Question

Question #3acdc

Answer 1

Here's how you can do that.

Explanation:

The idea here is that you need to work with a sample of this solution and figure out how many moles of propanol it contains.

As you know, molality is defined as the number of moles of solute present for every `"1 kg"` of solvent. In your case, propanol is the solute and water is the solvent.

To make the calculations easier, let's pick a sample of solution that contains exactly

`1 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 10^3"g"`

of water. You know that water has a molar mass of `"18.015 g mol"^(-1)`, which means that this solution will contain

`10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"`

Now, the mole fraction of propanol in this solution is defined as the ratio between the number of moles of propanol, let's say `n_p`, and the total number of moles present in solution.

In your case, this mole fraction is equal to

`(n_p color(red)(cancel(color(black)("moles"))))/((n_p + 55.51)color(red)(cancel(color(black)("moles")))) = 0.1538`

This means that you have

`n_p = (n_p + 55.51) * 0.1538`

`(1 - 0.1538) * n_p = 55.51 * 0.1538`

`0.8462 * n_p = 8.5374 implies n_p = 8.5374/0.8462 = 10.09`

Therefore, you know that a solution of propanol that contains `"1 kg"` of water and in which the solute has a `0.1538` mole fraction contains `10.09` moles of propanol.

You can thus say that the molality of the solution is equal to

`color(darkgreen)(ul(color(black)("molality = 10.09 mol kg"^(-1))))`

The answer is rounded to four sig figs, the number of sig figs you have for the mole fraction of propanol.