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Find the Molality, Molarity, and Mole Fraction when mass percent is given

An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm. Calculate the molality, molarity, and mole fraction of the ethylene glycol.

Assume a 100 gram sample.
This gives you 40 g of ethylene glychol dissolved in 100g of solution.
Determine the molar mass of e.g and then find the number of moles of the e.g.

Molality = moles of e.g. /kilograms of solution (be sure to convert 100 g to kg)

Molarity - to find the molarity you will need to convert the mass of the solution to volume in liters. This can be done using the density.
volume (in cm3) = density x mass of solution
Next convert from cm3 to liters ( 1 cm3 = 1mL = .001 L)
Molarity = moles of e.g. / volume of solution in liters

Mole Fraction - we need to know the moles of water present. Since the solution contains 40g e.g. the other 60 grams must be water. Find the moles of water present.
Mole Fraction = moles of e.g. / (moles of e.g. + moles of water)

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