question_answer

What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised? \[({{K}_{f}}. for water = 1.86 K kg mo{{l}^{-1}}):\]

A) 0.85 K

B) -3.53K

C) 0 K

D) -0.38K

Correct Answer: B

Solution :

Given mass of solute = 8.1 g

Mass of solvent = 100 g

For \[HBr\]

\[\alpha =90%\text{ }=0.9\]

\[i=-1+\alpha =1+0.9=1.9\]

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

\[=1.86\times \frac{moles\,of\,solute}{mass\,of\,solvent\,in\,kg}\times 1.9\]

\[=1.86\times \frac{8.1/81}{100/1000}\times 1.9\]

\[=1.86\times 1\times 1.9=3.534K\]

\[{{\operatorname{T}}_{f}}=T_{f}^{{}^\circ }-\Delta {{\operatorname{T}}_{f}}\]

or \[{{\operatorname{T}}_{f}}=0-3.534K\]

\[\therefore \,\,\,{{\operatorname{T}}_{f}}=-3.534K\]