what would have happened to your calculated molar mass if some lauric acid was spiled before finding the freezing point?

Answer 1

(1)delta T = Kf*m
Substitute delta T and Kf; solve for m

(2)m = mols/kg solvent
Substitute m and kg solvent; solve for mols

(3)mols = grams/molar mass. Rearrange to molar mass = grams/mols.

Now follow through with equation 1, 2, and 3.

If some of the lauric acid (but you have weighed a larger amout) is apilled delta T in 1 will be too small because you have too little to begin with. That makes m too small since k is a constant; i.e., m = delta T/Kf.

Too small m in equation 2 gives mols that is too small since mols = m x kg solvent; since m is too small then m x kg solvent will be too small.

Then go to equation 3. This one (rearranged form) is
molar mass = grams/mol. If you have mols too small that gives a molar mass that is too large. So your calculated molar mass is too large.