Question

The density of $3 \mathrm{M}$ sodium thiosulphate solution $\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)$ is $1.25 \mathrm{g} / \mathrm{ml}$. Calculate i) the percentage by mass of sodium thiosulphate, ii) the mole fraction of sodium thiosulphate and iii) molalities of $\mathrm{Na}^{+}$ and $\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}$ ions.

Answer 1

From density we will find out the mass of the solution. Then, mass% from it by dividing the mass found by total mass and multiplying with 100.
Given is 3 M.
Molality of $\mathrm{Na}^{+}$ ions $=$ (No. of moles of $\mathrm{Na}^{+}$ ions/ Mass of water in $\mathrm{kg}$ ) $\mathrm{X} 1000 .$ Mole fraction of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}=($ Mole of $\left.\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right) /\left(\right.$ Mole of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}+$ Mole of water $)$

Complete step by step solution:
-The molar mass to be calculated of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ is $158 \mathrm{g}$.
i) Mass of $1000 \mathrm{mL}$ of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ solution $=1.25 \times 1000=1250 \mathrm{g}$
Mass of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ in $1000 \mathrm{ml}$ of $3 \mathrm{M}$ solution $=3 \mathrm{X}$ Molar mass of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$

$$=3 \mathrm{X} 158 \mathrm{g}=474 \mathrm{g}$$

Mass percentage of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ in solution $=(474 / 1250) \mathrm{X} 100$

$$=37.92 \%$$

ii) No. of moles of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}=\left(3 \mathrm{X}\right.$ Molar mass of $\left.\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right) /$ molar mass of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$

$$=474 / 158$$

= 3.
​Mass of water = (1250−474) = 776g
Now, the molar mass of water as we know is 18g.
Thus, No. of moles of water = (776 / 18)g
= 43
Mole fraction of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}=\left(\right.$ Mole of $\left.\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right) /\left(\right.$ Mole of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}+$ Mole of water $)$

$$=3 /(43+3)=0.065$$

iii) Number of $\mathrm{Na}^{+}$ ions $=2 \mathrm{X}$ No. of moles of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$

$$=2 \times 3=6$$

Molality of $\mathrm{Na}^{+}$ ions $=\left(\right.$ No. of moles of $\mathrm{Na}^{+}$ ions $/$ Mass of water in $\left.\mathrm{kg}\right) \mathrm{X} 1000$

$$=(6 / 776) \times 1000=7.73 \mathrm{m}$$

Number of Moles of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ ions $=$ number of moles of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$

$$=3$$

Molality of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ ions $=\left(\right.$ No. of moles of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ ions $/$ Mass of water in $\left.\mathrm{kg}\right) \mathrm{X} 1000$

$$=(3 / 776) \times 1000=3.86 \mathrm{m}$$

The answers are i)37.92 %, ii)0.065 iii)7.73 m, 3.86 m.

Note: Molality of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ ions $=\left(\right.$ No. of moles of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ ions $/$ Mass of water in $\left.\mathrm{kg}\right) \mathrm{X}$
1000. Molality of $\mathrm{Na}^{+}$ ions $=$ (No. of moles of $\mathrm{Na}^{+}$ ions $/$ Mass of water in $\mathrm{kg}$ ) $\mathrm{X} 1000$