Answer
The amount of heat required for vaporization of 33.8 g of $H_{2}O$ is 76.3 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 water is 40.7 kJ/mole.
Given the mass of water is 33.8 g, and it vaporizes at 100c.
So, 1 mol of water = 18.02 g of water and dH vap of water = 40.7 kJ.
Hence conversion factors are :
(1mol $H_{2}O$ /18.02 g $H_{2}O$) and (40.7 kJ/ 1mol $H_{2}O$).
Hence for dH vap of 33.8 g of $H_{2}O$ can be calculated as follows:
= (33.8 g $H_{2}O$) x(1mol $H_{2}O$/18.02 g $H_{2}O$) x (40.7 kJ/ 1mol $H_{2}O$)
= 76.3 kJ.
Hence the amount of heat required for vaporization of 33.8 g of $H_{2}O$ is 76.3 kJ.
