Answer
49.6 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 the mass of isopropyl alcohol is 65.6 g.
dH vap of isopropyl alcohol at room temperature is 45.44 kJ./mol
We know:
60g = 1mol
1 g= 1/60 mol
So, 65.6 g contains 1.093 mol
For 1 mol of isopropyl, alcohol dH is 45.4 kJ/mol
For 1.093 mol dH for isopropyl alcohol is= 1.093x 45.44 kJ/mol= 49.6 kJ/mol.
