The vapor pressure of methanol is 84.0 mm Hg at 27.0°C. The ΔHvapor of methanol is 39.2 kJ/mol and R = 8.31 10-3 kJ/mol·K. Calculate the normal boiling point of methanol in Kelvin.

Never mind that one. I figured it out. But please help me with this one.

The vapor pressure of benzene, C6H6, is 60.0 mm Hg at 13.5°C. The molar heat of vaporization of benzene is 30.8 kJ/mol and R = 8.31 10-3 kJ/mol·K. What is the vapor pressure of benzene at 70.0°C?

You say you figured out the methanol problem. This one is done the same way.

p1 = 60 mm
p2 = ?
T1 = 13.5C + 273.15
T2 = 70 + 273.15
delta H vap = 30,800 J/mol.

I keep getting 469, but that's apparently not the answer...

To calculate the normal boiling point of methanol in Kelvin, you can use the Clausius-Clapeyron equation:

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

Where:
P1 = initial vapor pressure (in this case, 84.0 mm Hg)
P2 = final vapor pressure (1 atm, since it's the normal boiling point)
ΔHvap = enthalpy of vaporization (39.2 kJ/mol)
R = gas constant (8.31 * 10^-3 kJ/mol*K)
T1 = initial temperature (27.0°C + 273.15 K = 300.15 K)
T2 = final temperature (the boiling point we want to calculate)

Simplifying the equation:

ln(1) = -(39.2 / (8.31 * 10^-3)) * (1 / T2 - 1 / 300.15)

When P2 = 1 atm, the natural logarithm of 1 is 0, so:

0 = -(39.2 / (8.31 * 10^-3)) * (1 / T2 - 1 / 300.15)

Now, let's solve for T2 (the boiling point):

-(39.2 / (8.31 * 10^-3)) * (1 / T2 - 1 / 300.15) = 0

-(1 / T2 - 1 / 300.15) = 0

1 / T2 - 1 / 300.15 = 0

1 / T2 = 1 / 300.15

T2 = 300.15 K

Therefore, the normal boiling point of methanol is 300.15 K.