The correct option is A 169.5 torr
Effect of temperature on the vapour pressure of a liquid is given by Clausius – Clapeyron equation,
ln (P2P1)=ΔHR(1T1−1T2)
The normal boiling point of a liquid is the temperature at which its vapour pressure is equal to one atmosphere.
thus,
Vapour pressure at 342 K, P1=760 torr
Given data:
P1=x torrT1=25.00∘C=298 KP2=760 torrT2=69∘C=342 K
Molar enthalpy of vaporisation, ΔH=28.9 kJ/mol
Substituting the above values, we get,
ln (x760)=−(289008.314)(1298−1342)ln (x760)=−3476.06×0.000432ln (x760)≈−1.500x760=0.222x=169.5 torr
Vapour pressure at 25∘C,P1=169.5 torr