Question

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 1

(i) Mass of 1000 mL of `Na_(2)S_(2)O_(3)` solution
`=1.25 xx 1000 = 1250 g`
Mass of `Na_(2)S_(2)O_(3)` in 1000 mL of 3M solution
`= 3 xx` Mol. Mass of `Na_(2)S_(2)O_(3)`
`= 3xx 158 = 474g`
Mass percentage of `Na_(2)S_(2)O_(3)` in solution
`= (474)/(1250)xx100=37.92`
Alternatively, `M=(x xxd xx10)/(m_(A))`
`3=(x xx1.25xx10)/(158)`
`x = 37.92`
(ii) No. of moles of `Na_(2)S_(2)O_(3)=(474)/(158)=3`
Mass of water `=(1250-474)=776g`
No. of moles of water `= (776)/(18)=43.1`
Mole fraxtion of `Na_(2)S_(2)O_(3)=(3)/(43.1+3)=(3)/(46.1)=0.065`
(iii) No. of moles of `Na^(+)` ions
`=2xx` No. of moles of `Na_(2)S_(2)O_(3)`
`= 2xx 3=6`
Molality of `Na^(+)` ion `= ("No. of moles of "Na^(+)"ions")/("Mass of water in kg")`
`=(6)/(776)xx1000`
`=7.73m`
No. of moles of `S_(2)O_(3)^(2-)` ions = No. of moles of `Na_(2)S_(2)O_(3)`
`=3`
Molality of `S_(2)O_(3)^(2-)` ions `= (3)/(776) xx 1000 = 3.86 m`.