Question

How much energy is required to vaporize 48.7 g of dichloromethane (CH2Cl2) at its boiling point, if its ΔHvap is 31.6 kJ/mol?

Answer 1

Answer:

The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point

Explanation:

To solve the above question we have the given variable as follows

ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole

However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.

The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles

Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