How much water should be added to 55.0 mL of 0.201 M HCl to dilute the solution to 0.155 M?
i used M1V1 = M2V2 and i got 71.3mL
its wrong so im confused
71.3 mL is the final volume, not the amount of water added to achieve that volume. Does that help?
yeaa thanks!!
To solve this problem, you correctly used the dilution formula M1V1 = M2V2, where M1 is the initial molarity (0.201 M), V1 is the initial volume (55.0 mL), M2 is the final molarity (0.155 M), and V2 is the final volume (unknown). However, there seems to be a miscalculation. Let's go through the equation again to determine the correct answer.
M1V1 = M2V2
Initially, you have 55.0 mL of 0.201 M HCl, so:
M1 = 0.201 M
V1 = 55.0 mL
You want to dilute the solution to a final molarity of 0.155 M, so:
M2 = 0.155 M
V2 = unknown
Plugging in the given values, we have:
(0.201 M)(55.0 mL) = (0.155 M)(V2)
Multiply both sides of the equation:
11.055 = 0.155V2
11.055 รท 0.155 = V2
Therefore, V2 = 71.129 mL (rounded to three significant figures).
So, the correct answer is approximately 71.1 mL, not 71.3 mL.