Question

How much energy is required to vaporize 41. 4 g of dichloromethane (ch 2cl 2) at its boiling point, if its δh vap is 31. 6 kj/mol?

Answer 1

Answer: 15.4 kJ

Step-by-step explanation:

Molar mass of CH2Cl2,

MM = 1*MM(C) + 2*MM(H) + 2*MM(Cl)

= 1*12.01 + 2*1.008 + 2*35.45

= 84.926 g/mol

Mass(CH2Cl2) = 41.4 g

Use:

number of mol of CH2Cl2,

n = mass of CH2Cl2/molar mass of CH2Cl2

= (41.4 g)/(84.93 g/mol)

= 0.4875 mol

Given:

ΔH = 31.6 KJ/mol

= 31600 J/mol

Use:

Q = ΔH * number of mol

= 31.6 KJ/mol * 0.4875 mol

= 15.4 KJ

Therefore, it takes 15.4 kJ of energy to vaporize 41. 4 g of CH₂Cl₂ at its boiling point