ib exam question on titration, uncertainty calculation, ideal gas and open ended questions

http://lawrencekok.blogspot.com

Prepared by Lawrence Kok

IB Exam Questions on Titration, Ideal Gas, Uncertainty Calculation and Open Ended Questions.

http://lawrencekok.blogspot.com/

Equilibrium established when ethanoic acid and ethanol react together

in strong acid, using propanone as solvent. Eqn given.

CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O Density ethanoic acid is 1.05 g cm–3.

i.Find amt, mol, of acid presentii.Conc acid is 1.748 mol dm–3. Find % uncertainty of conc.

Titration performed on acid using a base. Result shown below

Find absolute uncertainty of titre for Titration 1 (27.60 cm3).

Liquid Vol/cm3 Ethanoic acid 5.00 ±

0.05

Ethanol 5.00 ± 0.05

Hydrochloric acid 1.00 ± 0.02

Propanone 39.0 ± 0.5

gmassvoldenstiymass

25.500.505.1

volmassDensity

molMol

RMMmassMol

0874.06025.5

RMM acid = 60

% uncertainty acid conc = % uncertainty in vol acid + % uncertainty in total vol

volmolacidConc .

(0.05/5.00) x 100 % = 1 %

(0.62/50) x 100% = 1.24 %

Total % uncertainty = (1 + 1.24) % = 2.24%

Uncertainty final – initial vol (28.80 ±0.05 – 1.20 ±0.05 ) = (27.60 ± 0.1)

Add absolute uncertainty together

i. Find total uncertainty, in vol of rxn mixtureMixture contained:

5.0 ± 0.1 cm3 of 2M H2O2

5.0 ± 0.1 cm3 of 1 % starch20.0 ± 0.1 cm3 of 1M H2SO4

20.0 ± 0.1 cm3 of 0.01 M Na2S2O3

50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KI

i. Add all vol together: Add all absolute uncertainty together.(5.0 ± 0.1) + (5.0 ± 0.1) + (20.0 ± 0.1) + (20.0 ± 0.1 ) + (50.0 ± 0.1) = (100 ± 0.5) cm 3

ii. Conc KI =Mass/ vol% uncertainty conc KI = % uncertainty mass + % uncertainty vol KI% ∆ mass = (0.0001/0.02) x 100% = 0.5 %% ∆ vol = (0.1/50) x 100% = 0.2 %% conc KI = (0.5 + 0.2)% = 0.7 %

iii. Final Conc KI = Conc KI in total mixture

ii. Find % uncertainty for KI conc in final rxn sol.

iii. Find % uncertainty for KI conc in overall rxn mixture

% ∆ final conc KI = % ∆ conc KI + % ∆ total vol KI

% ∆ conc KI = 0.7 %

% ∆ total vol = (0.5/100) x 100 % = 0.5% % conc KI

= (0.5 + 0.7) = 1.2 %

Mixture contained:5.0 ± 0.1 cm3 of 2M H2O2

5.0 ± 0.1 cm3 of 1 % starch20.0 ± 0.1 cm3 of 1M H2SO4

20.0 ± 0.1 cm3 of 0.01 M Na2S2O3

50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KItotal vol

Only vol/mass/conc KI

Two rxn kinetic investigated using iodine clock rxn.

Reaction A: H2O2 + 2I− + 2H+ → I2 + 2H2 OReaction B: I2 + 2S2O3

2− → 2I− + S4O62-

4.32 x 10-5 x 176.14 = 7.61 x10-3 g Vit C

KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O

3C6H8O6 + 3I2 → 3C6H6O6 + 6I- +

6H+

Iodometric titration on Vit C, (C6H8O6). Vit C titrated with 0.002M KIO3 , using excess KI

and starch.

Redox Titration – Vit C quantification

KIO3

M = 0.002M

Vit CAmt = ?

Mole ratio (1 :3)1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6

1 mol KIO3 3 mol C6H8O6

Amt = ?

transfer1g KI excess + starch

titrated

Vit C

5

3

1032.4)..(

31

)..(0072.0002.0

31

).()(

CVitMole

CVitMole

CVitMVKIOMV

i. Find mass, of KIO3, required to prepare 0.250 dm3 of 0.002M KIO3

ii Titration results shown in table belowFind % uncertainty in mean vol of KIO3 used.

