Equal volumes of ethanol (C2H5OH) and water are mixed. If the density of water is 1.00 g·cm–3 and that of ethanol is 0.789, what is the mole fraction of ethanol in the mixture?

a. 0.764

b. 0.309

c. 0.324

d. 0.236

e. 0.162

**Explanation:** Let's assume 1L of ethanol and 1L of water since we are told equal volumes are mixed. 1L=1000ml

1ml = 1cm^3

Density = mass /volume

We can use the volumes to find mass of each.

Water: 1.00 g·cm–3= mass/1000ml

mass of water = 1000g

Ethanol: 0.789g·cm–3 = mass/1000ml

mass of ethanol = 789g

Mole fraction of ethanol= Xethanol = moles of ethanol/ total moles

Let's find moles of each by dividing by the molar mass

Moles of water= 1000g/(18g/mol) = 55.56mol

Moles of ethanol = 789g/ (46.07 g/mol) = 17.13mol

Mole fraction of ethanol = 17.13mol/ ( 55.56mol+17.13mol) = 0.236

**Answer:** D

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