Answer to Question #144900 in General Chemistry for Vic

General Chemistry

Question #144900

How many milliliters of 11.5 M HCl(aq) are needed to prepare 405.0 mL of 1.00 M HCl(aq)

Expert's answer

We apply the relationship;

M₁V₁ = M₂V₂(Where M1= 11.5M, M2= 1.00M, V2 = 405.0mL, V1=?)

"V1 = \\dfrac{M2V2}{M1}"

"=\\dfrac{1.00M x 405.0mL}{11.5M} = 35.217mL"

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