Mean vol = (7.20 ± 0.10) cm3

Find amt of KIO3 used

Mol = M x V = 0.002 x 7.20 1000 = 1.44 x 10-5 mol

41000.5

250.0002.0250.0

002.0

.

mol

mol

molvolmolacidConc

Convert mole KIO3 → Mass/g

X RMM

5.00 x 10-4 x 214.00 = 0.107 g

% ∆ vol = (0.10/7.20) x 100 % = 1.4 %Find amt, Vit C in

sampleFind mass of Vit CConvert mole VIT C → Mass

RMM Vit C – 176.14

M x 0.0292 = 2.5 x 10-3

acid M = 2.5 x 10-3

0.0292 M = 0.0856M

Acid/Base Titration– Ethanoic acid in vinegar

CH3COOH M = ? V = 29.2ml

NaOH M = 0.1MV = 25.0ml

NaOH + CH3COOH → CH3COONa + H2O M = 0.1M M = ?V = 25ml V = 29.2ml

V = 250ml M = ?

Mole ratio (1 : 1) 1 mole NaOH - 1 mole acid 2.5 x 10-3 mole NaOH - 2.5 x 10-3 acid

Mole ratio – 1: 1

Diluted 10x

V = 25 ml M = ?

25ml of conc vinegar (ethanoic acid) was diluted to total vol 250 ml in a flask. Diluted vinegar was transfer to a burette. 25ml, 0.1M NaOH is pipette into a flask, with indicator added. End

point reached when average 29.2 ml of diluted vinegar added. Find its molarity.

mole ratio

Moles bef dilution = Moles aft dilution M1 V1 = M2V2M1 = Ini molarity M2= Final molarityV1 = Ini vol V2 = Final vol

Mole NaOH = MV = (0.1 x 0.025) = 2.5 x 10-3

0856.011

0292.0025.01.011

a

aa

bb

M

VMVM

formula

MMM

VMVM

856.02500856.025

1

1

2211

Acid/Base Titration - Empirical formula Na2CO3. x H2O

HCI M = 0.100 M V = 48.8ml

Na2CO3 M = ? MV = 25 ml

2HCI + Na2CO3 → 2NaCI + CO2 + H2O M = 0.1M M = ?V = 48.8ml V = 25.0ml

V = 1L M = ?

25 ml transfer

Mole ratio – 2: 1

Mass Na2CO3 . x H2O = 27.82 gMass Na2CO3 = 10.36 gMass of water = (27.82 – 10.36) g = 17.46 g

Diuted to 1L

27.82gNa2CO3. xH2O

27.82 g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25 ml of sol was neutralized by 48.8ml of 0.1 M HCI. Cal molarity and mass of Na2CO3 present in 1L of sol. Find x

Convert mol dm-3 → g dm-3 Empirical formula

Na2CO3 H2O

Mass/g

10.36 17.46

RMM 106 18.02

Mole 10.34/106= 0.09773

17.46/18.02= 0.9689

Lowest

ratio

0.09773/0.09733 1

0.9689/0.09733 10

Empirical formulaNa2CO3 . 10 H2O

MM

VMVM

b

bb

aa

0976.012

0250.00488.01.012

0.0976 x 106 = 10.36g/dm3

X RMM

Redox Titration - % Fe in iron tablet

Iron tablet contain FeSO4.7H2O. One tablet weigh 1.863 g crushed, dissolved in water and sol made up to 250 ml. 10ml of this sol added to

20 ml of H2SO4 and titrated with 0.002M KMnO4. Ave 24.5ml need to reach end point. Cal % Fe in iron tablet.

10ml transfer 20ml acid added

1.863 g250ml

KMnO4

M = 0.002MV = 24.5 ml

Fe2+

M = ? V = 30ml

MnO4- + 5Fe2+ + 8H+ → Mn2+ +

5Fe2+ + 4H2OM = 0.002M M = ? V = 24.5ml

Mole ratio – 1: 5

Mass (expt yield) = 1.703gMass (Actual ) = 1.863g% Fe = 1.703 x 100% 1.863 = 91.4%

6.125 x 10-3 x 278.05 = 1.703 g FeSO4

10ml sol contain - 2.45 x 10-4 Fe2+

250ml sol contain - 250 x 2.45 x 10-4

Fe2+

10 = 6.125 x 10-3 mole Fe2+

42

2

1045.2.51

.0245.0002.051

FeMoleFeMole

VMVM

bb

aa

Convert mole → Mass

X RMM

Mole bef dil = Mole aft dil M1 V1 = M2V2

M1 x 10 = 1.78 x 10-2 x 250 M1 = 1.78 x 10-2

x 250 10 M1 = 0.445M

2CIO- + 2I- + 2H+ → I2 + 2CI- + H2O

I2 + 2S2O32- → S4O6

2- + 2I-

10ml bleach (CIO-) diluted to a vol of 250 ml. 20ml is added to 1g of KI (excess) and iodine produced is titrated with 0.0206 M Na2S2O3.Using

starch indicator, end point was 17.3ml. Cal molarity of CIO in bleach.

