Post-lab questions: 1. Benzene, which freezes at 5.50 oC, has a Kf value of 5.10

Question

Post-lab questions:

1. Benzene, which freezes at 5.50 oC, has a Kf value of 5.10 oCkg/mol. How does this compare to that of t-butyl alcohol determined in this lab?

2. Would more or less solute be required to cause the same temperature change observed in this lab if the solvent were benzene?

3. If 0.10 mole of a nonelectrolyte solute were added to 50.0 g of benzene, what would the freezing point of the solution be?

4. How would the results in question 3 change if the solute were an ionic compound that dissociated to form two ions in solution? See van’t Hoff in your text book.

Explanation / Answer

Q1

Typical data on t-butyl alcohol is about 8-9 °C/molal

clearly, the benzene is not that effective lowering its depression point when solvents are added

Q2

we will need much more solute in benzene in order to achieve that of butyl alcohol

we need larger amount of solute per kg of solvent

Q3

dTf = Kf*m

dtf = 5.5*(0.1)/(50*10^-3)

dTf = 11°C

Tf = 5.5-11 = -5.5 °C

Q4

if it is ionic, then 2 ionis --> i = 2, insted of i = 1

dTf = i*Kf*m

dtf = 2*5.5*(0.1)/(50*10^-3)

dTf = 22°C

Tfmoix = 5.5-22 = -16.5°C

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