Explanation

(b)
$\bigtriangleup T_{f} = i \times k_{f} \times m$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ HBr\ \longrightarrow \ H^{+} + Br^{-}\ \$
Ions at equilibrium 1-$\alpha$ $\ \ \ \ \ \ \ \ \alpha\ \ \ \ \ \ \ \ \ \alpha\ \ \ \ \ \ \ \$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ Total\ \ ions = 1 - \alpha + \alpha + \alpha\ \ \ \ \$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 + \alpha\ \ \$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = 1 + \alpha$
Given, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k_{f} = 1.86\ K\ mol^{- 1}$
Mass of HBr = 8.1 g
Mass of $H_{2}O$ =100 g
($\alpha$) = degree of ionization = 90%
m(molality) =$\frac{mass\ of\ solute\ mol.wt.of\ solute\ }{mass\ of\ solvent\ in\ kg\ }$
$= \frac{8.1/81}{100/1000}$
$i = 1 + \alpha$
=1+90/100
=1.9
$\bigtriangleup T_{f} = i \times k_{f} \times m$
= 1.9$\times$ 1.86 $\times \frac{8.1\ /81}{100/1000}$
= 3.53$4^{\circ}C$
$\bigtriangleup T_{f} =$(depression in freezing point) = freezing point of water - freezing point of solution 3.534 = 0 – freezing point of solution.
$\therefore$ Freezing point of solution = -3.53$4^{\circ}C$