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.

Explanation

a. `Mw of Na_(2) S_(2) O_(3) = 23 xx 2 + 32 xx 2 + 16 xx 3`
`= 46 + 64 + 48 = 158 g`
`M = (% "by weight" xx 10 xx d)/(Mw_(2))`
`3 = (% "by weight" xx 10 xx 1.25)/(150)`
% by weight `= (3 xx 158)/(10 xx 1.25) = 37.92 %`
b. Weight of solute `= 37.92 g`
Weight of solution `= 100 g`
Weight of solvent `= 100 - 37.92 = 62.08 g`
`X_(2) = (W_(2)//Mw_(2))/((W_(1))/(Mw_(1))) + (W_(2))/(mw_(2)) = (37.92//158)/((62.08)/(18) + (37.92)/(158))`
`= (0.24)/(3.44 + 0.24) = (0.24)/(3.68) = 0.065`
c. `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (37.92 xx 1000)/(158 xx 62.08) = 3.86 m`
`Na_(2)S_(2)O rarr 2 Na^(o+) + S_(2) O_(3)^(2-)`
`1m` `2m` `1m`
`3.86 m` `3.86 xx 2 m` `3.86 m`
`m` of `Na^(o+) = 7.72 m`
`m` of `S_(2) O_(3)^(2-) = 3.86 m`