Redox Titration – CIO- in Bleach

Na2S2O3

M = 0.0206MV = 17.3ml

I2

M = ?

Mole ratio ( 1 : 1)2 mole CIO- : 1 mole I2 : 2 mole S2O3

2-

2 mole CIO- 2 mole S2O3

2-

10.0ml CIO-

transfer

V = 250mlM = 1.78 x 10-2 M

20ml transfer

1g KI excess added

M x V = Mol CIO- M x V = 3.56 x 10-4 M x 0.02 = 3.56 x 10-4 M = 3.56 x 10-4 002 M = 1.78 x 10-2 M diluted 25x

Diuted 25x

V = 10M = ?

titrated

Water added till 250ml

4

32

1056.3)..(22

0173.00206.0).(22

)()(

CIOMole

CIOMoleOSMV

CIOMV

KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O

3C6H8O6 + 3I2 → 3C6H6O6 + 6I- +

6H+

Iodometric titration on Vit C, (C6H8O6). 25 ml Vit C titrated with 0.002M KIO3 from burette,

using excess KI and starch. Ave vol KIO3 is 25.5ml. Cal molarity of Vit C.

Redox Titration – Vit C quantification

KIO3

M = 0.002MV = 25.5ml

Vit CM = ?V = 25ml

Mole ratio (1 :3)1 mol KIO3 : 3 mol I2 : 3 mol C6H8O61 mol KIO3 3 mol C6H8O6

V = 25mlM = ?

25ml transfer1g KI excess + starch

titrated

Vit C

4

3

1053.1)..(

31

)..(0255.0002.0

31

).()(

CVitMole

CVitMole

CVitMVKIOMV M x V = Mol Vit C

M x V = 1.53 x 10-4 M x 0.025 = 3.56 x 10-4 M = 3.56 x 10-4 0025 M = 6.12 x 10-3 M

2.82 x 10-3 x 63.5 = 0.179 g Cu in 25ml

1.79 g Cu in 250ml

% Cu = mass Cu x 100% mass brass = 1.79 x 100% 2.5 = 71.8%

2Cu2+ + 4I- → I2 + 2CuI

I2 + 2S2O32- → S4O6

2- + 2I-

2.5g brass react with 10ml HNO3 producing Cu2+ ion. Sol made up to 250ml using water. Pipette 25ml of sol to flask. Na2CO3 add to neutralize excess acid. 1g KI (excess) and starch added. Titrate with 0.1M S2O3

2- and end point, is 28.2 ml. Find molarity Cu 2+ and % Cu found in brass.

Redox Titration - % Cu in Brass

Na2S2O3

M = 0.1MV = 28.2ml

I2M = ?

Mole ratio (1 : 1)2 mol Cu2+ : 1 mol I2 : 2 mol S2O3

2-

2 mol Cu2+ 2 mol S2O3

2-

Pour into Volumetric flask

V = 250mlM = ?

25ml transfer

1g KI excess/ starch

10 ml HNO3

titrated

Water added 250ml

2.5g brass

32

2

232

2

1082.2).(22

0282.01.0).(

22

)()(

CuMole

CuMole

OSMVCuMV

M x V = Mol Cu 2+

M x V = 2.82 x 10-3 M x 0.025 = 2.82 x 10-3 M = 2.82 x 10-

3 0025 M = 1.13 x 10-3

M

Convert mole Cu → Mass Cu

X RMM

X 10

2Cu2+ + 4I- → I2 + 2CuI

I2 + 2S2O32- → S4O6

2- + 2I-

0.456 g brass react with 25ml HNO3 producing Cu2+ ions. Sol was titrate with 0.1M S2O3

2- and end point, reached when 28.5ml added. Cal mol, mass, molarity of Cu 2+ and %

Cu in brass.

Redox Titration - % Cu in Brass

Na2S2O3

M = 0.1MV = 28.5ml

I2

M = ?

