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? 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? 6.49 kj 15.4 kj 18.1 kj 55.1 kj 31.2 kj?

This energy required is the latent heat, which is required for physical transformations. Specifically, when it involves changing liquid to gas, this latent heat is called the heat of vaporization. The formula is:

Q = mΔHvap
The molar mass of the compound is
Q = (48.7 g)(31.6 kJ/mol)(1 mol/84.93 g)
Q = 18.12 kJ

Answer Link

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-12-06T13:14:56+01:00

The energy required to vaporize 48.7 g of dichloromethane is 18.1 KJ.

The heat of vaporization is the heat required to convert a liquid to gases at a particular temperature.

The molar mass of dichloromethane is 85 g/mol. The number of moles of dichloromethane = 48.7 g/85 g/mol = 0.57 mol.

The energy required to vaporize the solution = 31.6 KJ/mol (0.57 mol)

= 18.1 KJ

Therefore the energy required to vaporize 48.7 g of dichloromethane is 18.1 KJ.

Learn more: https://brainly.com/question/967776