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Answer 1

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Answer 2

» The density of a 3 M sodium thiosulphate...
The density of a 3 M sodium thiosulphate solution (Na2S2O3) is 1.25 g per ml. Calculated (i) the percentage by weight of sodium thiosulphate, (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of Na+ and S2O32- ions.
4 years ago
Answers : (2)
(i) Mole fraction = Moles of substance/Total moles
(ii) 1 mole of Na2S2O3 gives 2 moles of Na+ and 1 mole S2O32-
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 S2O2-3
Molality of Na2S2O3 = 3 * 1000/776 = 3.87
Molality of Na+ = 3.87 * 2 = 7.74m
Molality of S2O2-3 = 3.87m