Question

a solution is made by mixing 1.08 g of an unknown non-volatile non-electrolyte with 10.0 g of benzene. The freezing point of pure benzene is is 5.5 degrees Celsius. The molal freezing point depression constant, Kf, for benzene is 5.12 degrees Celsius per molale.

What is the value of the freezing point depression? What is the molality of the solution? What is the molar mass of the unknown?

Answer 1

To do that you need to use the following expression:

= Kf m

Where m is the molality, Kf the molal freezing point deppression constant and is the freezing point deppresion.

but = Tf° - Tf

Where Tf° is the freezing point of the pure solvent (in this case benzene) and Tf is the freezing point of the solution.

In this problem you are not providing the value of the freezing point of the solution, so you can't solve this problem with the data you are providing, however, let me guide you of how to solve it, once you got (or provide) the data for that freezing point missing:

Let's assume that the freezing point of the solution is 0.62 °C (I'm assuming this value below from the freezing point of the benzene because it's a non volatile and electrolyte compound as well as the freezing point of solution is usually lower from the freezing point of the pure solvent), the freezing point depression would be:

= 5.5 - 0.62 = 4.88 °C

With this value, and using the above expression, you may solve for molality "m":

= Kf m ----> m = /Kf

m = 4.88 °C / 5.12 °C/m = 0.9531 m

Finally, molality is: m = moles solute / kg solvent and moles = mass solute / MM

then: m = mass solute / MM * kg solvent ----> MM = mass solute / m * kg solvent

MM = 1.08 g / 0.9531 moles/kg * 0.010 kg

MM = 113.31 g/mol

As I told you before, please provide the missing data so I can give you the exact result. Hope this helps