JEE Main · 2007 · 29 Apr 2007 · Offline · Solutions

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A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K . The vapour pressure of propyl alcohol is 200 mm . If the mole fraction of ethyl alcohol is 0.6 , its vapour pressure (in mm ) at the same temperature will be

(A)350✓ Correct
(B)300
(C)700
(D)360

Answer: (A)

Solution

By Raoult's law the total vapour pressure is

P = P_{ethyl}^{\circ} x_{ethyl} + P_{propyl}^{\circ} x_{propyl}

. Substituting

290 = 0.6 P_{ethyl}^{\circ} + 200 \times 0.4

gives

0.6 P_{ethyl}^{\circ} = 210

. Hence the vapour pressure of pure ethyl alcohol is

\frac{210}{0.6} = 350

mm.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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