The vapour pressure of pure benzene at `50^(@)` is `268 mm` of `Hg`. How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure of `16.0 mm` of `Hg` at `50^(@)C`?

Explanation

Applying Raoult's law in the following form :
`(p_(0)-p_(s))/(p_(s))=(wM)/(Wm)=(w//m)/(W//M)`
= No. of moles of solute per mole of benzene
or `(n)/(N)=((268-167))/(167)=0.6047~~0.605`
Alternative method : We know that, `p_(s)`= Mole fraction of solvent `xx p_(0)`
or 167 = Mole fraction of solvent `xx 268`
So, Mole fraction of solvent `= (167)/(268)=0.623`
Mole fraction of solute `=1-0.623 = 0.377`
`(n)/(N)=("Mole fraction of solute")/("Mole fraction of solvent")=(0.377)/(0.623)=0.605`.