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?
Kf Benzene = 5.12 degrees celsius * kg/mol
Kf Benzene = 5.12 degrees celsius * kg/mol
1 Answer
Well, you could simply calculate the first thing that you can get, and go from there. I get about
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 listK_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
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
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