Answer to Question #209851 in Chemistry for lia

Question #209851

claculate the volume of 0.015M of HCL solution required to prepare 250ml of 5.25 x 10^-3

Expert's answer

C_M = n / V

n₁ = n₂

C_M1 x V₁ = C_M2 x V₂

0.015 x V1 = 0.00525 x 0.25

V₁ (HCl) = (0.00525 x 0.25)/0.015 = 0.0875 ml = 8.8 ml

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