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Vapour pressure of a solvent is the pres...

Vapour pressure of a solvent is the pressure exterted by vapour when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent of nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate. The vapour pressure of a mixture obtained by mixing two valatile liquids is given by `P_(M) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B)` where `P_(A)^(@)` and `P_(B)^(@)` are vapour pressures of pure components `A` and `B` and `X_(A), X_(B)` are their mole fractions in mixture. For solute-solvent system, the relatio becomes `P_(M) = P_(A)^(@).X_(A)` where `B` is non-volatile solute.
A mixture of two volatile liquids `A` and `B` `1` and `3` moels respectively has a `V.P` of `300 mm` at `27^(@)C`. IF one mole of `A` is further added to this solution, the vapour pressure becomes `290 mm` at `27^(@)C`. The vapour pressure of `A` is:

A

`250 mm`

B

`316 mm`

C

`220 mm`

D

`270 mm`

Text Solution

Verified by Experts

The correct Answer is:
A
`P_(T) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B)`
`therefore 300 = P_(A)^(@).(1)/(4)+P_(B)^(@).(3)/(4)`
`290 = P_(A)^(@).(2)/(5)+P_(B)^(@).(3)/(5)`
`therefore P_(A)^(@) = 250 mm`
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