Question

A solution is prepared by dissolving 0.5482 grams of iron(III) nitrate in enough water to make 100.0 mL of solution. A 10.00 mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the concentration of Fe3 ions (M) in the final solution

Answer 1

Answer:

$C_2=9.066x10^{-4}M$

Explanation:

Hello,

In this case, the first step is to compute the molarity of the first solution, for which we consider the molar mass of iron (III) nitrate that is 241.86 g/mol to compute the moles in 100.00 mL (0.1 L) of solution:

$M=\frac{0.5482g*\frac{1mol}{241.86 g} }{0.1L}=0.0227 M$

Which is actually the concentration of iron (III) ions. Therefore, for 10.00 mL of such solution, the concentration until a dilution to 250.0 mL results being:

$C_2=\frac{C_1V_1}{V_2} =\frac{0.0227M*10.00mL}{250.0mL} \\\\C_2=9.066x10^{-4}M$

Best regards.