Calculate the normality sulphuric acid solution if mole fraction of water in sulphuric acid solution is 0.85

We have mole fraction of solute + solvent = 1

Here, Mole fraction of water is 0.85 so mole fraction of sulphuric acid will be (1-0.85) = 0.15

Hence each mole of the solution, 0.15 mole of sulphuric acid is dissolved .

So mass of solvent (water) is ,

For 1 mole of water (H2O) = 18 gm

So, 0.85 mole of water = 18 x 0.85 = 15.3 gm =0.0153 Kg

Thus, Molality (m) = no. of moles of solute (Sulphuric acid)/ mass of solvent in Kg

Hence, Molality of the solution (m) = 0.15 / 0.0153 = 9.8 m (nearly equal to 10)

Hence each mole of the solution, 0.15 mole of sulphuric acid is dissolved .

So mass of solvent (water) is ,

For 1 mole of water (H2O) = 18 gm

So, 0.85 mole of water = 18 x 0.85 = 15.3 gm =0.0153 Kg

Thus, Molality (m) = no. of moles of solute (Sulphuric acid)/ mass of solvent in Kg

Hence, Molality of the solution (m) = 0.15 / 0.0153 = 9.8 m (nearly equal to 10)