Fraction of concentration of benzene in air water We have a 120 ml water sample with an unknown concentration of benzene. If the bottle size is 160 ml (with 40 ml of air) and benzene has reached equilibrium, what fraction of benzene is left in the water? 

We are only given Hcc (Henry constant - Cair/Cwater) 0.225 and Log Kow (to which I believe is needed for a later question) 2.53. 

I'm not sure where to start. We do not know mole fraction so we cannot find partial pressure using Henry's law. Also this rules out Raoult's Law. 

Also we can't use the Ideal gas law since we do not have a lot of the variables. I'm stumped. But we can make assumptions such as T for example.

Any suggestions?
You don't need the original concentration or any of the variables you are missing. Henry's Law gives you everything you need. The usual formulation of Henry's Law at equilibrium:

$$c_{i,s}=k_{H,i,s}c_{i,s}$$

Where $c_{i,air}$ is the concentration of solute $i$ in the vapor above the solution, $k_{H,i,s}$ is the Henry's Law constant for solute $i$ in solvent $s$, and $c_{i,sol}$ is the concentration of $i$ remaining in solution.

We can rearrange Henry's Law to give you what you need - a ratio of the amount of benzene in the air to the amount in the water:

$$k_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}$$

The other snippet of information you need is the relationship between moles and concentration. There is a reason you were given volumes. 

$$n_{air}=c_{air}*V_{air}$$
$$n_{sol}=c_{sol}*V_{sol}$$

You only need the ratio of $\dfrac{n_{air}}{n_{sol}}$, so you really only have one unknown. It's algebra time!

$$K_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}=\dfrac{\dfrac{n_{air}}{V_{air}}}{\dfrac{n_{sol}}{V_{sol}}}$$

$$\dfrac{K_{H,i,s}V_{air}}{V_{sol}}=\dfrac{n_{air}}{n_{sol}}$$