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For a dilute solution containing 2.5 g o...

For a dilute solution containing `2.5 g` of a non-volatile non-electrolyte solution in `100g` of water, the elevation in boiling point at `1` atm pressure is `2^(@)C`. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of `Hg)` of the solution is:
(take `k_(b) = 0.76 K kg mol^(-1))`

A

`724`

B

`740`

C

`736`

D

`718`

Text Solution

Verified by Experts

The correct Answer is:
A
We know that
`Delta T_(b) = K_(b) xx "molality"`
Also from Raoult's law,
`(P^(@)-P_(S))/(P_(S)) = (n)/(N) = (n xx M)/(W) = (n xx M xx 1000)/(W xx 1000)`
or `(P^(@) - P_(S))/(P_(S)) = ("molality" xx M)/(1000)`
`(P^(@)-P_(S))/(P_(S)) = (Delta T_(b))/(K_(b)) xx (M)/(1000)`
`(760-P_(S))/(P_(S)) = (2)/(0.76) xx (18)/(1000)`
`(760-P_(S))/(P_(S)) = 0.0474`
`P_(S) = 725.6 mm Hg`
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