The heat of vaporization of CH3CN is obtained from the question as 25 KJ/mol.
We can obtain the number of moles from;
PV = nRT
P = 1 atm
V = 1.17 L
n = ?
R = 0.082 atm LK-1mol-1
T = 81.6 °C + 273 = 354.6 K
n = PV/RT
Substituting values;
n = 1 atm × 1.17 L/0.082 atm LK-1mol-1 × 354.6 K
n = 0.04 moles
Using; q = n·ΔHv
q = Heat absorbed
n = number of moles
Hv = Heat of vaporization
ΔHv = q/n
ΔHv = 1.00 × 10^3/0.04
ΔHv = 25 KJ/mol
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