• # 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

 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 ${{\operatorname{T}}_{f}}=0-3.534K$ $\therefore \,\,\,{{\operatorname{T}}_{f}}=-3.534K$