Question

one litre solution of 0.5N is heated. the volume of the solution is reduced to 750cc and 2.675 gmof hcl is lost. Calculate: i) normality of the resultant solution ii) number of meq of hcl in 100cc of the original solution.

Answer 1

The original 0.5N HCl solution of 1 litre had 18.225 g of HCl. On heating the solution to a reduced volume of 750 cc and losing 2.675 g of HCl in the process, the normality of the solution was altered.
Initially, 18.225 g of HCl was dissolved in 1 litre of solution.
After the heating, mass of HCl dissolved in the reduced 750 mL of solution was found to be (18.225 - 2.675) g = 15.55 g
i) Hence, normality of this newly obtained solution = Equivalent weight of HCl mass / Volume of solution in litre
= (15.55 g/ Equivalent weight of HCl) / (750 mL/1000 mL per litre)
= (15.55 g/ 36.45 g per Mole Equivalent) / 0.75 litre
= 0.4266 MEq / 0.75 L
= 0.5688N

ii) Since 1 litre (or 1000 mL) of the original solution had 0.5 MEq of HCl, number of MEq of HCl in 100 cc (or 100 mL) of the same solution is (0.5/1000)*100
= 0.05