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Q.

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

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a

700

b

360

c

350 

d

300

answer is A.

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Detailed Solution

Using the expression  p=xApA+xBpB; (A is ethyl alcohol and B is propyl alcohol)

we get 290mmHg=0.6pA+0.4×200mmHg

This give pA=(290−0.4×200)mmHg0.6=350mmHg

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