# THE density of 3M aqueous solution of sodium thiosulphate is 1.25g/mL. Calculate 1. mole fraction of sodium thiosulphate2. molalities of Na+ and S2O32-

Calculate the mass of given solution using Density = mass /volume
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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