10 cm3 of a 2N HCl solution was brought to 50 cm3 with distilled water.

10 cm3 of a 2N HCl solution was brought to 50 cm3 with distilled water. The molar concentration of the equivalent of the prepared solution became (mol / dm3)?

Given:
V1 solution (HCl) = 10 cm3 = 0.01 l
CH1 (HCl) = 2 n
V2 solution (HCl) = 50 cm3 = 0.05 l

To find:
CH2 (HCl) -?

Decision:
1) Calculate the amount of hydrochloric acid substance:
n (HCl) = CH1 (HCl) * V1 solution / z = 2 * 0.01 / 1 = 0.02 mol;
2) Calculate the equivalent molar concentration (normality) of the prepared hydrochloric acid solution:
CH2 (HCl) = z * n (HCl) / V2 solution = 1 * 0.02 / 0.05 = 0.4 N.

Answer: The molar concentration of the equivalent (normality) of the prepared HCl solution is 0.4 N.