Answer to Question #144900 in General Chemistry for Vic

Question

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

Answer 1

We apply the relationship;

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

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

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