How to calculate the entropy change of benzene when it changes from vapour into liquid? Given information:
  
  
  The boiling point of benzene at atmospheric pressure is $353~\mathrm{K}$ 
  the enthalpy of vaporization of benzene is $30.8~\mathrm{kJ~mol^{−1}}$ at this temperature. 
  The molar heat capacities of the liquid and vapour are $136.1~\mathrm{J~K^{−1}~mol^{-1}}$ and $81.7~\mathrm{J~K^{−1}~mol^{-1}}$, 
  respectively, and may be assumed temperature independent.
  
  
  Calculate the entropy change of the system, the surroundings and hence the universe when $1~\mathrm{mol}$ of benzene vapour at $343~\mathrm{K}$ and atmospheric pressure becomes liquid benzene at $343~\mathrm{K}$. Also, will this process occur spontaneously?


I know that $\mathrm{d}S = \frac{\mathrm{d}H}{T}$ therefore,
$\displaystyle\mathrm{d}S = \frac{-30.8 \times 10^3~
\mathrm{J}}{343~\mathrm{K}}=-89.8~\mathrm{J~K^{-1}}$ which is the entropy of the system.

How do I continue from there, utilising the molar heat capacities given?
You just used $\mathrm{d}S=\mathrm{d}H/T$. Also you didn’t told whether the heat capacities are $C_{\mathrm{m},p}$ or $C_{\mathrm{m},V}$. Anyways I found that you are using $C_{\mathrm{m},p}$ from the benzene data page on Wikipedia. You need to do something like this:
$$\Delta S=\left(1\;\mathrm{mol}\cdot81.7\;\mathrm{\frac{J}{mol\cdot K}}\cdot\ln\frac{373\;\mathrm{K}}{343\;\mathrm{K}}\right)+\left(\frac{30.8\;\mathrm{\frac{kJ}{mol}}}{373\;\mathrm{K}}\right)+\left(1\;\mathrm{mol}\cdot136.1{\;\mathrm{\frac{J}{mol\cdot K}}}\cdot\ln\frac{343\;\mathrm{K}}{373\;\mathrm{K}}\right)\\\approx78.012\;\mathrm{J/K}$$