The density of 3M aqueous solution of sodium thiosulfate is 1.25g/mL. Calculate (1) mole fraction of sodium thiosulfate. (2) molalities of Na+ and S2O3 2- ions.

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Solution

(i) Mole fraction = Moles of substance/Total moles
(ii) 1 mole of Na₂S₂O₃ gives 2 moles of Na⁺ and 1 mole S₂O₃^2-
Molecular wt. of sodium thiosulphate solution (Na₂S₂O₃) = 23 * 2 + 32 * 2 + 16 * 3 = 158
(i) The percentage by weight of Na₂S₂O₃
= wt of Na₂S₂O₃/wt of solution * 100 = 3 * 158 * 100/100 * 1.25 = 37.92
[wt. of Na₂S₂O₃ = Molarity * Molwt]
(ii) Mass of 1 litre solution = 1.25 * 1000 g = 1250 g
[∵ density = 1.25g/l]
Mole fraction of Na₂S₂O₃
= Number of moles of Na₂S₂O₃/Total number of moles
Moles of water = 1250 – 158 3/18 = 43.1
Mole fraction of Na₂S₂O₃ = 3/3 + 43.1 = 0.065
(iii) 1 mole of sodium thiosulphate (Na₂S₂O₃) yields 2 moles
of Na⁺ and 1 mole of S₂O^2-₃
Molality of Na₂S₂O₃ = 3 1000/776 = 3.87
Molality of Na⁺ = 3.87 * 2 = 7.74m
Molality of S₂O^2-₃ = 3.87m

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