Question

the density of solution containing 13% by mass of sulphuric acid is 1.09g/ml calculate normality of solution

Answer 1

Given that  

13% by mass of sulphuric acid  

Let us aussme that 1000 g of solution.

13 % of 1000g solution = 130 g of H2SO4

number of moles = amount in g / molar mass

moles of H2SO4 =  130 g H2SO4 / 98 g/mole = 1.33 moles

density = mass / volume

so, volume = mass / density  

=1000 g solution / 1.09  g/mL = 917.4  mL of solution = 0.917 L

molarity = number of moles / volume in L

Molarity = 1.33  moles / 0.917 L

= 1.45 M

normality = molarity X n-factor = molarity X 2 = 1.45  X 2 = 2.9 N