Molarity, Molality, and Mole Fraction

Question

Home-distilled alcohol is tightly regulated in a city, and for good reason. Improper fermentation and distilling can result in significant concentrations of methanol (which is highly poisonous) in the final mixture. A 100 mL sample collected from a police raid of an illicit distilling operation was analyzed to contain relative volumes of 45 mL of ethanol (C2H6O, density = 0.789 g/mL), 5.0 mL of methanol (CH4O, density = 0.792 g/mL), and 60 mL of water (H2O, density = 1.000 g/mL). Determine to molarity, molality, and mole fraction of methanol and determine if it could have posed a health risk to anyone who may have consumed it. Consuming methanol solutions with a molarity of 0.3M or greater is considered instantly hazardous.

Answer 1

The first thing we will do is to find the grams (mass) of each ingredient. We will then convert these to moles and with this information, we can report molarity (moles/liter of solution), molality (moles/kg solvent) and mole fraction (mol of methanol / total mols).

mass ethanol: 45 ml x 0.789 g/ml = 35.5 g

mass methanol: 5.0 ml x 0.792 g/ml = 3.96 g

mass H2O: 60 ml x 1.00 g/ml = 60 g

mols ethanol: 35.5 g x 1 mol / 46 g = 0.772 mols

mols methanol: 3.96 g x 1 mol / 32 g = 0.124 mols

mols H2O: 60 g x 1 mol / 18 g = 3.33 mols H2O

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Molarity of methanol:

0.124 mols methanol / 100 ml x 1000 ml / L = 1.24 M

Molality of methanol:

0.124 mols methanol / 0.06 kg H2O = 2.07 m

Mole fraction methanol:

total moles = 0.772 mol + 0.124 mol + 3.33 mol = 4.23 mols

X_methanol = 0.124 / 4.23 = 0.0293

Yes, it would pose a health risk since the molarity of 1.24 M is greater than that which is considered hazardous