PLEASE HELP! If the presence of 2.00g of an unknown non-ionizing substance lowers the freezing temperature of a 10.00g sample of benzene by 6.33 degrees celsius, calculate the molar mass of the substance?

Well, you could simply calculate the first thing that you can get, and go from there. I get about `"161.8 g/mol"`.

The negative change in freezing point is given by

`DeltaT_f = -iK_fm < 0`

where:

• `i` is the van't Hoff factor, i.e. the effective number of dissociated solute particles per undissociated solute particle.
• `K_f = 5.12^@ "C"cdot"kg/mol"` (not `""^@ "C"`! What happens if you do not have the correct units?) is the freezing point depression constant for benzene. Here we list `K_f > 0`.
• `m` is the molality, i.e. `"mols solute"/"kg solvent"`.

Here we can solve for the molality first.

The change in freezing point is negative, i.e. we have freezing point depression. Furthermore, a non-ionizing solute has `i = 1`. Therefore:

`m = (DeltaT_f)/(-iK_f)`

`= (-6.33^@ "C")/(-(1)(5.12^@ "C"cdot"kg/mol")`

`=` `"1.236 mol solute/kg solvent"`

Therefore, if we have `"2.00 g solute"` and `"10.00 g benzene"`... well, benzene is the solvent. It has to be, because `K_f` can only be given for the solvent.

The molality, an intensive property, is fixed for whatever choice of solute mass and solvent mass we pick. So...

`"1.236 mol solute"/"kg solvent" = ("mol solute"/"??? g solute" xx "2.00 g solute")/("0.01000 kg solvent")`

Therefore:

`1/["1.236 mol solute"/cancel"kg solvent" xx (0.01000 cancel"kg solvent")/"2.00 g solute"] = "??? g"/"mol solute"`

And we have `color(blue)"161.8 g/mol"` for the molar mass.