# Question #f294a

##### 1 Answer

#### 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 theboiling point elevation, measured as the difference between the boiling point of the solution and the boiling point of the pure solvent#i# is thevan't Hoff factor, equal to#1# for non-ionizing solutes#K_b# is theebullioscopic constantof the solvent, equal to#"2.65 K kg mol"^(-1)# for benzene#-># seehere#b# is themolalityof the solution

Now, you know that when you dissolve

#DeltaT = "0.531 K"#

Rearrange the equation to solve for

#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 **moles** of solute for every

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.