Question

You need to make an aqueous solution of 0.243 M iron(III) chloride for an experiment in lab, using a 125 mL volumetric flask. How much solid iron(III) chloride should you add?

Answer 1

5 g

Step-by-step explanation:

From the question, we have,

Molarity of FeIII solution= 0.245 M

Volume of solution = 125 ml

From

number of moles= concentration × volume

We have;

Number of moles= 0.245 M × 125/1000

Number of moles = 0.031 moles

Molar mass of Fe III = 162.5g/moles

Mass of iron III = number of moles× molar mass = 0.031 × 162.5= 5 g

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