1, Assume that in the acetic acid experiment you obtained the following values:

Question

1, Assume that in the acetic acid experiment you obtained the following values:

Tf(acetic acid)=16.1^degC

Tf(with 3mL of unknown)= 11.3^degC

a. Calculate the moecular weight of the unknown.

b.Your instuctor shows you that you made a mistake in the interpolation and that the correct value is Tf (with 3mL of unknown)=11.0^degC. Calculate the new molecular weight of the unknown. What can you conclude about the importance of determining the tempeature accurately in this experiment?

2. A solution is prepared from 0.75g of an unknown nonelectrolyte and 4.0 grams of lauric acid. The freezing point of this solution is, 40.6^degC. Determine the molecular mass of the unknown. (Note that freezing point of pure lauric acid is 43.2^C and molality freezing point contsant is 3.9^degCkg/mol).

Explanation / Answer

Q1 Kf for acetic acid = 3.90 ; DeltaTf = 16.1-11.3 =4.8

Step2 Molality of unknown= 4.8/3.90 =1.2307; Let d be density of unknown liquid and M be its Molar mass

Step3 Moles of liquid = 3d/M = 1.2307 ; if we know d then we can find M = 3d/1.2307

(b) If correct value is 11 then Delta Tf = 5.1

Molality = 5.1/3.9 =1.3077

Molar Mass = 3d/1.2307 ; this shows the sensitivity of calculations towards temperature.

Q2 Let M be molar mass of unknown ; Moles of unknown = .75/M

Step2 Molality of unknown = (.75/M)x(1000/4)= 187.5/M

Step3 DeltaTf =2.6 ; Kf =3.90

Step4 3.90x187.5/M = 2.6 or M= 281.25

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