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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

`0.70`

B

`0.50`

C

`0.90`

D

`0.80`

Text Solution

Verified by Experts

The correct Answer is:
A
Relative lowering of vapour pressure = mole fraction of solute (Raoult.s law)
`(p-p_(S))/(p)=X_(2)`
`(p-p_(S))/(p)=(wM)/(mW)`
where w = wt. of solute
M = mol. wt. of solvent
m = mol. wt. of solute
W = wt. of solvent
`0.0125=(wM)/(mW)`
or, `(w)/(mW)=(0.0125)/(18)=0.00070`
Hence, molality = `(w)/(mW)xx1000`
`=0.0007xx1000`
`=0.70`
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