Mole ratio (1 : 1)2 mol Cu2+ : 1 mol I2 : 2 mol S2O3

2-

2 mol Cu2+ 2 mol S2O3

2-

transfer

1g KI excess starch

25ml HNO3

titrated

0.456g brass

32

2

232

2

1085.2).(22

0285.01.0).(

22

)()(

CuMole

CuMole

OSMVCuMV

M x V = Mol Cu 2+

M x V = 2.85 x 10-3 M x 0.025 = 2.85 x 10-3 M = 2.85 x 10-

3 0025 M = 1.14 x 10-3

M

Convert mole Cu → Mass Cu

2.85 x 10-3 x 63.5 = 0.18 g Cu

X RMM

% Cu = mass Cu x 100% mass brass = 0.18 x 100% 0.456 = 39.7 %

% Calcium carbonate in egg shell - Back Titration

250ml, 2M HNO3

Amt of HNO3 added

Amt of base (egg)

Amt of HNO3 left

Titrate NaOH M = 1.0 V = 17.0ml

Amt HNO3 react = Amt HNO3 – Amt HNO3 add left

HNO3 left

Transfer to flask

Left overnight in acid

added

25 g of egg shell (CaCO3) dissolved in 250 ml, 2 M HNO3. Sol titrated with NaOH.

17 ml, 1M NaOH needed to neutralize excess acid. Cal % CaCO3 by mass in egg shell.

NaOH + HNO3 → NaNO3 +

H2O M = 1.00M mol = ?V = 17 ml

Amt HNO3 add = M x V = 2.0 x 0.250 = 0.50 mol

Amt HNO3 react = Amt HNO3 add – Amt HNO3 left = 0.50 – 1.7 x 10-2 = 0.483 mol 2HNO3 + CaCO3 → (CaNO3)2 + H2O + 2CO2 Mole Mole 0.483 ?

Mole ratio (2 : 1) 2 mol HNO3 - 1 mol CaCO3 0.483 mol HNO3 - o.242 mol CaCO3

2107.1..

11

).(017.000.111

acidMole

acidMole

VMVM

aa

bb

25 g impureCaCO3 in egg shell

Convert mole CaCO3 → Mass /g

X RMM

0.242 x 100 = 24.2 g CaCO3

% CaCO3 = mass CaCO3 x 100% mass egg = 24.2 x 100% 25.0 = 96.8 %

% Calcium carbonate in egg shell - Back Titration

Amt of HCI added

Amt of base (egg)

Amt of HCI left

Titrate NaOH M = 0.10 V = 23.8 ml

Amt HCI react = Amt HCI – Amt HCI add left

HCI left

Transfer to flask


added

NaOH + HCI → NaCI + H2O M = 0.1 M mol = ?V = 23.8 ml

Amt HCI add = M x V = 0.2 x 0.272 = 0.0544 mol

Amt HCI react = Amt HCI add – Amt HCI left = 0.0544 – 2.38 x 10-3 = 0.00306 mol 2HCI + CaCO3 → CaCI3 + H2O + CO2 Mole Mole 0.00306 ?

Mole ratio (2 : 1) 2 mol HCI - 1 mol CaCO3 0.00306 mol HCI - o.00153 mol

CaCO3

31038.2..

11

).(238.01.011

acidMole

acidMole

VMVM

aa

bb

Convert mole CaCO3 → Mass /g

X RMM

0.00153 x 100 = 0.153 g CaCO3

% CaCO3 = mass CaCO3 x 100% mass egg = 0.153 x 100% 0.188 = 81.4 %

0.188g of egg shell (CaCO3) dissolved in 27.2 ml, 0.2 M HCI. Sol was titrated with NaOH. 23.8ml, 0.10M NaOH need to

neutralize excess acid. Cal mol, mass and % of CaCO3 by mass in egg shell.

0.188g impureCaCO3 in egg shell

27.20ml, 0.2M HCI

Amt of HCI added

Amt of base

Amt of HCI left

Titrate NaOH M = 0.1108 V = 33.64 ml

Amt HCI react = Amt HCI – Amt HCI add left

HCI left

Transfer to flask


added

NaOH + HCI → NaCI + H2O M = 0.1108 M mol = ?V = 33.64 ml

Amt HCI add = M x V = 0.250 x 0.05 = 0.0125 mol

Amt HCI react = Amt HCI add – Amt HCI left = 0.0125 – 3.727 x 10-3 = 0.008773 mol 2HCI + Ca(OH)3 → CaCI3 + H2O Mole Mole 0.008773 ?

