

mass of $H _{2} O =100 \,g$

$(\alpha)=$ degree of ionisation $=90 \%$

m(molality) $=\frac{\text { mass of solute } / mol \text { wt. of solute }}{\text { mass of solvent in } kg }$

$=\frac{8.1 / 81}{100 / 1000}$

$i=1+\alpha=1+\frac{90}{100}=1.9$

$\Delta T _{ f }= i \times K _{ f } \times m$

$=1.9 \times 1.08 \times \frac{8.1 / 81}{100 / 1000}=3.534^{\circ} C$

$\Delta T _{ f }$ (depression in freezing point) $=$ freezing point of water- freezing point of solution.

$3.534=0$ - freezing point of sol"

$\therefore $ freezing point $=-3.534$