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 the boiling point elevation, measured as the difference between the boiling point of the solution and the boiling point of the pure solventi is the van't Hoff factor, equal to1 for non-ionizing solutesK_b is the ebullioscopic constant of the solvent, equal to"2.65 K kg mol"^(-1) for benzene-> see hereb is the molality of 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
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.