Calculate the volume (mL) of 2.06 M magnesium phosphate solution require to prepare 46.20 mL of 2.3998 %w/v solution of Mg3(PO4)2
I would convert 2.3998 w/v% to M first.
2.l3998 g Mg3(PO4)2/100 mL soln
mols Mg3(PO4)2 = grams/molar mass = 2.13998/molar mass = approximately but you need to do it more accurately = 0.00814 (which is far short of the number of significant figures allowed based on the % in the problem).
0.00814 moles/100 mL = 0.0814 mols/L = 0.0814M( approx).
Then use the dilution formula of
c1v1 = c2v2
If I make 1 Liter of a 2M magnesium phosphate solution, how many grams of magnesium phosphate do I need
To calculate the volume of a 2.06 M magnesium phosphate solution required to prepare 46.20 mL of a 2.3998% w/v solution of Mg3(PO4)2, we need to use the following formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution (in M)
V1 = volume of the stock solution (in mL)
C2 = concentration of the desired solution (in %w/v)
V2 = volume of the desired solution (in mL)
In this case, we are given C1 = 2.06 M, V2 = 46.20 mL, and C2 = 2.3998% w/v. We need to find V1, the volume of the stock solution.
First, we need to convert C2 from %w/v to M.
To do this, we multiply C2 by 0.01 to convert it to decimal form:
C2 = 2.3998% * 0.01 = 0.023998
Now, we can plug the given values into the formula:
C1V1 = C2V2
2.06 M * V1 = 0.023998 * 46.20 mL
Now, we solve for V1:
V1 = (0.023998 * 46.20 mL) / 2.06 M
V1 ≈ 0.5399 mL
Therefore, approximately 0.5399 mL of the 2.06 M magnesium phosphate solution is needed to prepare 46.20 mL of a 2.3998% w/v solution of Mg3(PO4)2.