# An aqueous solution is 34.0 % by mass ethanol, CH_3CH_2OH, and has a density of 0.947 g/mL. What is the mole fraction of ethanol in the solution?

Nov 16, 2015

$0.168$

#### Explanation:

The idea here is that you need to pick a sample volume of this solution and use the given density to find its mass.

Once you know the sample's mass, you can use it to determine how much water and how much ethanol it contains.

To make the calculations easier, let's take a $\text{1.00-L}$ sample of this solution. The solution has a density of $\text{0.947 g/mL}$, which means that you get $\text{0.947 g}$ for every $\text{1 mL}$ of solution.

The mass of the $\text{1.00-L}$ sample will thus be

1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.947 g"/(1color(red)(cancel(color(black)("mL")))) = "947 g"

Now, you know that this solution is 34%"w/w" ethanol. In simple terms, this tells you that you get $\text{34 g}$ of ethanol for every $\text{100 g}$ of solution.

947color(red)(cancel(color(black)("g solution"))) * "34 g ethanol"/(100color(red)(cancel(color(black)("g solution")))) = "322 g ethanol"

This means that the solution will also contain

${m}_{\text{solution" = m_"water" + m_"ethanol}}$

${m}_{\text{water" = "947 g" - "322 g" = "625 g water}}$

To get the mole fraction of ethanol, you must determine how many moles of ethanol and of water the solution contains.

Do to that, use the two compounds' respective molar masses

625color(red)(cancel(color(black)("g"))) * "1 mole water"/(18.015color(red)(cancel(color(black)("g")))) = "34.7 moles water"

322color(red)(cancel(color(black)("g"))) * "1 mole ethanol"/(46.07color(red)(cancel(color(black)("g")))) = "6.99 moles ethanol"

The total number of moles in the solution will be

${n}_{\text{total" = 34.7 + 6.99 = "41.69 moles}}$

The mole fraction of ethanol will thus be

${\chi}_{\text{ethanol" = n_"ethanol"/n_"total}}$

chi_"ethanol" = (6.99color(red)(cancel(color(black)("moles"))))/(41.69color(red)(cancel(color(black)("moles")))) = color(green)(0.168)