Question #f294a

1 Answer
Jan 4, 2018

"120 g mol"^(-1)

Explanation:

The first thing that you need to do here is to calculate the molality of the solution.

To do that, you can use the equation

color(blue)(ul(color(black)(DeltaT = i * K_b * b)))

Here

  • DeltaT is the boiling point elevation, measured as the difference between the boiling point of the solution and the boiling point of the pure solvent
  • i is the van't Hoff factor, equal to 1 for non-ionizing solutes
  • K_b is the ebullioscopic constant of the solvent, equal to "2.65 K kg mol"^(-1) for benzene -> see here
  • b is the molality of the solution

Now, you know that when you dissolve "2.67 g" of this unknown solute into "100 g" of benzene, you get a boiling point elevation for the resulting solution equal to

DeltaT = "0.531 K"

Rearrange the equation to solve for b.

DeltaT = i * K_b * b implies b= (DeltaT)/(i * K_b)

Plug in your values to find

b = (0.531 color(red)(cancel(color(black)("K"))))/(1 * 2.65 color(red)(cancel(color(black)("K"))) "kg mol"^(-1)) = "0.2004 mol kg"^(-1)

This tells you that your solution contains 0.2204 moles of solute for every "1 kg" = 10^3 "g" of solvent.

You can thus say that your sample will contain

100 color(red)(cancel(color(black)("g benzene"))) * "0.2004 moles solute"/(10^3color(red)(cancel(color(black)("g benzene")))) = "0.02004 moles solute"

Finally, to find the molar mass of the solute, simply divide its mass by the number of moles it contains.

M_ ("M solute") = "2.67 g"/"0.02004 moles" = color(darkgreen)(ul(color(black)("120 g mol"^(-1))))

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the mass of benzene.