How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 123.5 mL of 0.48 M H2SO4?
5.2mL
How many mols do you need? That's mols = M x L = 0.48 x 0.1235 L = ?
Then M = mols/L soln. Substitute. Solve.
You know M = 5.00 and you know mols from above. Solve for L and convert to mL.
To calculate the amount of a concentrated stock solution needed to prepare a desired dilution, we can use the formula:
C1V1 = C2V2
where:
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = concentration of the desired dilution
V2 = volume of the desired dilution
In this case, we have:
C1 = 5.0 M (concentration of the stock solution)
V1 = unknown
C2 = 0.48 M (concentration of the desired dilution)
V2 = 123.5 mL (volume of the desired dilution)
Rearranging the formula, we can solve for V1:
V1 = (C2 * V2) / C1
Substituting the given values, we get:
V1 = (0.48 M * 123.5 mL) / 5.0 M
V1 = (59.28 mL) / 5.0
V1 ≈ 11.856 mL
Therefore, you would need to measure approximately 11.856 milliliters of the 5.0 M H2SO4 stock solution to prepare 123.5 mL of 0.48 M H2SO4.
To solve this problem, we can use the formula for dilution:
M1V1 = M2V2
Where:
M1 = initial concentration of the stock solution
V1 = volume of the stock solution
M2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
We are given:
M1 = 5.0 M (concentration of the stock solution)
V2 = 123.5 mL (final volume of the diluted solution)
M2 = 0.48 M (desired final concentration)
Let's calculate V1 (volume of the stock solution):
M1V1 = M2V2
5.0 M * V1 = 0.48 M * 123.5 mL
V1 = (0.48 M * 123.5 mL) / 5.0 M
V1 = 5.892 mL
Therefore, you would need to measure 5.892 mL of the 5.0 M H2SO4 stock solution to prepare 123.5 mL of 0.48 M H2SO4.