For the 6 questions I am offering 600 points, 100 for each. Since I cannot edit

Question

For the 6 questions I am offering 600 points, 100 for each. Since I cannot edit the points I just did 2 separate questions if you answer all of the I will give you 5 starts in both. Please show me the calculations, so I have an idea on how to solve it and study it for the test.

Thread for the other 300 points: http://www.chegg.com/homework-help/questions-and-answers/6-questions-offering-600-points-100--since-cannot-edit-points-2-separate-questions-answer--q7795065

Trial 2 44.5g 65.9 g Trial 1 44.7g 66.3 g Mass of empty test tube Mass of test tube + CH3COOH Mass of CH3COOH Freezing temperature of pure CH3COOH 21.6g 15.3 C 21.4g 16.1 C 1.21g 11.2 C 1.25g ass Of unknown Freezing Temperature of solution e of 10.9 C 10.9c

Explanation / Answer

1) we know that

depression in freezing point is given by

dTf = freezing point of pure substance - freezing point of solution

so

for Trail 1:

dTf = 15.3 - 10.9

dTf = 4.4 C

for Trail 2 :

dTf = 16.1 - 11.2

dTf = 4.9 C

2)

we know that

dTf = Kf x m

for acetic acid Kf is 3.9

so

for Trail 1 :

4.4 = 3.9 x m

m = 1.128

so the molality is 1.128m

for Trail 2 :

4.9 = 3.9 x m

m = 1.256

so the molality is 1.256m

3)

for Trail 1 :

mass of acetic acid = 21.6/1000 = 0.0216 kg

for Trail 2 :

mass of acetic acid = 21.4 / 1000 = 0.0214 kg

4)

we know that

molality = moles of unknown / mass of acetic acid (kg)

so

moles of unknown = molality x mass of acetic acid (kg)

now

for trail 1:

moles of unknown = 1.128 x 0.0216

moles of unknown = 0.02437

for Trail 2 :

moles of unknown = 1.256 x 0.0214

moles of unknown = 0.02689

5)

we know that

molecular weight of unknown = mass / moles

so

For trail 1 :

molecular weight of unknown = 1.25 / 0.02437

molecular weight of unknown = 51.29 g/mol

For Trail 2 :

molecular weight of unknown = 1.21 / 0.02689

molecular weight of unknown = 45 g/mol

6)

average molecular weight = (molecular weight in Trai1 + Trail 2) / 2

average molecular weight = ( 51.29 + 45 ) / 2

average molecular weight = 48.145 g/mol

so

the average molecular weight is 48.145 g/mol

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