Mole ratio (2 : 1) 2 mol HCI - 1 mol Ca(OH)2 0.008773 mol HCI - o.004386 mol

Ca(OH)2

310727.3..

11

).(03364.01108.0

11

acidMole

acidMole

VMVM

aa

bb

Convert mole Ca(OH)2 → Mass /g

X RMM

0.004386 x 74.1 = 0.325g Ca(OH)2

% Ca(OH)2 = mass Ca(OH)2 x 100% mass impure = 0.325 x 100% 0.5214 = 62.3 %

50 ml, 0.250M HCI

% Calcium hydroxide in antacid tablet - Back Titration

0.5214 g of impure Ca(OH)2 from antacid was dissolved in 50 ml, 0.250M HCI.

33.64ml, 0.1108 M NaOH need to neutralize excess acid. Cal % Ca(OH)2 in tablet.

0.5214g impureCa(OH)2

Amt of NaOH added

Amt of acid

Amt of NaOH left

Titrate HCI M = 0.5

V = 17.6 ml

Amt NaOH react = Amt NaOH – Amt NaOH add left

NaOH left

Transfer to flask


added

HCI + NaOH → NaCI + H2O M = 0.5 M mol = ?V = 17.6 ml

Amt NaOH add = M x V = 2 x 0.02 = 0.04 mol

Amt NaOH react = Amt NaOH add – Amt NaOH left = 0.04 – 8.8 x 10-3 = 0.0312 mol 2NaOH + H2A → Na3 A+ 2H2O Mole Mole 0.0312 ?

Mole ratio (2 : 1) 2 mol NaOH - 1 mol acid 0.0312 mol NaOH - 0.0156 mol acid

3108.8.

11

).(0176.05.011

baseMole

acidMole

VMVM

bb

aa

Molar mass of insoluble acid in tablet -Back Titration

2.04 g insoluble acid dissolve in 20 ml, 2M NaOH. Excess NaOH require

17.6 ml, 0.5M HCI to neutralize it. Find molar mass acid

2.04 g impureacid H2A

20 ml, 2M NaOH

Molar Mass Acid0.0156 mol acid - 2.04 g 1 mol acid - 2.04 0.0156 = 131

Element H B O

Step 1 Percentage/% 4.8% 17.7% 77.5%

RAM/RMM 1 11 16

Step 2 Number moles/mol

4.8/1 = 4.8

17.7/11 = 1.6

77.5/16

= 4.84

Step 3 Simplest ratio 4.8/1.6 = 3

1.6/1.6 = 1

4.84/1.6

= 3

Boric acid used to preserve food contains 4.8% hydrogen, 17.7% boron and rest is oxygen.

Find EF of boric acid.

Empirical formula - H3B1O3.

Empirical Formula Calculation

Empirical Formula CalculationStep 1: Write mass/ % of each elementStep 2: Find number of moles of each element (divide with RAM)Step 3: Obtain the simplest ratio

2.5 g of X combined with 4 g of Y to form compound formula XY2.

RAM of Y is 80, Find RAM of X.Element X Y

Step 1 Mass/g 2.5 4

RAM/RMM RAM 80


2.5/RAM= ?

4/80 = 0.05

Step 3 Simplest ratio

1 2 Empirical formula given as X1Y2. RAM = 100

100205.05.221

05.0/5.2

RAMRAM

RAM

CHO + O2 CO2 + H2O

X contain 85.7% of carbon by weight. 4.2 g of gas X occupy vol of 3.36 dm3 at stp.Find EF, RMM and MF of X

Element C H

Step 1 Percentage/% 85.7 14.3

RAM/RMM 12 1


85.7/12 = 7.14

14.3/1 = 14.3

Step 3 Simplest ratio 7.14/7.14

= 1

14.3/7.14 = 2

a) Empirical Formula = C1H2

b) Vol of 3.36 dm3 at stp – Mass, 4.2 g Vol of 22.4dm3 at stp – Mass 1 mol (RMM) 3.36 dm3 – 4.2 g 22.4 dm3 - (4.2 x 22.4)/3.36 = 28

c) Assume molecular formula of X - (CH2)n RMM of X is (12+2)n = 28 n = 2 Molecular formula X = C2H4

X contain carbon, hydrogen and oxygen. 0.50 g of compound on combustion, yield 0.6875 g of carbon dioxide and 0.5625 g of water. Find EF.

