The vapour pressure of two pure liquids (A) and (B) are 100 & 80 torr. The total pressure of the solution obtained by mixing 2 mol. of (A) and 3 mol of (B) would be

It is provided in question that:

${\mathrm{P}}_{\mathrm{A}}=100 \mathrm{torr}, {\mathrm{P}}_{\mathrm{B}}=80 \mathrm{torr}$

$\mathrm{moles} \mathrm{of} \mathrm{A}=2, \mathrm{moles} \mathrm{of} \mathrm{B}=3$

So, we can calculate the mole fraction of A as,

${\mathrm{x}}_{\mathrm{A}}=\mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{A}=\frac{\mathrm{no}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{A}}{\mathrm{total} \mathrm{moles}}$

$=\frac{2}{2+3}=0.4$

Similarly, we can calculate the mole fraction of B as,

${\mathrm{x}}_{\mathrm{B}}=\mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{B}=\frac{\mathrm{no}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{B} }{\mathrm{total} \mathrm{moles}}$

$=\frac{3}{2+3}=0.6$

$\mathrm{Total} \mathrm{pressure}={\mathrm{P}}_{\mathrm{A}}{\mathrm{x}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}{\mathrm{x}}_{\mathrm{B}}$

$=100\times 0.4+80\times 0.6=88 \mathrm{torr}$

Hence, the required answer is 88 torr.