Grade 12Physical Chemistry
Calculate the normality sulphuric acid solution if mole fraction of water in sulphuric acid solution is 0.85
4 Answers
Arun
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)
Regards
Arun
Gitanjali Rout
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)
Gitanjali Rout
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)
Abhiraj Pathak
Normality = gm equivalent of solute/Vol of solution (L)Considering 100g of water, i.e. 100/18 = nb =5.55 moles of water,xb = 0.85 = nb/(nb+na) On solving we get na = 0.979, i.e. 95.98 g of sulphuric acid.now gm equivalent of H2SO4 = given mass/equivalent mass = 95.98/49 = 1.95Now until we know the density of the solution we cannot proceed without taking certain assumptions
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