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)`

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`