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
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?
1 answer
1 answer