Determining boiling point from vapour pressure

[lee403]

Homework Statement

When benzene freezes at 5.5 C its density changes from 0.879 g cm^-3 to 0.891 g cm^-3. Its enthalpy of fusion is 10.59 kJ mol ^-1. Estimate freezing point of benzene at 1000 atm

Homework Equations

Solid liquid boundary: p=p*+ΔHfus/ΔV ln(T/T*)

The Attempt at a Solution

From the equation I can solve for T which I do not have a problem with. For ΔV fus I thought to take the the difference of the densities and then using the molar mass of benzene, 78.1 g/mol, get ΔVm in cm^-3.This is then converted to m^-3. I made the assumption that the first part is occurring under normal conditions, so 1 atm for the freezing point at 5.5 C given in the problem. I think assuming 1 atm is my biggest mistake when solving this problem because I think that everything else is just plugging in the values.

 

[mfb]

The 5.5 C probably refer to atmospheric pressure, but it doesn't change much to assume other low pressures.

ΔV in cm^-3 doesn't have correct units. You need volume per mole to make the units match.