The enthalpy of vaporization of the system is obtained as 23.6 kJ/mol.
We know that we can be able to obtain the enthalpy of vaporization from the vapor pressure of the system at two different temperatures that have been given in the question.
Given that;
ln(P2/P1) = -ΔHvap/R(1/T2 - 1/T1)
P2 = final pressure
P1 = initial pressure
ΔHvap = enthalpy of vaporization
T2 = final temperature
T1 = initial temperature
Then;
ln(135/24.3) = -ΔHvap/8.314 (1/325 - 1/273)
1.7 = -ΔHvap/8.314 (0.0031 - 0.0037)
1.7 = -ΔHvap/8.314 (-0.0006)
1.7 = 0.0006ΔHvap/8.314
ΔHvap = 1.7 * 8.314/0.0006
= 23.6 kJ/mol
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