Question

The value of ΔHvap for water at 100.°C is 40.7 kJ/mol. How much energy (in J) is required to completely vaporize 5.6 g of water at 100.°C? The molar mass of water is 18.02 g/mol.

Answer 1

Answer: $1.3\times 10^4J$    

Explanation:

Heat of vaporization is the amount of heat required to convert  1 mole of liquid substance to its vapor form.

Given :

Amount of heat required for 1 mol of water = 40.7 kJ/mol

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of $H_2O$ weighs = 18.02 g

Thus we can say:

18.02 g of $H_2O$ requires heat = 40.7 kJ

Thus 5.6 g of $H_2O$ requires heat  =$\frac{40.7}{18.02}\times 5.6=13kJ=1.3\times 10^4J$        (1kJ=1000J)

Thus the energy required to completely vaporize 5.6 g of water at 100.°C is $1.3\times 10^4J$