One litre of a solution of N/2 HCl was heated in beaker and it was observed that when the volume of solution got reduce to 600 mL, 3.25 g of HCl was lost. Calculate the normality of the resulting solution.

Explanation

Normality of solution `= ("Mass of HCl/Equivalent mass")/("Volume of solution litres")`
`("0.5 equiv.L"^(-1))=("Mass of HCl")/(("36.5 g equiv"^(-1))xx("1 L"))`
Mass of HCl `= ("0.5 equiv. L"^(-1))xx("36.5 g equiv"^(-1))xx("1 L")=18.25g`
Mass of HCl lost on heating = 3.25 g
Mass of HCl left `= 18.25 - 3.25 = 15.0 g`
Volume of resulting solution `= 600 mL = (600)/(1000) = 0.6 L `
Normality (N) `= ("Mass of acid/Gram equiv. of acid")/("Volume of solution in litres")`
`=((15.0g)//("36.5 g equiv"^(-1)))/(("0.6 L"))=0.685 "equiv. L"^(-1)=0.685N`.