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The relative lowering of vapour pressure...

The relative lowering of vapour pressure of an aqueous solution containing non - volatile solute is 0.0125 . The molality of the solution is

A

a)`0.70`

B

b)`0.50`

C

c)`0.60`

D

d)`0.80`

Text Solution

Verified by Experts

The correct Answer is:
A
According to Rault.s law , Relative lowering of vapour pressure `prop` mole fraction of solute Thus , mole fraction of solute = 0.0125
Mole fraction of a solute is related to the molality by the following expression .
`((1)/(X)-1)=(1000)/(m_(B)xxm)`
where , X= mole fraction of solute
`m_(B)=` molecular weight of solvent
m= molality
`((1)/(0.0125-1))=(1000)/(18xxm)`
`m=(12.5)/((1-0.0125)xx18)`
`=(12.5)/(17.775)`
`=0.70`
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