Question

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

Answer 1

c. As we know
`(P^(@) - P)/P^(@) = chi_(2)` = mole fraction of solute
The ration `(P^(@)-P)//P^(@)` is the relative lowering of vapour pressure, which is equal to `0.0125` here.
`:. chi_(2) = 0.0125`
The relation between `m` and `chi` is:
`m=(chi_(2) xx 1000)/(chi_(1) xx Mw_(1)) = (0.0125 xx 1000)/(( 1- 0.0125 xx 18))` `[{:(,Mw_((H_(2)O))),(=,18g mol^(-1)):}]`
` = (0.0125 xx 1000)/(0.9875 xx 18)`
=`0.70`
` :. m =0.70`