Question bank/JEE Main/Chemistry

JEE MainChemistryModeratePhysical Chemistry

Molar heat of vaporization of a liquid is 6 kmol−16\ k{mol}^{- 1}, If
the entropy change is 16Jmol−1K−116{Jmol}^{- 1}K^{- 1}, the boiling point of the
liquid is _______________ K

Show answer and solution
S=16 J mole−1K−1S = 16\ J\ {mole}^{- 1}K^{- 1}
Tb,=ΔHvapour ΔSvapour =6×100016=375KT_{b}, = \frac{\Delta H_{vapour\ }}{\Delta S_{vapour\ }} = \frac{6 \times 1000}{16} = 375K

Source: Grass-G / CH-11 Chemistry Paper-2 1Nov New Jee main.docx

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