THE density of 3M aqueous solution of sodium thiosulphate is 1.25g/mL. Calculate 1. mole fraction of sodium thiosulphate

2. molalities of Na⁺ and S₂O_3^2-

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 S₂O₃^2- So, molality of S₂O₃^2- is 3.865mol.kg^-1