(A) A student performed the freezing point depression experiment according to directions but mistakenly recorded the mass of lauric acid as 8.300 g instead of 8.003 g. The calculated molar mass of the unknown acid will be ___________ than the true value, because the mass of lauric acid appears in the denominator of the unknown acid's molecular weight determination

A. less
B. greater

(B) therefore, if it was mistakenly recorded too high, it would ___________ the calculated moles of the solute.

A. decrease
B. increase

thanx

delta T = Kb*m.

m = mols/kg solvent.
molar mass = grams/mols.

Use the threee equatin above to answer both questions. I will do the first.
equations 1 and 2 are not needed for the first one. Using equation 3, if the quantity grams is too large, then g/mols will be too large, which will make molar mass too large. (Note: I don't get the statement at the end that the the " mass of lauric acid appears in the denominator of the unknown acid's molecular weight determination ." Where the mass of lauric acid appears (numerator or denominator) depends upon how I set up the equations.

its less and increase

less and increase just did it right now

To determine the effect of the mistakenly recorded mass of lauric acid on the calculated molar mass of the unknown acid, we need to understand how the molar mass is calculated in the freezing point depression experiment.

In the experiment, the freezing point depression is measured when a known amount of solute (in this case, lauric acid) is dissolved in a known amount of solvent. The amount of solute can be determined using the formula:

moles of solute = (mass of solute) / (molar mass of solute)

Now, let's consider the effect of mistakenly recording a higher mass of lauric acid (8.300 g instead of 8.003 g) on the calculated moles of solute.

Since the mass of lauric acid appears in the denominator of the molar mass calculation, a higher recorded mass of lauric acid will result in a higher calculated molar mass for lauric acid.

And because the molar mass of lauric acid is higher, the calculated moles of solute (lauric acid) will be lower. This is because with a higher molar mass, the same mass of lauric acid will correspond to fewer moles.

So, to answer the questions:

(A) The calculated molar mass of the unknown acid will be **less** than the true value because the mistakenly recorded mass of lauric acid is higher, resulting in a higher calculated molar mass for lauric acid.

(B) Therefore, if the recorded mass of lauric acid is mistakenly too high, it would **decrease** the calculated moles of the solute (lauric acid) since a higher molar mass corresponds to fewer moles for the same mass.