Question

A 75.0 mL sample of water is heated to its boiling point. How much heat (in kJ ) is required to vaporize it?

Answer 1

Given :

A 75.0 mL sample of water is heated to its boiling point.

To Find :

The heat (in kJ ) is required to vaporize it .

Solution :

Heat of vaporization of water , $\Delta H_{vap}=40.65\ kJ/mol$.

Molar mass of $H_2O$ , $M=18\ g/mol$ .

Density of water , $\rho=1\ g/ml$ .

Mass of 75 ml of water sample , $m=75\ g$ .

Moles of 75 g water = $\dfrac{75}{18}=4.17\ mol$ .

Now , heat required to vaporize 4.17 mol of water :

$H=4.17\times 40.65\ kJ\\\\H=169.51\ kJ$

Therefore , heat required to boil 75 ml sample of water is 169.51 kJ.

Hence , this is the required solution .