The vapour pressure of pure benzene `C_(6)H_(6)` at `50^(@)C` is `268 "torr"`. How many moles of non-volatile solute per mole of benzene is required to prepare a solution of benzene having a vapour pressure of `167 "torr"` at `50^(@)C`?

Options

A. `0.377`
B. `0.605`
C. `0.623`
D. `0.395`

Answer

D

Explanation

`(268-167)/(268) = X_("solute") = 0.377`
`0.623` mole wt benzene contains `0.377` mole of solute.
So 1 mole will contain `= (0.377)/(0.623) = 0.605`