At certain temperature, the solution of benzene in toulene exihibits the vapour pressure (in m bar) representes as $P = 150 x + 65$, where 'x' is mole fraction of benzene, the vapour pressure of pure benzene is
 

At certain temperature, the solution of benzene in toulene exhibits the vapour pressure (in m bar) represents as$P = 150 x + 65$, where 'x' is mole fraction of benzene,

x=1 for pure butane's mole fraction, substituting this value of x in the given equation,

P=150(1)+65=215 bar. Thus the vapour pressure of pure benzene is 215 m bar.

Hence, the correct option (D).215 bar