Question

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?

Answer 1

`15.5` `"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

• `DeltaT_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` `""^"o""C/"m` (from Wikipedia).

We're given that the final boiling point is `159.2` `""^"o""C"`, and according to Wikipedia, the normal boiling point of acetic acid is `117.9` `""^"o""C"`, so the change in boiling point `DeltaT_b` is

`DeltaT_b = 159.2` `""^"o""C"` `- 117.9` `""^"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` `"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

`"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"" ")|)`