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 apply Raoult's law for an ideal binary solution. Raoult's law states that the partial vapour pressure of each volatile component in a solution is directly proportional to its mole fraction in the solution.

Let benzene be represented as component A and octane as component B.
The vapour pressure of pure benzene at 45°C is given as:
$P_A^{\circ }=280\text{mm Hg}$
The vapour pressure of pure octane at 45°C is given as:
$P_B^{\circ }=420\text{mm Hg}$

The molar ratio of benzene to octane in the solution is 3 : 2.
This means for every 3 moles of benzene, there are 2 moles of octane. Let the number of moles of benzene be $n_A=3$ and the number of moles of octane be $n_B=2$.

First, we calculate the mole fraction of each component in the solution:
The mole fraction of benzene (${\chi }_A$) is:
${\chi }_A=\frac{n_A}{n_A+n_B}=\frac{3}{3+2}=\frac{3}{5}=0.6$
The mole fraction of octane (${\chi }_B$) is:
${\chi }_B=\frac{n_B}{n_A+n_B}=\frac{2}{3+2}=\frac{2}{5}=0.4$

According to Raoult's law, the total vapour pressure of the solution ($P_{\text{total}}$) is the sum of the partial vapour pressures of the individual components:
$P_{\text{total}}=P_A+P_B=(P_A^{\circ }\times {\chi }_A)+(P_B^{\circ }\times {\chi }_B)$

Substituting the values into the formula:
$P_{\text{total}}=(280\times 0.6)+(420\times 0.4)$
$P_{\text{total}}=168+168$
$P_{\text{total}}=336\text{mm of Hg}$

Thus, the vapour pressure of the solution is 336 mm of Hg, which matches the option 336 mm of Hg.