Question

The density of $3M$ solution of sodium thiosulphate $\left(N{a}_2{S}_2{O}_3\right)$ is $1.25$ g/mL.
Which of the following option(s) is/are correct?

• A. Weight % of $N{a}_2{S}_2{O}_3$ is $37.95\%$
• B. Molality of $N{a}_2{S}_2{O}_3$ is $3.865m$
• C. Molality of $N{a}^{+}$ is $3.865m$
• D. Mole fraction of solute is $0.063$

Answer 1

Answer: (A), (B), (D)

Mole fraction of solute is $0.063$

(i) Let us consider one litre of sodium thiosulphate solution.
$\therefore$ Wt. of the solution = density $\times$ volume (mL)
$=1.25\times 1000=1250g$
Wt. of $N{a}_2{S}_2{O}_3$ present in $1$ L of the solution
$=$ molarity $\times$ mol. wt.
$=3\times 158=474g.$
$\text{Wt. \% of}N{a}_2{S}_2{O}_3=\frac{474}{1250}\times 100=37.95\%$
(ii) Wt. of solute $\left(N{a}_2{S}_2{O}_3\right)=474g.$
Moles of solute $=\frac{474}{158}=3$
$\text{Wt. of solvent}\left(H_{2}O\right)=1250-474=776g$
$\text{Moles of solvent}=\frac{776}{18}=43.11$
$\therefore \text{Mole fraction of}N{a}_2{S}_2{O}_3=\frac{3}{3+43.11}=\mathrm{0.063.}$
(iii) Molality of $N{a}_2{S}_2{O}_3=\frac{\text{moles of}N{a}_2{S}_2{O}_3}{\text{wt. of solvent in grams}}\times 1000$
$=\frac{3}{776}\times 1000=3.865m$
$\because$ 1 mole of $N{a}_2{S}_2{O}_3$ contains $2$ moles of $N{a}^{+}$ ions and $1$ mole of $S_2{O}_3^{2-}$ ions.
$\therefore \text{molality of}N{a}^{+}=2\times 3.865=7.76m$
$\text{Molality of}{S}_2{O}_3^{2-}=3.865m$