How many mL of 6.00 M alcohol are needed to prepare 200. mL of 2.50 M alcohol solution?
since the amount of alcohol present is the same,
6.00x = 2.50*200
To find the volume of 6.00 M alcohol needed to prepare 200. mL of a 2.50 M alcohol solution, we can use the formula:
(M1)(V1) = (M2)(V2)
Where:
M1 = initial molarity (6.00 M)
V1 = initial volume (unknown)
M2 = final molarity (2.50 M)
V2 = final volume (200. mL)
Let's plug in the values we know:
(6.00 M)(V1) = (2.50 M)(200. mL)
Now we can solve for V1:
V1 = (2.50 M)(200. mL) / (6.00 M)
V1 ≈ 83.33 mL
Therefore, approximately 83.33 mL of 6.00 M alcohol is needed to prepare 200. mL of a 2.50 M alcohol solution.