Question

The density of a $3 \,M$ sodium thiosulphate solution $(Na_2S_2O_3 )$ is $1.25 \,g$ per mL. Calculate (i) the percentage by weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of $Na^+$ and $S_2O_3^{2-}$ ions.

Answer 1

(a) Let us consider $1.0 \,L$ solution for all the calculation
(i) Weight of $1\, L$ solution $= 1250\, g$
Weight of $Na_2S_2O_3 = 3\times 158 = 474\, g $
$\Rightarrow $ Weight percentage of $ Na_2S_2O_3 = \frac{ 474}{1250} \times 100 =37.92 $
(ii) Weight of $H_2O$ in $1\, L$ solution $= 1250-474 = 776\, $g
Mole fraction of $ Na_2S_2O_3 = \frac{3}{ 3+ \frac{776}{18} } = 0.065$
(iii) Molality of $Na^+ = \frac{3 \times 2}{776} \times 100 = 7.73\, m $