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Thd AHyap of a certain compound is 11.51 kJ - mol- and its A Svapº is 69.92 J. mol-.K-!! What is the normal boiling point of
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Answer #1

Given, Hvap = 11.51 kJ/mol = 11.51x103 I/mol as wap = 69.92 I/molok T= ? (where T = normal boiling point ). boiling point theT = Dhivap = 11.51 x 103 /mol Devap 69.92 J/molok IT = 164.61kany query please comment.

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