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.

Answer

(i) 37.92 (ii) 0.065 (iii) 7.73m

Explanation

(i) 37.92 (ii) 0.065 (iii) 7.73m
Let us consider 1.0 L solution for all the calculation.
(i) Weight of 1 L solution = 1250 g
Weight of `Na_(2)S_(2)O_(3) = 3 xx 158 = 474 g rArr` Weight percentage of `Na_(2)S_(2)O_(3) = (474)/(1250) xx 100 = 37.92`
Weight of `H_(2)O` in 1 L solution = 1250 - 474 = 776 g
Mole fraction of `Na_(2)S_(2)O_(3) = (3)/(3+(776)/(18))=0.065`
(iii) Molality of `Na^(+) = (3 xx 2)/(776) xx 1000 = 7.73 m`