What volume of 6.0 M sulfuric acid is required for the preparation of 500.0 mL of 0.30 M solution?
500mL of .3M contains .15 moles acid
.15moles ÷ 6.0moles/L = 0.025L
Well, sulfuric acid sure knows how to make a big entrance! Now, let's calculate how much of this acid we need.
To find the volume, we can use the formula:
C1V1 = C2V2
Where C1 is the concentration of the concentrated solution (6.0 M), V1 is the volume of the concentrated solution (unknown), C2 is the concentration of the diluted solution (0.30 M), and V2 is the volume of the diluted solution (500.0 mL).
Plugging in the numbers, we have:
(6.0 M)(V1) = (0.30 M)(500.0 mL)
Now we need to solve for V1:
V1 = (0.30 M)(500.0 mL) / 6.0 M
V1 = 25.0 mL
So, you'll need 25.0 mL of 6.0 M sulfuric acid to prepare 500.0 mL of 0.30 M solution. Just remember to handle it with care and use appropriate safety protocols!
To calculate the volume of a concentrated solution needed to prepare a diluted solution, you can use the equation:
C1V1 = C2V2
Where:
C1 = concentration of the concentrated solution (in this case, 6.0 M)
V1 = volume of the concentrated solution needed
C2 = concentration of the diluted solution (in this case, 0.30 M)
V2 = volume of the diluted solution (in this case, 500.0 mL or 0.500 L)
Rearranging the equation to solve for V1, we have:
V1 = (C2 * V2) / C1
Substituting the values into the equation:
V1 = (0.30 M * 0.500 L) / 6.0 M
V1 = 0.025 L
Therefore, the volume of 6.0 M sulfuric acid needed to prepare 500.0 mL of 0.30 M solution is 0.025 L (or 25 mL).
To find the volume of 6.0 M sulfuric acid required for the preparation of a 0.30 M solution, we can use the equation:
C1V1 = C2V2
Where:
C1 = concentration of the initial solution (6.0 M)
V1 = volume of the initial solution (unknown)
C2 = concentration of the final solution (0.30 M)
V2 = volume of the final solution (500.0 mL)
Rearranging the equation to solve for V1:
V1 = (C2 * V2) / C1
Substituting the given values:
V1 = (0.30 M * 500.0 mL) / 6.0 M
Now, we can calculate V1:
V1 = (0.15 mol / 6.0 M) * 500.0 mL
V1 = 7.5 mL
Therefore, 7.5 mL of 6.0 M sulfuric acid is required for the preparation of 500.0 mL of 0.30 M solution.