If at a particular temperature, the density of `18MH_(2)SO_(4)` is `1.8g cm^(-3)`, calculate (a) molality, (b) `%` concentrating by weight of solute and solvent (c) mole fraction of water and `H_(2)SO_(4)`.

Answer

(a) `500` molal (b) `2%` (c) `0.1`

Explanation

(a) Molarity of the solution `M=18` mol `L^(-1)` solution
hence, moles of solute `(H_(2)SO_(4))n_(1)=18` mol,
mass of solute `w_(1)=18xx98g`
mass of solution `(w_(1)+w_(2))=1000xx1.8=1800g`
mass of water (solvent `w_(2)`) `=(1800-18xx98)=36g`
We know molarity (conc. in mol `kg^(-1)` solvent)
`=(1000w_(1))/(m_(1)w_(2))=(1000xx18xx98)/(98xx36)=500` molal
(b) `%` concentration by weight of solute
`=(18xx98)/(18xx98+36)xx100=(18xx98)/(1800)xx100=98%`
`%` concentration by weight of solvent `(36)/(1800)xx100=2%`
(c) `X_(H_(2)SO_(4))=(18)/(18+2)=(18)/(20)=0.9`
`X_(H_(2)SO_(4))=(2)/(0.20)=0.1`