Calculate Mass of Non-Volatile Solute in Octane to Reduce V.P. 80%

[Vriska]

I've got this, calculate the mass of a non volatile solute (molar mass 40) that should be dissolved in 114 g of octane to reduce its vapour pressure to 80 percent

I do this, v.p = v.p_pure * mole fraction of solvent in solution

we have 1 mole of solvent and let's take n moles of substance so .8 = 1/(n+1) doing all this we'll get our mass to be 10 grams

is this right? some people are getting 8.

annd one more thing, there's isn't actually 1 mole in the solution is there? some would exist in the gas state, say p =1, how would I theoretically get the exact number of moles I've got to add to this? Do I need to know the volume and temperature too?

 

[mjc123]

Your answer is right. I think some people are assuming there is 1 mol total solution, so 0.8 mol octane and 0.2 mol solute. But there is 1 mol octane, so you need 0.25 mol solute to get a solvent mole fraction of 0.8.
You're meant to assume that the amount of solvent in the gas phase is negligible compared to the liquid. If this were not true, you would indeed have to make corrections for the gas phase fraction.