Find the boiling point of a compound given a pressure
You could use the Clausius Clapeyron equation:

$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}}{\Delta V \cdot T}$$

Since the change in pressure is rather small, it is fair to assume the molar enthalpy of evaporation to be constant and use the value given for standard conditions, which should be easy to obtain.

The other simplification you can make is to assume that the volume of the liquid is very small compared to the volume of the gas:

$$ V_\mathrm{l} &lt;&lt; V_\mathrm{g}$$ 
so that
$$\Delta V = V_\mathrm{g}-V_\mathrm{l} \sim V_\mathrm{g}$$

Now by using the ideal gas law and expressing volume as molar quantity,

$$v=\frac{p}{RT}$$

you get the ODE

$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}\cdot p}{T^2}$$

from whose solution you get a nice equation to solve your problem.