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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.
The amount of solute `("mol. wt. 60")` required to dissolve in `180 g` water to reduce the vapour pressure to `4//5` of the pure water:

A

`120 g`

B

`150 g`

C

`200 g`

D

`60 g`

Text Solution

Verified by Experts

The correct Answer is:
B
`(P^(@)-P_(S))/(P_(S)) = (n)/(N)` or `(P^(@)-(4P^(@))/(5))/((4P^(@))/(5)) = (w xx 18)/(60 xx 180)`
`therefore w = 150 g`
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