Answer :
The energy of vaporizing one mol is 40.79. 1 mol of water is 18g so the number of moles you have is 30/18. Once you know the number of moles you will need to multiply by the heat of vaporization per mole. Also water is not exactly 18g/mol so you might have to calculate it to a couple decimal places depending on what your teacher wants.
The amount of heat required to vaporize 30 g of water sample has been 69.98 kJ.
Standard molar heat of vaporization has been given as the amount of heat required to vaporize 1 mole of sample.
The molar mass of water has been 18 g. Thus, the standard heat of vaporization for water has been the amount of heat required to vaporize 18 g of water.
Computation for Heat of vaporization
The heat of vaporization of water has been 40.79 kJ/mol. It has been required to vaporize 18 g of sample.
The amount of heat required to vaporize 30 g of sample has been:
[tex]\rm 18\;g=40.79\;kJ\\30\;g=\dfrac{40.79}{18}\;\times\;30\;kJ\\30\;g=67.98 \;kJ[/tex]
The amount of heat required to vaporize 30 g of water sample has been 69.98 kJ.
Learn more about heat of vaporization, here:
https://brainly.com/question/15140455
