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JEE MainChemistryModeratePhysical Chemistry

Water is added to the solution M such that the mole fraction of
water in the solution becomes 0.9. The boiling point of this solution is

Show answer and solution
ΔTb=Kb⋅m\Delta T_{b} = K_{b} \cdot m
=0.52×0.10.9×18×1000=5209×18=3.20= 0.52 \times \frac{0.1}{0.9 \times 18} \times 1000 = \frac{520}{9 \times 18} = 3.20
Tb=373+3.20=376.2KT_{b} = 373 + 3.20 = 376.2K

Source: Grass-G / CH-27 Chemistry Paper 2 15 Jan.docx

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