The enthalpy of vaporization of liquid water is 40.65 kJ/mol. Calculate the energy required to vaporize 12.5 g of liquid water.
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Ответ:
The energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid into its vapor state without change in its temperature.
Given : The enthalpy of vaporization of water is 40.65 kJ/mol.
n = number of moles =![\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{12.5g}{18g/mole}=0.694mole](/tpl/images/0507/6465/19b67.png)
Thus 1 mole of water requires heat = 40.65 kJ
0.694 moles of water requires heat =![\frac{40.65}{1}\times 0.694=28.2kJ](/tpl/images/0507/6465/773e1.png)
Thus the energy required to vaporize 12.5 g of liquid water is 28.2 kJ
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