what would have happened to the calculated molar mass if some benzoic acid was spilled before adding it to the lauric acid?

Question

also, what would have happened to the calculated molar mass if the thermometer used actually read 1.4C too high

I NEED HELP PLEASE

THANKS

Answer 1

Which was the solute and which the solvent? I'm assuming from the question I answered last night about lauric acid that the lauric acid is the solute and benzoic acid is the solvent.
(1)delta T = Kf*m
(2) m = mols/kg solvent
(3) mols = grams/molar mass

Follow from the equations above.
If the benzoic acid is the solvent and you spilled some of it then the m of the solution will be too large from equation 2; i.e., m = mols/kg solvent. If kg solvent is smaller the m is larger from equation 2 and that makes delta T larger from equation 1. Now we go forward.
larger dT in 1 gives larger m, larger m in 2 gives larger mols, larger mols in 3 gives smaller molar mass.

If the thermometer read too high by 1.4 or too low by 1.4 (actually too high or too low by essentially any number) the molar mass is not affected. Why? Because you read 1.4 degrees too high BOTH times so delta T remains the same. Here is an example. Let's say the reading on a good thermometer is 0C for water as the normal freezing point and -10 for the new freezing point so delta T is 0-(-10) = 10 C.
Now lets use the "bad" thermometer which reads too high by 2C (make it easier to calculate). So instead of reading 0 for the normal freezing point it will read +2. Then the NEW freezing point for the solution will be -10+2 = -8. Now delta T = +2-(-8) = 10. Voila!
As long as you measure DELTA T it makes little difference whether it is all that accurate BUT you want it to be "off" by the same amount for every reading you make..

Answer 2

thanks