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`        The density of a 3 M sodium thiosulphate solution (Na2S2O3) 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 S2O_3^(2-) ions.`
3 years ago

396 Points
```										Sol. Let us consider 1.0 L solution for all the calculation.
(i) Weight of 1 L solution = 1250 g
Weight of Na2S2O3 = 3 × 158 = 474 g
⇒ Weight percentage of Na2S2O3 = 474/1250 × 100 = 37.92
(ii) Weight of H2O in 1 L solution = 1250 – 474
= 776 g
Mole-fraction = 3/(3+ 776/18) = 0.065
(iii) Molality of Na+ = (3 ×2)/776 × 100 = 7.73 m
Molality of S2O_3^(2-) = (3 )/(776 ) × 1000 = 3.86 m

```
3 years ago
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