Element C H O

Step 1 Mass/g 0.1875 0.0625 0.25

RAM/RMM 12 1 16


0.1875/1 2=

0.01562

0.0625/1= 0.0625

0.25/16 = 0.01562


0.015620.01562

= 1

0.06250.01562

= 4

0.015620.01562

= 1

Conservation of massMass C atom before = Mass C atom afterMass H atom before = Mass C atom after

Mol C atom in CO2= 0.6875 = 0.0156 mol 44

Mass C = mol x RAM C atom = 0.015625 x 12 = 0.1875 g

Mol H atom in H2O= 0.5625 = 0.03125 x 2 = 0.0625 mol 18

Mass H = mol x RAM H atom = 0.0625 x 1 = 0.0625 g

0.6875g 0.5625g0.50g 0.75g

Mass of O = (Mass CHO – Mass C – Mass H) = 0.5 – 0.1875 - 0.0625 = 0.25 g

Empirical formula – C1H4O1


Element C H O

Step 1 Mass/g 0.731 0.0730 0.195

RAM 12.01 1.01 16.01


0.731/12.01

= 0.0609

0.0730/1.01

= 0.0730

0.195/16.01

= 0.0122


0.06090.0122

= 5

0.07300.0122

= 6

0.01220.0122

= 1

CHO + O2 CO2 + H2O2.68 g 0.657 g1.00g


X contain elements carbon, hydrogen and oxygen. Find EF of X, if 1 g X form 2.68 g of carbon dioxide and 0.657 g water when

combust with O2.Find MF of X, if 0.3 mol has mass of 98.5 g.

Empirical Formula = C5H6O1 Mole → Mass 0.3 mol → 98.5 g 1 mol → 98.5/0.3 RMM = 328 gmol-1

Assume MF - (C5H6O1)n = 328RMM = ( (5 x 12.01) +(6 x 1.01) + ( 1 x 16.01) )n = 328 82.11 x n = 328 = 4

MF = (C5H6O1)4 C20H24O4

Find MF, given 0.3 mol X has mass of 98.5 g.




Mass C = mol x RAM C = 0.0609 x 12 = 0.731 g

Mol H atom in H2O= 0.657 x 2 = 0.0729 mol 18

Mass H = mol x RAM H = 0.0729 x 1 = 0.0736 g

Mass O = (Mass CHO – Mass C – Mass H) = 1.0 – 0.731 - 0.0736 = 0.195 g

Find EF for X with composition by mass. S 23.7 %, O 23.7 %, CI 52.6 %

Given, T- 70 C, P- 98 kNm-2 density - 4.67g/dm3 What molecular formula?

Empirical formula - SO2CI2

Density ρ = m (mass) V (vol)

Ideal Gas Equation

Element S O CI

Composition 23.7 23.7 52.6

Moles 23.732.1

= 0.738

23.716.0

= 1.48

52.635.5

= 1.48

Mole ratio 0.738 0.738

1

1.48 0.738

2

1.48 0.738

2PRTM

PRT

VmM

RTMmPV

nRTPV

Density = 4.67 gdm-3

= 4.67 x 10-3 gm-3

M = (4.67 x 10-3) x 8.31 x (273 +70) 9.8 x 104

M = 135.8135.8 = n [ 32 + (2 x 16)+(2 x 35.5) ]135.8 = n [ 135.8]n = 1MF = SO2CI2

P = 98 kN-2

= 9.8 x 104 Nm-2

3.376 g gas occupies 2.368 dm3 at T- 17.6C, P - 96.73 kPa.

Find molar massPV = nRTPV = mass x RT MM = mass x R x T PV = 3.376 x 8.314 x 290.6 96730 x 2.368 x 10-3 = 35.61

Vol = 2.368 dm3

= 2.368 x 10-3 m3

P – 96.73 kPa → 96730Pa

T – 290.6K

6.32 g gas occupy 2200 cm3, T- 100C , P -101 kPa.

