(a) A solution is prepared by dissolving `3.65 g` of `HCl` in 500 mL of the solution. Calculate the normality of the solution
(b) Calculate the volume of this solution required to prepare 250 mL of 0.05 N solution.

Explanation

Step I. Calculation of normality of solution
Mass of `HCl = 3.65 g`
Molar mass `HCl = 1+35.5 = 36.5 u = 36.5 "g mol"^(-1)`
Basicity of `HCl` from the formula =1
`:.` Equivalent mass `HCl=("Molar mass")/("Bascity")=((36.5"g mol"^(-1)))/(1)=(("3.65 g"))/(("36.5 g equiv"^(-1)))=0.1 "equiv"`
Volume of solution in litres `= 500 mL = (500)/(1000)=0.5 L`
Normality of solution `(N)=("No. of equivalent of HCl")/("Volume of solution in litres")=("0.1 equiv.")/(("0.5 L"))="0.2 equiv L"^(-1)=0.2N`
Step II. Calculation of volume of concentrated `HCl` solution required
Normality of conc. `HCl = (N_(1))=0.2 N`
Let the volume of conc. HCl needed `= V_(1)`
Normality of dilute `HCl (N_(2))=0.05 N`
Volume of dilute `HCl (V_(2))=250 mL`
Applying normality equation : `N_(1)V_(1) -= N_(2)V_(2)`
`(0.2 N)xxV_(1) = (0.05 N) xx ("250 mL")`
`V_(1)=((0.05N)xx(250 mL))/((0.2 N))=62.5mL`.