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pochemuha
3 years ago
10

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. Ho

w much solid iron(III) chloride should you add?
Chemistry
1 answer:
irakobra [83]3 years ago
3 0

Answer:

5 g

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

n

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