1.5 g of chalk was treated with 10 " mL of " 4 N HCl. The chalk was dissolved and the solution was made to 100 mL. 25 " mL of " this solution required 18.75 " mL of " 0.2 N NaOH solution for comjplete neutralisation. Calculate the percentage of pure `CaCO_3` in the sample of chalk.

Explanation

Excess of HCl in `25 mL=18.75xx0.2mEq`
Excess of HCl in 100 mL`=18.75xx0.2xx4=15mEq`
Total HCl used `=10xx4=40mEq`
HCl used `=40-15=25mEq=25mEq` of `CaCO_3`
`=25xx10^(-3)xx50g of CaCO_3=1.25g`
`% of CaCO_3=(1.25xx100)/(1.5)=83.35%`