Question #f294a

1 Answer
Jan 4, 2018

Answer:

#"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.