Question

Calculate the molality, molarity, and mole fraction of FeCl3 in a 29.5 mass % aqueous solution (d = 1.280g/mL)

Answer #1

**let us assume we had 100 g solution**

**volume fo solution = mass /density = 100/1.28 = 78.125
ml = 0.078125 L**

**29.5 % mass fraction means we have 29.5 g FeC3 in 100 g
solution**

**FeCl3 moles = mass of FeCl3 / Molar mass of FeCl3 = 29.5
/162.2 = 0.1819**

**solvent mass = 100-29.5 = 70.5 g = 0.0705
kg**

**water moles = mass of water / molar mass of water =
70.5/18 =3.917**

**Mol fraction of FeCl3 = moles of FeCl3 / total moles of
solution = ( 0.1819/0.1819+3.917) = 0.04438**

**FeCl3 molality = moles of solute / solvent mass in kg =
0.1819/0.0705 = 2.58**

**Molarity = moles of solute / solution vol in L = 0.1819
/0.078125 = 2.33**

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Please show work

An aqueous solution contains 4.3% NaCl by mass.
Part A
Calculate the molality of the solution.
Express your answer using two significant figures.
Part B
Calculate mole fraction of the solution.
Express your answer using two significant figures.

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