Describe the preparation of 100 mL 0.50 M solution of Fe(NO3)3 from a stock solutiion that is 60%(w/w)Fe(NO3)3 solution with a specific gravity of 1.110 g/ml. Calculate the molar NO3 concentration in the resulting solution. Molar Mass of Fe(NO3)3: 241.86
To prepare a 0.50 M solution of Fe(NO3)3, you will need to calculate the amount of stock solution and the amount of solvent (usually water) required.
First, let's calculate the amount of Fe(NO3)3 present in the stock solution:
Given:
- Mass% of Fe(NO3)3 in the stock solution: 60%
- Specific gravity of the stock solution: 1.110 g/mL
- Volume of stock solution needed: 100 mL
To calculate the mass of Fe(NO3)3 in 100 mL of the stock solution:
1. Calculate the mass of the stock solution:
Mass of stock solution = Volume of stock solution × Specific gravity
Mass of stock solution = 100 mL × 1.110 g/mL = 111.0 g
2. Calculate the mass of Fe(NO3)3 in the stock solution:
Mass of Fe(NO3)3 = Mass of stock solution × Mass% of Fe(NO3)3 / 100
Mass of Fe(NO3)3 = 111.0 g × 60 / 100 = 66.6 g
Next, let's calculate the molar NO3 concentration in the resulting solution:
1. Calculate the moles of Fe(NO3)3:
Moles of Fe(NO3)3 = Mass of Fe(NO3)3 / Molar mass of Fe(NO3)3
Moles of Fe(NO3)3 = 66.6 g / 241.86 g/mol ≈ 0.2756 mol
2. Calculate the volume of the resulting solution:
Volume of resulting solution = 100 mL = 0.100 L
3. Calculate the molar NO3 concentration:
Molar NO3 concentration = Moles of Fe(NO3)3 / Volume of resulting solution
Molar NO3 concentration = 0.2756 mol / 0.100 L = 2.756 M
Therefore, the molar NO3 concentration in the resulting solution is approximately 2.756 M.
First I would calculate the molarity of the iron(III) nitrate solution.
1.110 g/mL x 1000 mL x 0.60 = g/L
Then g/molar mass = mols and this is the M of the soln.
Then dilute that stock soln using the dilution formula.
c1v1 = c2v2 where
c = concn
v = volume.