Question

Question #f294a

Answer 1

`"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.