What volume of a 0.400 M HCl solution is required to prepare 3.00 L of 0.0872 M HCl?

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What volume of a 0.400 M HCl solution is required to prepare 3.00 L of 0.0872 M HCl?

Expert Solution
Step 1

From neutralization of solution,

V1S1 = V2S2

V2 = v1×s1v2

   

volume of HCl solution (V1) =3.00 L 

concentration of HCl (S1) =0.0872 M 

 

concentration of HCl (S1) = 0.400 M 

volume of HCl solution (V2) = 3.00 L ×0.0872 M 0.400 M

                                             = 0.654 L

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