A certain weight of pure CaCO3 is made to react completely with 200 ml of an HCl solution to give 224 ml of CO2 gas at STP. The normality of the HCl is:- 0.05 N 0.1 N 1.0 N 0.2 N.
The reaction for the process will be as : CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + CO2(g) + H2O(l)
The molar ratio of HCl :CaCO3 = 2:1
Now, moles of CO2 = volume in liters / 22.4 Lmol-1 = 0.224 L / 22.4 Lmol-1 = 0..01 moles CO2
So, moles of HCl will be = 2 x moles of CO2 = 2 x 0.01 moles = 0.02 moles HCl ( since molar ratio is 2:1)
Molarity of HCl can be calculated as = moles of solute / volume of solution in liters = 0.02 moles / 0.200 L = 0.1 M HCl
Finally, Normality = molarity x nequivalents ( n = number of hydrogen ions in the acid HCl which is only one i.e. H+)
=> Normality= 0.1 x 1 = 0.1 N
Hence, the required answer is 0.1 N
The molar ratio of HCl :CaCO3 = 2:1
Now, moles of CO2 = volume in liters / 22.4 Lmol-1 = 0.224 L / 22.4 Lmol-1 = 0..01 moles CO2
So, moles of HCl will be = 2 x moles of CO2 = 2 x 0.01 moles = 0.02 moles HCl ( since molar ratio is 2:1)
Molarity of HCl can be calculated as = moles of solute / volume of solution in liters = 0.02 moles / 0.200 L = 0.1 M HCl
Finally, Normality = molarity x nequivalents ( n = number of hydrogen ions in the acid HCl which is only one i.e. H+)
=> Normality= 0.1 x 1 = 0.1 N
Hence, the required answer is 0.1 N