The density of a 3 M sodium thiosulphate solution $(N a_{2} S_{2} O_{3})$ is $1.25 g / m L$ . Calculate (i) the percentage by mass of sodium thiosulphate, (ii) the mole fraction of sodium thiosulphate and (iii) molalities of $N a^{+}$ and $S_{2} O_{3}^{2 -}$ ions.

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Solution

(i) Mass of 1000 mL of $N a_{2} S_{2} O_{3}$ solution = $1.25 \times 1000 = 1250 g$
Mass of $N a_{2} S_{2} O_{3}$ in 1000 mL of 3 M solution
$= 3 \times M o l m a s s o f N a_{2} S_{2} O_{3}$
$= 3 \times 158 = 474 g$
Mass percentage of $N a_{2} S_{2} O_{3}$ in solution
$= \frac{474}{1250} \times 100 = 37.92$
Alternatively, $M = \frac{x \times d \times 10}{m_{A}}$
$3 = \frac{x \times 1.25 \times 10}{158}$
$x = 37.92$
(ii) No. of moles of $N a_{2} S_{2} O_{3} = \frac{474}{158} = 3$
Mass of water $= (1250 - 474) = 776 g$
No. of moles of water $= \frac{776}{18} = 43.1$
Mole fraction of $N a_{2} S_{2} O_{3} = \frac{3}{43.1 + 3} = \frac{3}{46.1} = 0.065$
(iii) No. of moles of $N a^{+}$ ions
$= 2 \times N o . o f m o l e s o f N a_{2} S_{2} O_{3}$
$= 2 \times 3 = 6$
Molality of $N a^{+}$ ions $= \frac{N o . o f m o l e s o f N a^{+} i o n s}{M a s s o f w a t e r i n k g}$
$= \frac{6}{776} \times 1000$
$= 7.73 m$
No. of moles of $S_{2} O_{3}^{2 -}$ ions = No. of mole of $N a_{2} S_{2} O_{3}$
$= 3$

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