`29.2% (w//w) HCl` stock, solution has a density of `1.25 g mL^(-1)`. The molecular weight of `HCl` is `36.5 g mol^(-1)`. The volume `(mL)` of stock solution required to prepare a `200 mL` solution of `0.4 M HCl` is :

Explanation

Stock solution of HCl `=29.2%(w//w)`
Thus, 29.2 g of HCl are present in 100 g of the solution
As density of the solution `="1.25 g mL"^(-1)`,
`"volume of 100 g of the solution "=(100)/(1.25)"mL"`
As molar mass of HCl `="36.5 g mol"^(-1)`, molarity of the solution
`=(29.2)/(36.5)xx(1.25)/(100)xx1000=10M`
`underset("(stock solution)")(M_(1)V_(1))=underset("(solution required)")((M_(2)V_(2)))`
`10xxV_(1)=0.4xx200" or "V_(1)="8 mL"`