Question

What is the molality of a solution prepared by dissolving 30.57 g of iron (III) chloride in 284.7 g of water?

Report your answer to THREE significant figures.

Answer 1

Answer:

Given mass of Iron(III) chloride, FeCl3=30.57 g and

Molar mass of FeCl3=162.2 g/mol

Therefore moles of FeCl3=mass/molar mass=30.57 g/162.2 g/mol=0.18847 mol.

Mass of water=284.7 g =0.2847 Kg. (Since 1 Kg=1000 g)

We know that molality=moles of solute/mass of solvent.

Here solute is FeCl3 and solvent is water.

Therefore molality=0.18847 mol/0.2847 Kg=0.6619 mol/Kg

Molality ~ 0.662 m.

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