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
A. `8 mL`
B. `6 mL`
C. `7 mL`
D. `5 mL`

Answer 1

Correct Answer - A
Molarity of stock solution (soln.) of `HCl`
`= ((% "mass")("Density of soln")(10))/(("Molar mass")`
`((29.2)(1.25)(10))/(36.5) = 10M`
According to dilution equation `M_(i) V_(i) = M_(f) V_(f)`,
`V_(i) = (M_(f) V_(f))/(M_(i)) = ((0.4M)(200 mL))/((10M)) = 8 mL`