Chemistry · General chemistry II · Worked example
Find molality, mass percent and mole fraction
A solution is made by dissolving 10.0 g of NaCl in 250.0 g of water. Find its molality, its mass percent of NaCl and the mole fraction of NaCl.
Moles of solute
Divide by the molar mass of NaCl, 58.44 g/mol.
Molality
Divide by the solvent’s mass in kilograms: 250.0 g is 0.2500 kg.
Mass percent
The solution weighs 10.0 g + 250.0 g = 260.0 g.
Moles of water
Divide by the molar mass of water, 18.02 g/mol.
Mole fraction
χ(NaCl) is the moles of NaCl over the total moles of NaCl and water.
Result
The solution is 0.684 m, 3.85% NaCl by mass, and the mole fraction of NaCl is 0.0122.
Your turn
Find the molality of a solution of 18.0 g of glucose, C₆H₁₂O₆ (180.16 g/mol), in 150.0 g of water.
Show the answer and explanation
0.666 m.
18.0 g ÷ 180.16 g/mol = 0.0999 mol, and 0.0999 mol ÷ 0.1500 kg = 0.666 mol/kg.
Keep exploring
Solutions & concentration opens in molality with 10.0 g of NaCl and 250.0 g of water and gives 0.684 mol/kg. Switch the calculation to mass percent or mole fraction to check the other two answers.
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