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

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Solution

ate the mass of given s

Mass of 1000 ml of 3M solution will be given by density x volume = 1.25 g ml-1 x 1000 ml = 1250 g
Mass of Na2S2O3 in 1000 ml solution = molar mass of Na2S2O3 x number of moles of solute
= 158 g/mol x 3 mol
= 474 g
Mass of the solvent = mass of solution - mass of Na2S2O3
= 1250 g - 474 g =776 g
Moles of Na2S2O3 = 3 mol ( 3 M) given
Moles of solvent water = mass of water/molar mass of water = 776 g/18 gmol-1 = 43.1 mol
Total number of moles in solution = 3 + 43.1 = 46.1 mol
Mole fraction of Na2S2O3 = moles of Na2S2O3 /total moles = 3/46.1 = 0.065

Molality of solution = moles of solute/solvent in kg
= 3 mol / 0.776 kg = 3.865 mol⋅kg-1

1 mole of Na2S2O3 gives 2 moles of Na+ So, molality of Na+ is 2 x 3.865 = 7.73 mol.kg-1
1mole of Na2S2O3 gives 1 moles of S2O32-So, molality of S2O32- is 3.865mol.kg-1

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