1.

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

Answer»

`0.70`
`0.50`
`0.60`
`0.80`

Solution :Relative LOWERING of vapour pressure of an aqueous solution CONTAINING nonvolatile solute is EQUAL to mole fraction of solute.
`THEREFORE (P^(@)-P_(s))/(P^(@))=(n)/(n+N)=0.0125`
`rArr (n+N)/(n)=(1)/(0.0125)rArr (N)/(n)=(1)/(0.0125)-1=(0.9875)/(0.0125)`
`therefore (N)/(n)=(0.9875)/(0.0125)`
Now, MOLALITY `=(0.0125xx1000)/(0.9875xx18)=0.70`.


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