Answer
10.4 kJ/mol
Work Step by Step
The amount of heat required to vaporize one mole of liquid is defined as the heat of vaporization. Heat of vaporization is represented by dH vap.
Given mass of sweat is 4.25g.
dH vap of sweat at room temperature is 44 kJ./mol
We know:
18g = 1mol
1 g= 1/18 mol
So, 4.25g contains 0.23611 mol
For 1 mol of sweat dH is 44 kJ/mol
For 0.23611mol dH is= 0.23611x 44 kJ/mol= 10.4 kJ/mol
