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

Explanation

(i) 3 moles `Na_(2)S_(2)O_(3)=3xx(2xx23+2xx32+3xx16)g=3xx168=474g`
1 litre solution = 1000 cc = 1250 g.
Hence, wt `%=(474//1250)xx100=37.92`
(ii) Water = `1250-776g=776//18 " mole"=43.11" moles"`
`"Mole fraction"=3//(3+43.11)=0.065`
(iii) `"Molality of "Na_(2)S_(2)O_(3)=("3 mol")/(0.776kg)=3.866`
`"Molality of "Na^(+)=2xx3.866," Molality of "S_(2)O_(3)^(2-)=3.866`