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Question

The molar enthalpy of vaporization of hexane (C6H14) is 28.9 kJ/mol, and its normal boiling point is 69C. What is the vapour pressure of hexane at 25.00C in torr? [e1.5=0.222]

A
169.5 torr
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B
195.4 torr
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C
134.6 torr
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D
202.2 torr
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Solution

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(1T11T2)
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=25.00C=298 KP2=760 torrT2=69C=342 K
Molar enthalpy of vaporisation, ΔH=28.9 kJ/mol

Substituting the above values, we get,
ln (x760)=(289008.314)(12981342)ln (x760)=3476.06×0.000432ln (x760)1.500x760=0.222x=169.5 torr
Vapour pressure at 25C,P1=169.5 torr

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