Based on these results, what is the Kf of biphenyl?
0.20 grams of benzene (C6H6) is dissolved into 5.0 grams of melted biphenyl. The freezing point of the mixture was determined to be 4.4 degrees below the freezing point of pure biphenyl.
(Use standard units in your calculation (C/m) but do not include the units in your answer. Also note these are experimental values so they may not equal the literature value for Kf).
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Dissolving a solute into a solvent lowers the freezing point of the solvent. The best example of this is spreading salt on roads in the winter since NaCl dissolved in water lowers the freezing point of water and keeps the roads from being icy. The equation for the freezing point depression of a solution is:
where deltaT is the change in freezing point temperature, Kf is the freezing point constant for the particular solvent, m is the concentration of the solution in terms of molality (moles solute per kg solvent), and i is the van't Hoff factor which accounts for ionic species. In this case the solute is benzene which is not ionic so i=1. Let's determine the molality of the solution:
m=[0.2 grams benzene*(1 mole/78.12 grams)]/0.005 kg biphenyl=0.512 m (mol/kg).
Since the deltaT is given as 4.4 degrees, we can now solve for Kf:
Kf=deltaT/m=4.4 deg C/0.512 m=8.59
The Kf of biphenyl is 8.59.
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