What is the freezing point of a solution containing 8.1 g HBr in 100 g of water. Assuming the acid to be 90% ionised? (K_f for water = 1.86 K kg mol^-1)
 

$∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m};$
         $\underset{1-\mathrm{\alpha }}{\mathrm{HBr}}\to \underset{\mathrm{\alpha }}{{\mathrm{H}}^{+}}+\underset{\mathrm{\alpha }}{{\mathrm{Br}}^{-}}$
$$\begin{gathered} \mathrm{Total} \mathrm{ions} =1-\mathrm{\alpha }+\mathrm{\alpha }+\mathrm{\alpha }=1+\mathrm{\alpha } \\ \therefore \mathrm{i}=1+\mathrm{\alpha } \end{gathered}$$
Given,       
   $$\begin{gathered} {\mathrm{K}}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1} \\ \mathrm{Mass} \mathrm{of} \mathrm{HBr}=8.1 \mathrm{g} \\ \mathrm{Mass} \mathrm{of} {\mathrm{H}}_2\mathrm{O}=100 \mathrm{g} \\ \left(\mathrm{\alpha }\right)=\mathrm{degree} \mathrm{of} \mathrm{ionisation} =90\% \end{gathered}$$
$\mathrm{molality}( \mathrm{m}) =\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}/\mathrm{molecular} \mathrm{weight} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solvent}(\mathrm{in} \mathrm{kg})}$
$$\begin{gathered} \mathrm{m}\left(\mathrm{molality}\right)=\frac{8.1/81}{100/1000} \\ \mathrm{i}=1+\mathrm{\alpha }=1+\frac{90}{100}=1.9 \\ ∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m} \\ =1.9\times 1.86\times \frac{8.1\times 81}{100\times 1000}=3.{534}^{\mathrm{o}}\mathrm{C} \end{gathered}$$
$∆{\mathrm{T}}_{\mathrm{f}}$(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.