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.

Explanation

(i) Mass of 1000 mL of `Na_2S_2O_3` solution
= 1.25 `xx` 1000 = 1250 g
Mass of `Na_2S_2O_3` in 1000 mL of 3M solution
`474/1250xx100=37.92`
`"Alternatively, "M=(x xxd xx10)/m_A`
`3=(x xx1.25xx10)/158` `x=37.92`
(ii) No. of moles of `Na_2S_2O_3=474/158=3`
Mass of water = (1250 – 474) = 776 g
No. of moles of water = `776/18` = 43.1
Mole fraction of `Na_2S_2O_3`
`=3/(43.1+3)=3/46.1=0.065`
(iii) No. of moles of `Na^+` ions
`=2xx"No. of moles of "Na_2S_2O_3`
`2xx3=6`
Molality of `Na^+` ions
`=("No. of moles of "Na^+ions)/("Mass of water in kg")=6/76xx100=7.73m`
No. of moles `S_2O_3^(2-)`ions = No. of moles of `Na_2S_2O_3=3`
`"Molarity of "S_2O_3^(2-)ions=3/776xx1000=3.86m`