The correct option for the value of vapour pressure of a solution at 45⁰C with benzene to octane in molar ratio 3 : 2 is : [At 45⁰C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

To find the vapour pressure of the solution, we can use Raoult's Law for an ideal solution. According to Raoult's Law, the total vapour pressure of a mixture of two volatile liquids ($P_{\text{total}}$) is equal to the sum of the partial vapour pressures of its individual components.

Let benzene be component A and octane be component B.
The formula for the total vapour pressure is:
$P_{\text{total}}=P_{\text{A}}^0{\chi }_{\text{A}}+P_{\text{B}}^0{\chi }_{\text{B}}$
where:
$P_{\text{A}}^0$ is the vapour pressure of pure benzene = 280 mm Hg
$P_{\text{B}}^0$ is the vapour pressure of pure octane = 420 mm Hg
${\chi }_{\text{A}}$ is the mole fraction of benzene in the liquid phase
${\chi }_{\text{B}}$ is the mole fraction of octane in the liquid phase

The molar ratio of benzene to octane is given as 3 : 2.
Thus, we can calculate the mole fractions as follows:
${\chi }_{\text{benzene}}=\frac{3}{3+2}=\frac{3}{5}=0.6$
${\chi }_{\text{octane}}=\frac{2}{3+2}=\frac{2}{5}=0.4$

Now, substitute these values into Raoult's Law formula:
$P_{\text{total}}=(280\text{mm Hg}\times 0.6)+(420\text{mm Hg}\times 0.4)$
Calculate each term:
$P_{\text{benzene}}=280\times 0.6=168\text{mm Hg}$
$P_{\text{octane}}=420\times 0.4=168\text{mm Hg}$

Sum the partial pressures to get the total vapour pressure:
$P_{\text{total}}=168+168=336\text{mm Hg}$

Therefore, the value of the vapour pressure of the solution is 336 mm of Hg.