2011實作試題解答

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Name:

Code:

43 rd

_{International}

Chemistry Olympiad

Practical Tasks

Answer Sheets

Grading

12 July 2011

Ankara, Turkey

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Task 1

12% of the total

Analysis of Chloride Mixtures

A

B

1

2

3 Task 1

x%

16

2

6

42

12 A. Determination of total chloride by Fajans Method

Exact concentration of AgNO3 in standard solution = 0.09968 mol/L

Volume of standard AgNO3 solution used

V = mL,

1 point is for correct number of significant figures from reading volumes.

(15 +1 points)

V-V

_true 0 15 points Volume of AgNO3 0 pnts 0 pnts

pnts = -27.27 (|V-V

_true

|) + 19.09

+0.15 mL -0.7 mL -0.15 mL +0.7 mL Vtrue (mL) Unknown 1 18.00 Unknown 2 20.00 Unknown 3 22.00 Unknown 4 24.00

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B. Determination of Mg

2+

by direct titration with EDTA

Exact concentration of EDTA in standard solution = 0.01019 mol/L

Volume of standard EDTA solution used V = mL

1 point is for correct number of significant figures from reading volumes.

(15+1 points)

Treatment of Data

1.

Total amount of Cl- ion in 100.0 mL unknown solution = mmol (2 pt, within ±10%) Unknown 1 17.94 Unknown 2 19.94 Unknown 3 21.93 Unknown 4 23.92

V-V

_true 0 15 points Volume of EDTA 0 pnts 0 pnts

pnts = -15 (|V-V

_true

|) + 18.75

+0.25 mL -2.25 mL -0.25 mL +1.25 mL Vtrue (mL) Unknown 1 24.00 Unknown 2 22.00 Unknown 3 20.00 Unknown 4 18.00

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2.

Total amount of Mg2+ ion in 100.0 mL unknown solution = mmol (2 pt, within ±10%)

Unknown 1 0.9782 Unknown 2 0.8967 Unknown 3 0.8152 Unknown 4 0.7337

3. For

MgCl

2

=

g/100 mL

(2 pt, within ±10%) Unknown 1 0.09313 Unknown 2 0.08537 Unknown 3 0.07762 Unknown 4 0.06986

For

NaCl =

g/100 mL

(4

pt, within ±10%)

Unknown 1 0.9341 Unknown 2 1.061 Unknown 3 1.186 Unknown 4 1.312

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Task 2

12% of the total

Hydrogen generation from ammonia borane

Part A. Reaction of ammonia borane in the absence of catalyst

1. The gas volume versus time data and the graph for the reaction of ammonia borane in the absence of catalyst

Time (min)

Vgas

(mL) Part A. 4 in the absence of catalyst (mL) 0 1 2 3 4 5 6 7 8 9 10

(3 points)

1 point for the graph

If the volume of gas produced stays within the range of 0-2 mL 2 points

>2 mL 1 points

>3 mL 0 points

A

B

Task 2

x%

1

2

1

2

3

4 i

ii

3

1

12

2

24

12

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2. Report the volume of gas evolved, Vuncatalyzed.

V

uncatalyzed = 0 mL

(1 point)

Part B. Reaction of ammonia borane in the presence of catalyst

1. The gas volume versus time data and the graph for the reaction of ammonia borane in the presence of catalyst

Time (min)

Vgas

(mL) Part B. Gas volume versus time graph in the presence of catalyst

0

1

17

2

31

3

45

4

57

5

68

6

73

7

73

8

73

9

73

10

(12 points)

6 points for V

measured

> 0.92 V

max

, Points for volume = 30 (V

measured

- 0.8V

max

)/V

max

6 points for rate

measured

> 0.92 rate

max

, Points for rate = 30 (rate

measured

- 0.8rate

max

)/rate

max

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2. Calculate the maximum number of moles and the maximum volume (mL) of hydrogen gas which will be evolved theoretically from the hydrolysis of 29.5 mg ammonia borane with a purity of 97% w/w at 25 °C. The atmospheric pressure is 690 torr.

V(H2)max = mL

(2 points)

3. Calculate the rate of hydrogen generation in your experiment. i) in mL H2/ min

Rate = mL H2·min-1

(2 points)

ii) in mmol H2/ min by assuming that the temperature is 25 °C. The atmospheric pressure is 690 torr.

Rate = mmol H2·min-1

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purity of potassium tetrachloropalladate(II) is 98% w/w.

The rate of hydrogen production per mole of palladium = mol H2·(mol Pd)-1·min-1

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Task 3

16% of the total

1

2

3

4 Task 3

X%

i

ii

5

6

3

12

38

16 Synthesis, purification and separation of a

diastereomeric mixture

1. Copy (sketch) the TLC1 plate in bag “TLC1” on your answer sheet.

1) For the appearance of three spots loaded on the base line,

1 point.

2) For the well developed and separated spots on the TLC.

2 points

3) For the solvent front line and the base line

1 point

4) For the accurate representation of the sketch of the TLC plate.

1 point

5 points

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2. Copy (sketch) the TLC2 plate in bag “TLC2” on your answer sheet.

1) For the appearance of three spots loaded on the base line.

1 point.

2) For the well developed and separated spots on the TLC.

2 points

3) For the solvent front line and the base line

1 point

4) For the accurate representation of the sketch of the TLC plate.

1 point

5) For the absence of fraction B in fraction A and vice versa.

1 point

6 points

TLC2

3. Determine and record the Rfvalues of the spots on the TLC2 plate in bag “TLC2.”

4. Measure and record the volume and absorbance values for fraction A and fraction B.

Spots Rf value Fraction A Fraction B Starting material (SM) 0.40-0.50 0.20-0.30 0.65-0.75

3 points

Sample

Volume

Absorbance

Fraction A Fraction B mL 3 points mL 3 points 3 points 3 points

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If 20.0 mL ≤ volume 0.0 point If 20.0 mL ≤ volume < 30.0 mL, 0.0-3.0 points If 30.0 mL ≤ volume < 45.0 mL, 3.0 points If 45.0 ≤ volume < 55.0 mL, 3.0-0.0 points If 55.0 ≤ volume 0.0 points If absorbance < 0.30, 0 points If 0,30 ≤ absorbance < 0.40, 0-3 point If 0,40 ≤ absorbance < 0.60, 3.0 points If 0,60≤ absorbance < 0.70, 3-0 points If absorbance ≥ 0.75, 0 points If %yield < 35, 0.0 points If 35.0 ≤ %yield < 45.0, 0-6 points If 45.0 ≤ %yield < 53.0, 6 points If 53.0 ≤ %yield < 56.0, 6-0 points If %yield ≥ 56, 0 points If 80.0 mL ≤ volume 0.0 points If 80.0 mL < volume ≤100.0 mL, 0.0-3.0 points If 100.0 ≤ volume < 140.0 mL, 3.0 points If 140.0 ≤ volume < 150.0 mL, 3.0-0.0 points If volume ≥ 150.0 mL, 0.0 points If absorbance < 0.45, 0 points If 0.45 ≤ absorbance < 0.55, 0-3 point If 0.55 ≤ absorbance < 0.70, 3.0 points If 0,70 ≤ absorbance < 0.80, 3.0-0.0 points If absorbance ≥ 0.8, 0.0 points If %yield < 23, 0.0 points If 23.0 ≤ %yield < 33.0, 0-6 points If 33.0 ≤ %yield < 41.0, 6 points If 41.0 ≤ %yield < 44, 6-0 points If %yield ≥ 45, 0 point

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