`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 :

Answer

8

Explanation

29.2% (w/w) HCl means that 29.2 g of HCl is present in 100 g of solution.
Volume of HCl solution `=100/1.25 = 80`g
Molarity of solution `=(29.2 //36.5)/80 xx 1000`
= 10 M
Applying `M_(1)V_(1) = M_(2)V_(2)`
`10 xx V_(1) = 200 xx 0.4`
`V_(1) = (200 xx 0.4)/10 =8`