Question

The density of a $3M$ sodium thiosulphate solution $(N{a}_2{S}_2{O}_3)$ is $1.25g$ per mL. Calculate (i) the percentage by weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of $N{a}^{+}$ and $S_2{O}_3^{2-}$ ions.

Answer 1

(a) Let us consider $1.0L$ solution for all the calculation
(i) Weight of $1L$ solution $=1250g$
Weight of $N{a}_2{S}_2{O}_3=3\times 158=474g$
$\Rightarrow$ Weight percentage of $N{a}_2{S}_2{O}_3=\frac{474}{1250}\times 100=37.92$
(ii) Weight of $H_{2}O$ in $1L$ solution $=1250-474=776$g
Mole fraction of $N{a}_2{S}_2{O}_3=\frac{3}{3+\frac{776}{18}}=0.065$
(iii) Molality of $N{a}^{+}=\frac{3\times 2}{776}\times 100=7.73m$