Calculate mole fraction of solute in an aqueous 4m solution assuming the density of the solution as `1.0g//mL`

Answer

`0.0674`

Explanation

(i) Let us consider one litre of sodium thiosulphate solution
`:.` wt of the solution `=` density `xx` volume (mL)
`=1.25xx1000=1250g`
Wt. of `Na_(2)S_(2)O_(3)` present in 1L of the solution
`=` molarity `xx` mol.wt.
`=3xx158=474g`
Wt `%` of `Na_(2)S_(2)O_(3)=(474)/(1250)xx100=37.92%`
(ii) Wt. of solute `(Na_(2)S_(2)O_(3))=474g`
Moles of solute `=(474)/(158)=3`
Wt. of solvent `(H_(2)O)=1250-474=776g`
Moles of solvent `=(776)/(18)=43.11`
`:.` mole fraction of `Na_(2)S_(2)O_(3)=(3)/(3+43.11)=0.063`
(iii) Molality of `Na_(2)S_(2)O_(3)= ("moles of " Na_(2)S_(2)O_(3))/("wt.of solvent in grams")xx1000`
`=(3)/(776)xx1000=3.865m`
`:.1` mole of `Na_(2)S_(2)O_(3)` contains 2 moles of `Na^(+)` ions and 1 mole of `S_(2)O_(3)^(2-)` ions
`:.` molality of `Na^(+)=2xx3.865=7.73m`
Molality of `S_(2)O_(3)^(2-)=3.865m`