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?

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"" ")|)#