CBSE Class 12 Chemistry 2016 Outside Delhi Set 1 Solved Paper

Question : 15

Total: 26

Calculate the boiling point of solution when 4g of MgSO4(M=120gmol−1) was dissolved in 100 g of water, assuming MgSO4 undergoes complete ionization.
(Kb for water =0.52K‌kg‌mol−1)

Solution:

∆Tb‌=iKb⋅m
i‌=2
‌=i×Kb×‌

W2×1000

M×W1

‌=2×0.52K‌kg‌mol−1 ×‌

4g×1000g∕kg

120g∕mol×100g

‌=‌

2×0.52

‌=0.346K
Boiling point of water ‌=‌373.15K∕373K
Tb‌=Tb0+∆Tb
‌=373.15K+0.346K‌ or ‌ 373K+0.346K
‌=373.496K‌ or ‌373.346K

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