Calculate RMM of gas

PV = nRT n = PV RT n = (101 x 103) (2200 x 10-6) 8.31 x ( 373 )

n = 7.17 x 10-2 mol

Vol = 2200 cm3

= 2200 x 10-6 m3 RMM = mass

nRMM = 6.32 7.17 x 10-2

= 88.15

P = 101 kNm-2 = 101 x 103

Nm-2

Calculate RMM of gas Mass empty flask = 25.385 gMass flask fill gas = 26.017 g

Mass flask fill water = 231.985 gTemp = 32C, P = 101 kPa

Find molar mass gas by direct weighing, T-23C , P- 97.7 kPaMass empty flask = 183.257 gMass flask + gas = 187.942 gMass flask + water = 987.560 gMass gas = (187.942 – 183.257) = 4.685 gVol gas = Vol water = Mass water = (987.560 – 183.257) = 804.303 cm3

RMM determination

PV = nRTPV = mass x R x T MM = mass x R x T PV = 4.685 x 8.314 x 296 97700 x 804.303 x 10-6

= 146.7

Vol gas = 804.303 cm3 = 804.303 x 10-6 m3

P = 97.7 kPa = 97700 Pa

Density water = 1g/cm3

M = m x RT PV = 0.632 x 8.314 x 305 101 x 103 x 206 x 10-6 = 76.8

m gas = (26.017 – 25.385) = 0.632 g

vol gas = (231.985 – 25.385) = 206 x 10-6

m3

X contain C, H and O. 0.06234 g of X combusted,

0.1755 g of CO2 and 0.07187 g of H2O produced.

Find EF of X

Element C H O

Step 1 Mass/g 0.0479 0.00805 0.006384

RAM/RMM 12 1 16


0.0479/1 2= 0.00393

0.00805/1=

0.00797

0.006384/16 = 0.000393


0.003930.000393

= 10

0.007970.00039

3= 20

0.0003930.000393

= 1


CHO + O2 CO2 + H2O


Mass C = mol x RAM C = 0.00393 x 12 = 0.0479 g

Mol H atom in H2O= 0.07187 = 0.0039 x 2 = 0.00797 mol 18

Mass H = mol x RAM H = 0.00797 x 1.01 = 0.00805 g

Mass of O = (Mass CHO – Mass C – Mass H) = 0.06234 – 0.0479 - 0.00805 = 0.006384 g

0.06234 g 0.1755 g 0.07187 g


Sodium azide, undergoes decomposition rxn to produce N2 used in air bag

2NaN3(s) → 2Na(s) + 3N2(g)Temp, mass and pressure was collected in table below

i. State number of sig figures for Temp, Mass, and Pressure

i. Temp – 4 sig fig Mass – 3 sig fig Pressure – 3 sig fig

Temp/C Mass NaN3/kg Pressure/atm

25.00 0.0650 1.08

ii. Find amt, mol of NaN3 present

ii.

iii. Find vol of N2, dm3 produced in these condition

RMM NaN3 – 65.02

molMol

RMMmassMol

00.102.600.65

PnRTV

nRTPV

n = 1.50 mol

P – 1.08 x 101000 Pa = 109080 Pa

2NaN3(s) → 2Na(s) + 3N2(g)

T – 25.00 + 273.15 = 298.15K

2 mol – 3 mol N2

1 mol – 1.5 mol N2

33 1.340341.0109080

15.29831.850.1

dmmV

V

PnRTV

Sodium azide, undergoes decomposition rxn to produce N2 used in air bag

2NaN3(s) → 2Na(s) + 3N2(g)Temp, mass and pressure was collected in table below

Temp/C Volume N2/L Pressure/atm

26.0 36 1.15

Find mass of NaN3 needed to produce 36L of N2

RMM NaN3 – 65.02

RTPVn

nRTPV

1.1 x 65.02 = 72 g NaN3

P – 1.15 x 101000 Pa = 116150 Pa

2NaN3(s) → 2Na(s) + 3N2(g)

T – 26.0 + 273.15 = 299.15K

3 mol N2 – 2 mol NaN3

1.7 mol N2 – 1.1 mol NaN3

moln

n

7.115.29931.81036116150 3

Vol = 36 dm3

= 36 x 10-3 m3

Convert mole NaN3 → Mass /g

Copper carbonate, CuCO3, undergo decomposition to produce a gas.

Determine molar mass for gas X

CuCO3(s) → CuO(s) + X (g)Temp, mass, vol and pressure was collected in table below

Temp/K Vol gas/ cm3

Pressure/kPa

Mass gas/g

293 38.1 101.3 0.088 Find Molar mass for gas X

P – 101300 Pa

T – 293 K

Vol = 38.1 cm3

= 38.1 x 10-6 m3PV

mRTM

RTMmPV

nRTPV

5.55101.3810130029331.8088.0

6

M

M

Potassium chlorate, KCIO3, undergo decomposition to produce a O2.

