# The boiling point of a solution containing 10.44g og an unknown nonelectrolyte in 50g of acetic acid is 159.2 degree celsius. What is the molar mass of the solute?

Aug 17, 2017

$15.5$ $\text{g/mol}$

#### Explanation:

We're asked to find the molar mass of the solute, given some known information about boiling-point elevation.

To do this, we'll be using the equation

ul(DeltaT_b = imK_b

where

• $\Delta {T}_{b}$ is the change in boiling point of the solution

• $i$ is the van't Hoff factor of the solute, which is $1$ since it is a nonelectrolyte

• $m$ is the molality of the solution

• ${K}_{b}$ is the molal boiling-point elevation constant for the solvent (acetic acid), which is $3.07$ $\text{^"o""C/} m$ (from Wikipedia).

We're given that the final boiling point is $159.2$ $\text{^"o""C}$, and according to Wikipedia, the normal boiling point of acetic acid is $117.9$ $\text{^"o""C}$, so the change in boiling point $\Delta {T}_{b}$ is

$\Delta {T}_{b} = 159.2$ $\text{^"o""C}$ $- 117.9$ $\text{^"o""C}$ = ul(41.3color(white)(l)""^"o""C"

Plugging in known values:

41.3color(white)(l)""^"o""C" = (1)m(3.07color(white)(l)""^"o""C/"m)

The molality (moles of solute per kilogram of solvent) is

color(red)(ul(m = 13.5color(white)(l)"mol/kg"

Now that we know the molality of the solution, we can use the given value of $50$ $\text{g acetic acid}$ and the molality equation to find the number of moles of the solute present in solution:

"mol solute" = ("molality")("kg solvent")

So

$\text{mol solute" = (color(red)(13.5color(white)(l)"mol/")cancel(color(red)("kg")))(0.050cancel("kg")) = color(green)(ul(0.673color(white)(l)"mol solute}$

Finally, we can use the given number of grams of solute and this to find its molar mass:

color(blue)("molar mass") = (10.44color(white)(l)"g")/(color(green)(0.673color(white)(l)"mol")) = color(blue)(ulbar(|stackrel(" ")(" "15.5color(white)(l)"g/mol"" ")|)