Q9 / 91 · KVPY 2020 (SB/SX) · Chemistry

The boiling point (in ${}^{\circ}C$ ) of $0.1$ molal aqueous solution of ${CuSO}_{4} . 5 H_{2} O$ at 1 bar is closest to: [Given: Ebullioscopic (molal boiling point elevation) constant of water, $K_{b} = 0.512 {KKgmol}^{- 1}] : -$

Solution

$i = 2$

$∆ T_{b} = i \cdot K_{b} \cdot m = 2 \times 0.512 \times 0.1 = 0.1024$
$T_{b}^{'} = T_{b}^{0} + Δ T_{b} = 100 + 0.1024 = 100.10$

Hence, the correct answer is option: 'C'.