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