Question

What is the freezing point of a solution containing 8.1 g HBr in 100 g water, assuming that the acid to be 90% ionized?
$(K_f (H_2O) = 1.86 \,K \,kg\, mol^{-1})$

• A. $0.85^{\circ}C$
• B. $-3.8^{\circ}C$
• C. $-0^{\circ}C$
• D. $-3.5^{\circ}C$

Answer 1

Answer: (D)

Consider the dissociation of 1 mol HBr. The degree of dissociation is 90%, i.e., $\alpha = 0.90$
$\begin{matrix}&HBr_{\left(aq\right)}&\to&H^{+}\left(aq\right)&+&Br^{-}\left(aq\right)\\ Initial\, conc.&1\, mol&&0&&0\\ Final \, conc.&1 - 0.90 = 0.10\, mol&&0.90\, mol&&0.90\, mol\end{matrix}$
$\therefore \quad$ Total number of particles $= 0.10 + 0.90 + 0.90$
$= 1.90\, mol$
van’t Hoff factor $\left(i\right) = \frac{1.90}{1.00} = 1.90$
$\Delta T_{f} = iK_{f} \frac{W_{B}}{M_{B}\cdot W_{A}} = \frac{1.90 × 1.86 \,K\,kg\,mol^{-1} × 8.1 \,g}{81 \,g \,mol^{-1} \times 100 \times10^{-3} \,kg}$
$= 3.5 \,K = 3.5^{\circ}C$
F.P. of solution = F.P. of solvent - depression in F.P.
$= 0^{\circ}C - 3.5^{\circ}C = -3.5^{\circ}C$