Question

What is the freezing point of a solution containing $ \text{8}\text{.1 g HBr} $ in 100 g water assuming the acid to be 90% ionised? ( $ {{k}_{f}} $ for water $ =1.86\,\text{K}\,\text{mo}{{\text{l}}^{-1}} $ )

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

Answer 1

The Correct Option is B

$ \Delta {{\Tau }_{f}}=i\times {{k}_{f}}\times m $ $ \text{Ions}\,\text{at}\,\text{equilibrium}\overset{HBr}{\mathop{1-\alpha }}\,\xrightarrow{{}}\underset{\alpha }{\mathop{{{H}^{+}}}}\,+\underset{\alpha }{\mathop{B{{r}^{-}}}}\, $ $ \therefore $ $ \text{Total}\,\text{ions}=1-\alpha +\alpha +\alpha $ $ =1+\alpha $ $ \therefore $ $ i=1+\alpha $ Given, $ {{k}_{f}}=1.86\,\text{K}\,\text{mo}{{\text{l}}^{-1}} $ mass of $ \text{ }\!\!~\!\!\text{ HBr =}\,\text{8}\text{.1 g} $ mass of $ \text{ }\!\!~\!\!\text{ }{{\text{H}}_{\text{2}}}\text{O}\,\text{=100 g} $ $ \text{( }\!\!\alpha\!\!\text{ )}\,\,\text{=} $ degree of ionization $ =90% $ $ m $ (molality) $ \text{= }\frac{\text{mass}\,\text{of}\,\text{solute/mol}\text{.wt}\text{.of}\,\text{solute}}{\text{mass of solventin kg}} $ $ =\frac{8.1/81}{100/1000} $ $ i=1+\alpha $ $ =1+90/100 $ $ =1.9 $ $ \Delta {{T}_{f}}=i\times {{k}_{f}}\times m $ $ =1.9\times 1.86\times \frac{8.1/81}{100/1000} $ $ =3.534{{\,}^{o}}C $ $ \Delta {{T}_{f}}= $ (depression in freezing point) = freezing point of water - freezing point of solution 3.534 = 0-freezing point of solution $ \therefore $ Freezing point of solution $ =-3.534{{\,}^{o}}C $