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?

1 Answer
Aug 17, 2017

Answer:

#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"" ")|)#