QAIPMT-2005-Ch-2507: NEET 2005 Chemistry - Class 12 AIPMT-2005-Ch-2507 Solution: A solution has a 1:4 mole ratio of pentane to hexane. The…

Step-by-Step Solution (Raoult's Law) 1. Find the mole fractions in the liquid phase ($x$): The problem states that the mole ratio of pentane to hexane is $1:4$. � Total parts = $1 + 4 = 5$ � Mole fraction of pentane ($x_{\text{pentane}}$) = $\frac{1}{5} = 0.20$ � Mole fraction of hexane ($x_{\text{hexane}}$) = $\frac{4}{5} = 0.80$ 2. Calculate the partial vapour pressures using Raoult's Law: The partial vapour pressure of a component is its liquid mole fraction multiplied by its pure vapour pressure ($P = x \cdot P^\circ$): � Partial pressure of pentane ($P_{\text{pentane}}$) = $0.20 \times 440\text{ mm Hg} = 88\text{ mm Hg}$ � Partial pressure of hexane ($P_{\text{hexane}}$) = $0.80 \times 120\text{ mm Hg} = 96\text{ mm Hg}$ 3. Calculate the total vapour pressure ($P_{\text{total}}$): $P_{\text{total}}=P_{\text{pentane}}+P_{\text{hexane}}=88\text{\ mm\ Hg}+96\text{\ mm\ Hg}=184\text{\ mm\ Hg}$ 4. Find the mole fraction of pentane in the vapour phase ($y_{\text{pentane}}$): The mole fraction of a component in the gas/vapour phase is the ratio of its partial pressure to the total pressure: $y_{\text{pentane}}=\frac{P_{\text{pentane}}}{P_{\text{total}}}=\frac{88}{184}\approx 0.478$