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The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 1.62...

The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 1.62 g of water boils at atmospheric pressure?

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Answer #1

Answer: given that, Heat of vapourization of wate is 40.66 KJ/mol.

Heat absorbed when 1.62 g of Water boils at atmosperic pressure

= 1.62 g H2O ( 1mole H2O / 18.0 g H2O ) ( 40.66 KJ / 1mole H2O )

   = 1.62 ( 18 / 18 ) ( 40.66 / 18 )

   = 1.62 1 2.2588

   = 3.6594 KJ

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