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Question: The elevation in boiling point of a solution of \(9.43\)g of \(MgCl_{2}\) in 1 kg of water is \((K_{...

The elevation in boiling point of a solution of 9.439.43g of MgCl2MgCl_{2} in 1 kg of water is (Kb=0.52Kkgmol−1)(K_{b} = 0.52Kkgmol^{- 1}) molar mass of MgCl2=94.3gmol−1MgCl_{2} = 94.3gmol^{- 1}

A

0.1560.156

B

0.520.52

C

0.170.17

D

0.940.94

Answer

0.1560.156

Explanation

Solution

MgCl2MgCl_{2} Mg2++2Cl−,i=3Mg^{2 +} + 2Cl^{-},i = 3

ΔTb=iKbm=3×0.52×9.4394.3×1=0.156\Delta T_{b} = iK_{b}m = 3 \times 0.52 \times \frac{9.43}{94.3 \times 1} = 0.156