A binary solution is prepared by mixing 2 moles of liquid P and 3...

Mole fraction of P, $x_P = \dfrac{2}{2+3} = 0.4$

Mole fraction of Q, $x_Q = \dfrac{3}{2+3} = 0.6$

According to Raoult's law, the ideal vapour pressure of the solution is:

$P_{\text{ideal}} = P_P^\circ x_P + P_Q^\circ x_Q$

$P_{\text{ideal}} = 120 \times 0.4 + 160 \times 0.6 = 48 + 96 = 144 \text{ mm Hg}$

The observed vapour pressure is $155 \text{ mm Hg}$, which is greater than the ideal vapour pressure ($155 > 144$). This indicates that the solution shows a positive deviation from Raoult's law.

For a solution showing positive deviation:

1. The intermolecular attractive forces between the different molecules (P-Q) are weaker than those between the similar molecules (P-P and Q-Q).

2. $\Delta H_{\text{mix}} > 0$ (endothermic process).

3. $\Delta V_{\text{mix}} > 0$ (volume increases on mixing).

4. The solution can form a minimum boiling azeotrope.

Thus, the statement about weaker intermolecular attractive forces between P and Q is correct.

Answer: The intermolecular attractive forces between P and Q are weaker than those in pure P and pure Q.