# 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?

##### 1 Answer

#### 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 **for every**

The mass of the

#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 **for every**

Your sample will thus contain

#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_"solution" = m_"water" + m_"ethanol"#

#m_"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_"total" = 34.7 + 6.99 = "41.69 moles"#

The mole fraction of ethanol will thus be

#chi_"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)#