Answer
The amount of heat required for vaporization of 43.9 g of acetone is 22.0 kJ.
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.
dH vap of acetone is 29.1 kJ/mole.
Given the mass of water is 43.9 g, and it vaporizes at 100c.
So, 1 mol of acetone = 58.09 g of water and dH vap of acetone = 29.1 kJ.
Hence conversion factors are :
(1mol $C_{3}H_{6}O$/58.09 g $C_{3}H_{6}O$) and (29.1 kJ/ 1mol $C_{3}H_{6}O$).
Hence for dH vap of 43.9 g of $C_{3}H_{6}O$ can be calculated as follows:
= (43.9 g $C_{3}H_{6}O$) x(1mol $C_{3}H_{6}O$/58.09 g $C_{3}H_{6}O$) and (29.1 kJ/ 1mol $C_{3}H_{6}O$)
= 76.3 kJ.
Hence the amount of heat required for vaporization of 43.9 g of acetone is 22.0 kJ