Answer:
the quantity required can go from 117 ml (for maximum concentration) up to 2900 ml ( if the concentrated solution has molarity =0.420 M)
Explanation:
the amount of water required to dilute a solution V₁ liters of Molarity M₁ to V₂ liters of M₂
moles of hydrochloric acid = M₁ * V₁= M₂ * V₂
V₁ = V₂ * M₂/M₁
where
M₂ = 0.420 M
V₂ =2.90 L
Since the hydrochloric acid can be concentrated up to 38% p/V ( higher concentrations are possible but the evaporation rate is so high that handling and storage require extra precautions, like cooling and pressurisation)
maximum M₁ =38% p/V = 38 gr/ 0.1 L / 36.5 gr/mol = 10.41 M
then
min V₁ = V₂ * M₂/ max M₁ = 2.90 L* 0.420 M/ 10.41 M= 0.117 L = 117 ml
then the quantity required can go from 117 ml up to 2900 ml ( if M₁ = M₂)
