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