Question

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?

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 H_{2}O (
1mole H_{2}O / 18.0
g H_{2}O )
( 40.66 KJ /
1mole H_{2}O )

= 1.62 ( 18 / 18 ) ( 40.66 / 18 )

= 1.62 1 2.2588

= 3.6594 KJ

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1.
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