okay so in lab we made up 2 solutions of FeCl3 and KI. We varied the volumes...I was just wondering how to calculate a new concentration for those solutions. We weighed out .825g of the FeCl3 and 1.35 g of the KI. For trial one we used 20 mL of both. For trial 2 we used 20.0 mL Fe and 10.0 mL KI, skip to trial 4, 15 mL Fe and 10 mL KI. The reaction eq = 2I+2Fe3+-->I2 + 2Fe2+.
I know that M=mol/L but the listed concentrations in the answers were .020 M Fe and .020 M KI (Trial 1), .020 Fe, .010M KI (trial 2), trial 4 i don't remember.
You don't give enough information to calculate it but here is how you do it.
mols FeCl3 = 0.825/162.2 = 0.00509 mols.
How what was the volume you diluted this to? I will call it 250 mL but use what you did in the experiment.
So for Fe in trial 1, you take 20 mL of the 250. You have removed 0.00509 x 20/250 = 4.07E-4 mols. You're placing that with 20 mL of the KI solution which gives 40 mL total. So (Fe) in trial 1 is 4.07E-4 mol/0.040 L = ?
KI is done similarly.
ahh thank you!
To calculate the concentration of a solution, you need to know the moles of solute and the volume of the solution. The molar concentration formula is:
Molarity (M) = moles of solute / volume of solution in liters
Let's go through each trial and calculate the concentrations:
Trial 1:
FeCl3: Weighed 0.825g of FeCl3
To find the moles of FeCl3, divide the weight by its molar mass. The molar mass of FeCl3 can be calculated by adding the atomic masses of iron (Fe) and chlorine (Cl).
Molar mass of FeCl3 = (55.85 g/mol) + (3 * 35.45 g/mol) = 162.2 g/mol
moles of FeCl3 = 0.825g / 162.2 g/mol = 0.00508 mol
Volume of solution = 20 mL = 0.02 L
Molarity of FeCl3 = 0.00508 mol / 0.02 L = 0.254 M
KI: Weighed 1.35g of KI
Using the same procedure, the molar mass of KI can be calculated:
Molar mass of KI = 39.10 g/mol + 126.9 g/mol = 166 g/mol
moles of KI = 1.35g / 166 g/mol = 0.00813 mol
Volume of solution = 20 mL = 0.02 L
Molarity of KI = 0.00813 mol / 0.02 L = 0.407 M
Therefore, the concentrations for Trial 1 are:
FeCl3 = 0.254 M
KI = 0.407 M
Trial 2:
FeCl3: Volume = 20.0 mL = 0.02 L (same as before)
Molarity remains the same at 0.254 M
KI: Volume = 10.0 mL = 0.01 L
moles of KI = 1.35g / 166 g/mol = 0.00813 mol
Molarity of KI = 0.00813 mol / 0.01 L = 0.813 M
Therefore, the concentrations for Trial 2 are:
FeCl3 = 0.254 M
KI = 0.813 M
For trial 4, you did not provide the volumes used for FeCl3 and KI. Please provide the volumes so that I can calculate the concentrations for trial 4 as well.