Answer: 137 kJ
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{100g}{17g/mol}=5.9moles[/tex]
1 mole of ammonia requires heat = 23.3 kJ
Thus 5.9 moles of ammonia require heat =[tex]\frac{23.3}{1}\times 5.9=137kJ [/tex]
Thus the energy is required to evaporate 100 g of ammonia at this temperature is 137 kJ
