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.

Options

A. The `%` weight of `Na_(2)S_(2)O_(3) is 37.92`
B. The mole fraction of `Na_(2)S_(2)O_(3) is 0.065`
C. The molalityof `Na^(+) is 8.732`
D. The molality of `S_(2)O_(3)^(2-) is 3.866`

Answer

A, B, D

Explanation

Given, `M_(NaS_(2)O_(2))=3`
`:.` Moles of `Na_(2)S_(2)O_(3)=3`
wt.of `Na_(2)S_(2)O_(3)=3xx158=474g`
Wt. of solution `1000xx1.24=1250 g`
volume of solution `=1000mL`
`:.` wt.of.water`=1250-474=776 g`
Now `%` by weight of `Na_(2)S_(2)O_(3)`
`=(wt.of Na_(2)S_(2)O_(3))/("wt.of solution")xx100`
`=(474)/(1250)xx100`
Mole fraction of `Na_(2)S_(2)O_(3)`
`=("mole of"Na_(2)S_(2)O_(3))/("moles of"Na_(2)S_(2)O_(3)+"moles of"H_(2)O)`
`=(3)/(3+((776)/(18))=0.065)`
Molality of `Na^(+)=("moles of"Na^(+))/("wt.of water"("in"kg))`
`=(6)/(776//1000)=7.732 m`
Molality of `S_(2)O_(3)^(2-)=(3)/(776//1000)`
`=3.886 m`