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Chemistry Question on Solutions

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? (Kf(H2O)=1.86 K kg mol−1)(K_f (H_2O) = 1.86 \,K \,kg\, mol^{-1})

A

0.85∘C0.85^{\circ}C

B

−3.8∘C-3.8^{\circ}C

C

−0∘C-0^{\circ}C

D

−3.5∘C-3.5^{\circ}C

Answer

−3.5∘C-3.5^{\circ}C

Explanation

Solution

Consider the dissociation of 1 mol HBr. The degree of dissociation is 90%, i.e., α=0.90\alpha = 0.90 HBr(aq)→H+(aq)+Br−(aq) Initial conc.1 mol00 Final conc.1−0.90=0.10 mol0.90 mol0.90 mol\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= 0.10 + 0.90 + 0.90 =1.90 mol= 1.90\, mol van?? Hoff factor (i)=1.901.00=1.90\left(i\right) = \frac{1.90}{1.00} = 1.90 ΔTf=iKfWBMB⋅WA=1.90?1.86 K kg mol−1?8.1 g81 g mol−1×100×10−3 kg\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∘C= 3.5 \,K = 3.5^{\circ}C F.P. of solution = F.P. of solvent - depression in F.P. =0∘C−3.5∘C=−3.5∘C= 0^{\circ}C - 3.5^{\circ}C = -3.5^{\circ}C