Question

3. Molar heat of vaporisation of a liquid is 6 kJ mol-1. If the entropy change is 16 J mol-1 K-1, the boiling point of the liquid is 1) 375°C 2) 375 K 3) 273 K 4) 102°C​

Answer 1

Answer:

273k

Explanation:

maine Use hi likha hai ispe bharosa na kare

Answer 2

Answer:

357 K

explanation:

Since boiling occurs at equilibrium conditions so deltaG =0

Now

T= delta H/ delta S

T = 6000J/mol÷ 16 J/mol/K

T = 357 K

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