Question

`29.2% (W//W) HCl` stock solution has density of `1.25g mL^(-1)`. The molar mass of `HCl` is `36.5g mol^(-1)`. The volume `(mL)` of stock solution required to prepare a `200 mL` solution of `0.4M HCl` is

Answer 1

Correct Answer - `(8mL)`
Mass of `HCl` in `1.0 mL` stock solution
`= 1.25 xx (29.2)/(100) = 0.365 g`
Mass of `HCl` required for `200 mL 0.4 M HCl`
`= (200)/(1000) xx 0.4 xx 36.5 = 0.08 xx 36.5 g`
`:. 0.365 g` of `HCl` is present in `1.0 mL` stock solution.
`0.08 xx 36.5 g HCl` will be present in `(0.08 xx 36.5)/(0.365) = 8.0 mL`