The elevation in boiling point of a solution of 9.43 g of MgCl2 in 1 kg of water is (Kb = 0.52 K kg mol-1, Molar mass of MgCl2 = 94.3 g mol-1)

Question

The elevation in boiling point of a solution of 9.43 g of MgCl2 in 1 kg of water is (Kb = 0.52 K kg mol-1, Molar mass of MgCl2 = 94.3 g mol-1) (A) 0.156 (B) 0.52 (C) 0.17 (D) 0.94

Tardigrade Admin · Accepted Answer

MgCl2 leftharpoons Mg2+ + 2Cl-, i = 3 Δ Tb = iKbm = 3 ×0.52 × (9.43/94.3 × 1) = 0.156

Q. The elevation in boiling point of a solution of $9.43 g$ of $M g C l_{2}$ in $1 k g$ of water is $(K_{b} = 0.52 K k g m o l^{- 1}$ , Molar mass of $M g C l_{2} = 94.3 g m o l^{- 1})$

$0.156$

$0.52$

$0.17$

$0.94$

$M g C l_{2} ⇌ M g^{2 +} + 2 C l^{-}$ , $i = 3$
$Δ T_{b} = i K_{b} m$
$= 3 \times 0.52 \times \frac{9.43}{94.3 \times 1} = 0.156$

Q. The elevation in boiling point of a solution of 9.43g of MgCl2​ in 1kg of water is (Kb​=0.52Kkgmol−1, Molar mass of MgCl2​=94.3gmol−1)

0.156

0.52

0.17

0.94