How man kilojoules of energy are needed to vaporize a 28.6g sample of liquid benzene at its normal boiling point of 80C?
q = mass benzene x heat vaporization = ?
heat vaporization must be available in tables.
To calculate the amount of energy required to vaporize a sample of liquid benzene, we need to use the formula: Q = m * ΔHvap, where Q is the energy required (in joules), m is the mass of the liquid (in grams), and ΔHvap is the molar enthalpy of vaporization (in J/g).
Step 1: Convert the mass of the liquid from grams to kilograms.
Given mass = 28.6 g
Mass in kilograms = 28.6 g / 1000 = 0.0286 kg
Step 2: Determine the molar enthalpy of vaporization of benzene.
The molar enthalpy of vaporization of benzene is ΔHvap = 30.8 kJ/mol (at its normal boiling point).
Step 3: Convert the molar enthalpy of vaporization from kilojoules per mole to joules per gram.
To convert the molar enthalpy of vaporization from kJ/mol to J/g, we need to divide by the molar mass of benzene (78.11 g/mol).
ΔHvap = 30.8 kJ/mol * (1000 J/1 kJ) / (78.11 g/1 mol) = 395 J/g
Step 4: Calculate the energy required to vaporize the liquid benzene.
Q = m * ΔHvap
Q = 0.0286 kg * 395 J/g = 11.287 J
Therefore, approximately 11.287 kilojoules (kJ) of energy are needed to vaporize a 28.6g sample of liquid benzene at its normal boiling point of 80°C.