A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300K. 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

Options

A. 350
B. 300
C. 360
D. 700

Answer

A

Explanation

Mole fraction of ethyl alcohol = 0.6
Mole fraction of methyl alcohol = 0.4
A rarr Ethyl alcohol & B rarr Methyl alcohol
Then `P_(T) = P_(A)^(@) X_(A) + P_(B)^(@) X_(B)`
`290 = (P_(A)^(@) xx 0.60)+(200 xx 0.40) `
` P_(A)^(@)`= (290-80)/0.60=350 mm