`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

`100 g` solution contains `29.2 g HCl`,
`m_(HCl)=36.5 g mol^(-1)`
Volume of solution `=(100)/(1.25)=80 mL`
`therefore "Molarity"`
`=("mass of solute")/("mol.wt.of solute"xx"Volume"("in"mL))xx1000`
`=(29.2)/(36.5xx80)xx1000=10M`
Let `V mL` of this `HCl` are used to prepare
`200 mL` of `0.4 M HCl`, then
milli-mole of conc. `HCl= "milli-mole of dil". HCl`
( milli-mole does not change on dilution)
`Vxx10=200xx0.4`
or `V=8`