Question

31.65 mL volume of 1.07 M Fe(NO3)3 is mixed with 44.96 mL of 0.79M Fe(NO3)3 solution. Calculate the molar concentration of the final solution. Save rounding until the end.

Answer 1

The molar concentration of the final solution is 0.9057 M

Step-by-step explanation:

Step 1: Data given

Volume of a 1.07 M Fe(NO3)3 solution = 31.65 mL = 0.03165 L

Volume of a 0.79 M Fe(NO3)3 solution = 44.96 mL = 0.04496 L

Step 2: Calculate moles of 1.07 M Fe(NO3)3 solution

Moles = molarity * volume

Moles = 1.07 M * 0.03165 L

Moles = 0.0338655 moles

Step 3: Calculate moles of 0.79 M Fe(NO3)3 solution

Moles = 0.79 M * 0.04496 L

Moles = 0.0355184 moles

Step 4: Calculate total moles

Total moles = 0.0339 + 0.0355

Total moles = 0.0693839 moles

Step 5: Calculate total volume

Total volume = 31.65 mL + 44.96 mL

Total volume = 76.61 mL = 0.07661 L

Step 6: Calculate molar concentration of the final solution

Concentration = moles / volume

Concentration = 0.0693839 moles / 0.07661 L

Concentration = 0.9057 M

The molar concentration of the final solution is 0.9057 M

Answer 2

[Fe(NO₃)₃] = 0.90 M

Step-by-step explanation:

Molarity = Mol/L

Total volume of solution: 31.65 mL + 44.96 mL = 76.61 mL

Let's determine the moles of salt ([M] . volume)

0.03165 L . 1.07M = 0.0338 moles

0.04496L . 0.79M = 0.0355 moles

Total moles Fe(NO₃)₃ = 0.0338 + 0.0355 = 0.0693 moles

Volume of solution = 76.61 mL = 0.07661 L

0.0693 mol / 0.07661 L = 0.90 mol/L