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(D) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{S} = P_{A}^{o} \cdot X_{A} + P_{B}^{o} \cdot X_{B}$.Given: $n_

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$600 \ mm \ Hg, A$

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(D) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{S} = P_{A}^{o} \cdot X_{A} + P_{B}^{o} \cdot X_{B}$.Given: $n_{A} = 1 \ mole$,$n_{B} = 3 \ moles$,$P_{A}^{o} = 200 \ mm \ Hg$,$P_{S} = 500 \ mm \ Hg$.Mole fractions are $X_{A} = \frac{1}{1+3} = 0.25$ and $X_{B} = \frac{3}{1+3} = 0.75$.Substituting the values: $500 = (200 	imes 0.25) + (P_{B}^{o} 	imes 0.75)$.$500 = 50 + 0.75 \cdot P_{B}^{o}$.$450 = 0.75 \cdot P_{B}^{o} \Rightarrow P_{B}^{o} = \frac{450}{0.75} = 600 \ mm \ Hg$.Since $P_{A}^{o} (200 \ mm \ Hg) < P_{B}^{o} (600 \ mm \ Hg)$,component $A$ is the least volatile.

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