Posted by Anonymous on Sunday, February 10, 2013 at 10:08pm.
If you are talking about the freezing point method there will be no effect. WHY? Because the reading of the normal freezing point is 1 degree too high and the reading for the freezing point of the solution containing the solute is alo 1 degree too high; however, the DIFFERENCE reads as it should.
Example:
normal freezing point = 5.0
freezing point soln = 4.5
Difference = 5.0-4.5 = 0.5
Now suppose the thermometer reads too high by 3 degrees.
Normal freezing point = 8.0
freezing oint soln = 7.5
Difference = 8.0-7.5 = 0.5
Voila.
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