1. What is the freezing point depression of the benzoic/lauric acid solution? 2.

Question

1. What is the freezing point depression of the benzoic/lauric acid solution?

2. What is the Molal concentration of benzoic/lauric acid solution?

3. Amount of Benzoic acid solute

4. Molar mass of benzoic acid(experimental value)

5. Molar mass of benzoic acid(expected value)

6. Molar mass percent difference

There is 7.08 grams of lauric acid in the mixture and 1.71 grams of benzoic acid in the mixture.

freezing point of pure lauric acid is 44.1 degrees celcius

freezing point of benzoic/lauric acid mixture is 36.6 degrees celcius

The molal freezing point depression constant of lauric acid has been determined Kf=3.9 degrrees celcius.

Please let me know if more info is needed and please show detailed steps so I can wrap my head around these concepts. Thankyou

Explanation / Answer

1) DTf= To-Ts
   DTf= 44.1-36.6
   DTf=7.5 degree celicius

2)DTf = i*Kf*m
7.5=1/2*3.9*m
m=7.5*2/3.9
m=3.84 molal

3)molality= (wt.of solute*1000)/(m.wt of solute*wt.of solvent)
3.84=(wt.of benzoic acid*1000)/(122.17*7.08)
wt .of benzoic acid=3.326 gtams

4) DTf= (Kf *wt.of solute*1000)/(m.wt of benzoic acid*wt.of solvent)
     7.5=*3.9*1.71*1000)/(m.wt of benzoic acid*7.08)
      m.wt of benzoic acid = 244.28gr/mole

5) the expected value= 122.14
because the chemical formula of benzoic acid is c6H5COOH
M.WT OF BENZOIC ACID=12*7+6*1+2*16=122gr/mole
to get the exact value we need to multiply with a vantoffs factor (i)

6) the percentage difference=100%
the expected value is 122.14 but expermental value is double the expected value

for benzoic acid vantoffs factor (i)=1/2 (because of association)

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