How much 0.85 M HCl solution can be made by diluting 250 mL of 10 M HCl?

Answer is: 2940 mL of the HCL solution.

c₁ (HCl) = 10.0 M.

V₂ (AgNO₃) = ?.

c ₂ (AgNO₃) = 0.85 M.

V ₁ (AgNO₃) = 250 mL : 1000 mL/L = 0.25 L.

c ₁ - original concentration of the solution, before it gets diluted.

c ₂ - final concentration of the solution, after dilution.

V ₁ - volume to be diluted.

V ₂ - final volume after dilution.

c ₁ · V₁ = c₂ · V₂.

V ₂ (HCl) = c₁ · V₁ : c₂.

V ₂ (HCl) = 10 M · 0.25 L : 0.85 M.

V ₂ (HCl) = 2.94 L · 1000 mL = 2940 mL.

The amount of 0.85M HCl solution that can be made by diluting 250 ml of 10M HCl is calculated using

M1V1=M2V2 formula

M1=0.85M

V1=?

M2 = 10M

V2 = 250 ml

by making the V1 the subject of the formula

V1 = M2V2/M1

V1 = (10 Mx250 ml) / 0.85 M = 2941.2 ml