Question

The density of a solution containing 13% by mass of sulphuric acid is 1.09g/ml.calculate the molarity and normality of the solution.

Answer 1

• Molarity = moles of solute in 1 litre of solution .
• Density of solution = 1.09g/ml

This means 1.09 g of solution is present in 0.001 L .

1.09 in 0.001L

so 1.09 * 1000 in 1L

Mass of solution in its 1 L = 1.09 * 1000 g

• mass of sulphuric acid = 13% of mass of solution

thus, mass of sulphuric acid present in 1L of solution =0.13*1.09*1000= 13*10.9g

• moles of sulphuric acid in 1L of solution= mass/molar mass = 13*10.9/98 = 1.446

So, molarity = 1.446 M

normality = molarity X n-factor = molarity X 2 = 1.446 X 2 = 2.892 N

n-factor for an acid = no.of ionizable H+ ....in case of H2SO4 = 2

Hope you understood