what volume of concentrated 18M H2SO4 is required to prepare 250mL of a 6.0M solution?
how would you solve this problem? i know it has something to do with the molarity formula but no idea how to apply it
well, if I had one liter of the final stuff I would have 6 mols of H2SO4 in it
so in 250 mL I need 6/4 or 1.5 mols H2SO4
so the real question is:
What volume of 18M H2SO4 contains 1.5 mols of H2SO4
well there are 18 in a liter
so
I need 1.5/18 liter =.08333...... Liter which is 83 mL
You are welcome :)
so v would be 83mL?
ohh okay now i understand. thank you
To solve this problem, you can use the principle of dilution to calculate the volume of concentrated H2SO4 needed. The formula you mentioned, molarity formula, will indeed be used.
The formula for dilution is:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution (18M)
V1 = volume of the concentrated solution used (unknown in this case)
C2 = final concentration of the solution (6.0M)
V2 = final volume of the solution (250 mL or 0.250 L)
1. Rearrange the formula to solve for V1:
V1 = (C2V2) / C1
2. Substitute the given values into the formula:
V1 = (6.0M * 0.250 L) / 18M
3. Calculate the volume of concentrated H2SO4 required:
V1 = 0.0833 L or 83.3 mL
So, you will need approximately 83.3 mL of concentrated 18M H2SO4 to prepare a 6.0M solution with a final volume of 250 mL.
Use the dilution formula of
c1v1 = c2v2
18M*vmL = 6M*250mL
v = ?mL.