Question

The density of $3M$ solution of sodium thiosulphate (${Na}_2{S}_2{O}_3)$ is $1.56g/mL$. Calculate
(i) amount of ${Na}_2{S}_2{O}_3$ in % w/w
(ii) mole fraction of ${Na}_2{S}_2{O}_3$
(iii) Molality of ${Na}^{+}$ and $S_2{O}_3^{2-}$ ions

Answer 1

$3$ moles of $N{a}_2{S}_2{O}_3=3\times 158=474g$

Volume= $1000$

$\rho =1.56g/ml$, weight= $1560g$

(i) Weight %= $\frac{474}{1560}=30.38$

(ii) Mole fraction, moles of solvent= $\frac{W}{18}=\frac{1560-474}{18}=60.83$ moles

Mole fraction of $N{a}_2{S}_2{O}_3=\frac{3}{3+60.83}=0.046$

(iii) Molality= $\frac{3}{1.086}=2.76$

Molality of $N{a}^{+}=2\times 2.76=5.52m$

Molality of $S_2{O}_3^{2-}=2.76m$