Question

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

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

Answer 1

Answer: (B)

$\Delta T_f=i\times K_f\times m$

$\underset{1-\alpha }{\mathrm{HBr}}\to \underset{\alpha }{H^{+}}+\underset{\alpha }{B{r}^{-}}$

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

$\therefore i=1+\alpha$

Given; $K_f=1.86Kmo{l}^{-1}$

mass of $HBr=8.1g$

mass of $H_{2}O=100g$

$(\alpha )=$ 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 $=-3.534$