Question

Calculate the molality, molarity, and mole fraction of FeCl3 in a 23.0 mass % aqueous solution (d = 1.230 g/mL). molality=____m molarity=____M mole fraction=___

Answer #1

Given that 23.0 % by mass FeCl3, therefore

mass FeCl3 in 100 g solution = 23 g

mass water in 100 g solution = 77 g = 0.077 kg

mass solution = 100 g = 0.100 kg

volume solution = mass / density = 100 g / 1.230 g/ml = 81.3008 ml
= 0.0813 L

moles FeCl3 = mass / molar mass = 23 g / 162.2 g/mol = 0.1418
mol

moles H2O = 77 g / 18.016 g/mol = 4.2739 mol

total moles in solution = 0.1418 mol + 4.2739 mol = 4.4157
mol

molarity = moles FeCl3 / litres solution = 0.1418 mol / 0.0813 L =
1.744 M

molality = moles solute / kg solvent = 0.1418 mol / 0.077 kg =
1.8415 m

mole fraction FeCl3 = moles FeCl3 / total moles = 0.1418 mol /
4.4157 mol = 0.03211

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Please show work

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