Question

A Freezing-Point Depression experiment is done using 2 different test tubes to find the molecular weight of an unknown mixture. Tube 1 consists of pure Lauric acid, and Tube 2 consists of Lauric acid + an unknown solute mixture. From the calculations of this experiment, the molecular weight of the unknown acid can be found.

A) How would the loss of some Lauric acid from Test Tube 1 have affected the calculation of a molecular weight? Would the calculated molecular weight have been higher, lower, or the same amount?

B) How would the loss of some Lauric acid from Test Tube 2 have affected the results? Would the calculated molecular weight have been too high, too low, or unchanged?

Answer 1

In Freezing point experiment, firstly freezing point of pure lauric acid is measured i.e T1 than the freezing point of a mixture of lauric acid and an unknown acid is measured i.e T2. Here ΔT is calculated using expression ΔT = T1-T2. Further, using values of ΔT and Kf, molality is calculated using equation ΔT = Kf × m. Now, we have molality and weight of lauric acid used, so, we can calculate moles using an equation, molality = moles of solute/kg of solvent. Now, we have moles of solute hence the molecular weight of an unknown acid can be easily calculated using an equation; molar mass = moles/grams of an unknown acid.

1) If some amount of lauric acid is lost from test tube 1 then it results in lower molecular weight of an unknown acid. When some amount of lauric acid is lost means there will be a change in weight of lauric acid compared to the weight of lauric acid taken for the experiment which further results in lower moles of solute. Hence, lower molecular weight is observed.

2) If some amount of lauric acid is lost from a mixture of lauric acid and unknown acid that is from test tube 2 then it results in higher molecular weight of an unknown acid. Values of ΔT, molality, and moles of solute will remain the same. But calculating molecular weight, moles are divided by grams of the mixture is considered. Here, the calculation will be affected due to loss of lauric acid from test tube 2. Therefore, increases the molecular weight of an unknown acid.