250-mL volumetric flask, 125 mL of a 6.0 M hydrochloric acid was added. This was then diluted with more water to reach the 250-mL mark. What is the molarity of the acid at this time?

Question

250-mL volumetric flask, 125 mL of a 6.0 M hydrochloric acid was added. This was then diluted with more water to reach the 250-mL mark. What is the molarity of the acid at this time?

Initial: M1 = 6.0 M hydrochloric
acid V1 = 125 mL = 0.125 L

Diluted: M2 = ? M hydrochloric acid
V2 = 250 mL = 0.250 L

M1V1 = M2V2
M1V1 = M2V2
V2

M2 = M1 x V1
V2

= 6.0 M x 0.125 L
0.250L

3M =

3 M hydrochloric acid is left at this time (diluted solution).

Did I do this right?

Answer 1

Yes you did it correctly and the answer is correct; however, you didn't account for the boards not putting in the spaces where they belong. These boards count one space as one space; however, two spaces, three spaces, 4 spaces, etc etc just get one space. Therefore, you can space all you wish on the space bar and you get ONLY once space. What you must do is write

M1V1 = M2V2, then to solve for M2 we do it this way.
M2 = (M2V2)/V2
Then M2 = (6.0 x 0.125)/0.250 =
6.0/2.00 = 3.0 M

Answer 2

Yes, you did the calculation correctly. The dilution formula M1V1 = M2V2 can be used to find the molarity of a solution after dilution. In this case, the initial concentration (M1) of the hydrochloric acid is 6.0 M, the initial volume (V1) is 0.125 L, and the final volume (V2) is 0.250 L. By substituting these values into the dilution formula, you correctly calculated that the molarity after dilution (M2) is 3.0 M hydrochloric acid.