As the ratio of pentane to hexane = 1 : 4

$$ \therefore $$ Mole fraction of pentane = 1/5

Mole fraction of hexane = 4/5

Total vapour pressure

= $$\left( {{1 \over 5} \times 440 + {4 \over 5} \times 120} \right)$$

= 184 mm of Hg

$$ \therefore $$ Vapour pressure of pentane in mixture

= (Vapour pressure of mixture pentane $$ \times $$ Mole fraction of in vapour phase)

$$ \Rightarrow $$ 88 = 184 × mole fraction of pentane in vapour phase

$$ \Rightarrow $$ Mole fraction of pentane in vapour phase

= $${{88} \over {184}}$$ = 0.478

One molal solution means one mole solute present in 1 kg (1000 g) solvent.

$$ \therefore $$ mole of solute = 1

Mole of solvent (H₂O) = $${{1000} \over {18}}$$

X_solute = $${1 \over {1 + {{1000} \over {18}}}}$$ = 0.018

Hence total vapour pressure

= [(Mole fraction of P) × (Vapour pressure of P)] + [(Mole fraction of Q) × Vapour pressure of Q)]

= $$\left( {{3 \over 5} \times 80 + {2 \over 5} \times 60} \right)$$

= 48 + 24 = 72 torr

From molarity equation

M₁V₁ + M₂V₂ = MV

$$ \Rightarrow $$ 1× 2.5 + 0.5 × 3 = M × 5.5

$$ \Rightarrow $$ M = $${4 \over {5.5}}$$ = 0.73M