`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 has density `=1.25` g/mL
Now, mole of HCl required in 0.4 M HCl
`=0.4xx0.2` mole `=0.08` mole
If x mol of original HCl solution is taken then mass of solution `=1.25 x`
mass of `HCl=(1.25x xx0.292)`
mole of `HCl=(1.25x xx0.292)/36.5=0.08`
So, `x=(36.5xx0.08)/(0.29xx1.25)` mol `=8` mL.