The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3 : 2 is : [At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

Given data at 45 °C:

$\mathrm{P}⁎$_benzene = 280 mm Hg and $\mathrm{P}⁎$_octane = 420 mm Hg.

The molar ratio of benzene to octane is 3 : 2, so the total number of moles in the mixture is 3 + 2 = 5.

The mole fractions are

$x$_benzene = 3/5 = 0.6

$x$_octane = 2/5 = 0.4

Assuming an ideal solution, Raoult’s law gives the total vapour pressure

$P$_total = _{benzene · 

_{benzene

_{octane

Substituting the values:

$P$_total = 0.6 · 280 + 0.4 · 420

$0.6 · 280 = 168$
$0.4 · 420 = 168$

Therefore,

$P$_total = 168 + 168 = 336 mm Hg

The vapour pressure of the solution at 45 °C is 336 mm Hg.