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

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

Grade:12

4 Answers

Arun
25750 Points
5 years ago

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
184 Points
5 years ago
 

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
184 Points
5 years ago

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
82 Points
5 years ago
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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