The density of a `3 M Na_(2) S_(2) O_(3)` (sodium thiosulphate) solution is `1.25 g mL^(-1)`. Calculate:
a. % by weight of `Na_(2) S_(2) O_(3)`
b. Mole fraction of `Na_(2) S_(2) O_(3)`
c. Molalities of `Na^(o+)` and `S_(2) O_(3)^(2-)` ions.

Answer

(i)`30%,(ii)0.046,(iii)Na^(+)=5.42m,S_(2)O_(3)^(2-)=2.71m`

Explanation

(i) Let us consider one litre of `Na_(2)S_(2)O_(3)` solution.
`therefore wt.` of the solution `="density"xx"volume"(ml)`
`=1.58xx1000=1580gram`
`wt`. Of `Na_(2)S_(2)O_(3)` present in `1` litre of the solution
`="molarity"xx"mol". Wt`
`=3xx158=474 gram`
`wt %` of `Na_(2)S_(2)O_(3)=(474)/(1580)xx100=30%`
(ii)`wt.` of solute `(Na_(2)S_(2)O_(3))=474 gram`
mole of solute`=(474)/(158)=3`
`wt.` of solvent `(H_(2)O))=1580-474=1106` gram moles of solvent `=(1106)/(18)=61.444`
`:.` mole fraction of `Na_(2)S_(2)O_(3)=(3)/(3+61.44)=0.046`
(iii) Molality of `Na_(2)S_(2)O_(3)=("moles of "Na_(2)S_(2)O_(3))/("wt. of solvent in grams")xx1000=(3)/(1106)xx1000=-2.71`
`because 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.
`therefore` molality of `Na^(+)=2xx2.71=5.42m`
molality of `S_(2)O_(3)^(2-)=2.71m`