Question

In the laboratory you dissolve 16.7 g of iron (III) sulfate in a volumetric flask and add water to a total volume of 250

What is the molarity of the solution?

What is the concentration of the iron(III) cation?

What is the concentration of the sulfate anion?

Answer 1

To calculate the molarity of the solution, we need to first calculate the number of moles of iron (III) sulfate dissolved in the solution.

Number of moles = mass ÷ molar mass

Molar mass of iron (III) sulfate = 55.85 + 3(32.06) + 4(16.00) = 399.88 g/mol

Number of moles = 16.7 g ÷ 399.88 g/mol = 0.0418 mol

Now we can calculate the molarity of the solution:

Molarity = number of moles ÷ volume in liters

Volume in liters = 250 mL ÷ 1000 mL/L = 0.250 L

Molarity = 0.0418 mol ÷ 0.250 L = 0.167 M

The iron (III) cation concentration is equal to the concentration of the iron (III) sulfate, which is 0.167 M.

The sulfate anion concentration is also 0.167 M, since each mole of iron (III) sulfate dissociates to give one mole of sulfate ions.

Step-by-step explanation: