What volume of 6.0 M sulfuric acid is required for the preparation of 500.0 mL of 0.30 M solution?
Steve answered this for you below.
To calculate the volume of a solution required for a specific concentration, you can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case, we have:
C1 = 6.0 M
V1 = unknown
C2 = 0.30 M
V2 = 500.0 mL
Using this formula, we can solve for V1:
C1V1 = C2V2
6.0 M (V1) = 0.30 M (500.0 mL)
Now, let's convert the volumes to the same units. Since 1 mL = 0.001 L, we can convert the mL to L:
C1V1 = C2V2
6.0 M (V1) = 0.30 M (500.0 mL x 0.001 L/mL)
Simplifying further:
6.0 M (V1) = 0.30 M (0.500 L)
Dividing both sides of the equation by 6.0 M:
V1 = (0.30 M x 0.500 L) / 6.0 M
V1 = 0.025 L
Now, let's convert the volume back to mL:
V1 = 0.025 L x 1000 mL/L
V1 = 25 mL
Therefore, to prepare a 0.30 M solution, you would need 25 mL of 6.0 M sulfuric acid.