Find amt O2 collected and mass of KCIO3 decomposed

KCIO3

Temp/K Vol gas/ dm3

Pressure/kPa

299 0.250 101.32KCIO3(s) → 2KCI(s) + 3O2 (g)

RTPVn

nRTPV

2

3

.010.029931.8

10250.0101300

Omoln

n

Vol = 0.250 dm3

= 0.250 x 10-3 m3

P – 101300 Pa

Convert mole KCIO3 → Mass

2KCIO3 → 2KCI + 3O2

2 mol – 3 mol O2

0.0066 mol – 0.01 mol O2

0.0066 x 122.6 = 0.81 g KCIO3

RMM KCIO3 – 122.6

Biological Oxygen Demand

2Mn2+ + O2 + 4OH- → 2MnO2 + 2H2O

2MnO2 + 4I- + 4H+ → 4I2 + 2Mn2- + 4H2O

4I2 + 4S2O3 2- → 4I- + 2S4O6 2-

Dissolve O2 reacts with alkaline manganese (Mn2+) to form (Mn4+)

4Mn2+ + 4OH- → 2Mn(OH)2

1 mol 2 mol2Mn(OH)2 + O2 → 2MnO(OH)2

2MnO(OH)2 + 8H+ + 6I- → 2I3- +

6H2O2 mol 2 mol

2I3- + 4S2O3 2- → 6I- +

2S4O6 2-

2 mol 4 mol

DO bottle

Mn2+ salt

1g KI excessalkaline/OH-

shake

White ppt Mn(OH)2

Conc H2SO4

White ppt dissolve in acid

Na2S2O3

M = 0.05MV = 12.5ml

titrated S2O32-

1 O2 + 4 S2O32- →

products M = ? M = 0.05M V = 12.5ml

I- oxidized to I2 by Mn2+

O2M = ?V = 500ml

2 mol 4 mol

4 mol 4 mol

1 mol O2 : 4 mol I2 : 4 mol S2O3

2-

1 mol O2 4 mol S2O3

2-

Brown I2 sol form

Starch added

Water sample added

1 mol O2 : 4 mol S2O32-

Iodometric titration I2/thiosulphate/starch

↓Mn2+ oxidized by O2 to

Mn4+ ↓

Mn4+ oxidized I- to I2

I2 react with starch (blue black colour)

↓S2O3

2- added to reduce I2

↓I2 used up – blue black

disappear

1 mol 2 mol

41

)()(

32

2 OSMVOMV

Redox titration Winkler Method

Biological Oxygen Demand

Dissolve O2 reacts with alkaline manganese (Mn2+) to form (Mn4+)

DO bottle

Mn2+ salt

1g KI excessalkaline/OH-

shake

White ppt Mn(OH)2

Conc H2SO4

White ppt dissolve in acid

Na2S2O3

M = 0.05MV = 12.5ml

titrated S2O32-

1 O2 + 4 S2O32- →

products M = ? M = 0.05M V = 12.5ml

I- oxidized to I2 by Mn2+

O2M = ?V = 500ml

1 mol O2 : 4 mol I2 : 4 mol S2O3

2-

1 mol O2 4 mol S2O3

2-

Brown I2 sol form

Starch added

Water sample added

1 mol O2 : 4 mol S2O32-

41

)()(

32

2 OSMVOMV

500ml water tested for dissolve oxygen by adding Mn2+ in alkaline sol, followed by addition of KI and acid. I2 produced is reduced by titrating with 0.05M S2O3

2-. Ave vol S2O32- used is

12.50ml. Cal dissolved oxygen in g/dm3.

molOMole

OSMVOMole

OSMVOMV

42

322

32

2

1056.10125.005.041.

)(41.

41

)()(

Mass O2 = (1.56 x 10-4 x 32.0) g = 0.005 g in 500ml = 0.01 g in 1000ml = 0.01 g/dm3

Convert mole O2 → Mass /g

RMM O3 – 32

Redox titration Winkler Method

Student A

Weigh 1.00 g X and add 100ml water – 1.00%Perform 10 x serial dilution

Transfer 1 ml of 1 % to 9 ml = 0.1%Transfer 1 ml from 0.1% to 9 ml = 0.01%

Transfer 1 ml from 0.01% to 9ml = 0.001%

How both student overcome this problem.

Using serial dilution technique. Dilution factor 10x.

Given conc of 1.00 % = 1.00 g in 100 ml water.Student A told to prepare 0.001% of X = (0.001g X in

100 ml)Student B told to prepare 0.001% of Y = (0.001g Y in

100 ml)Electronic balance has only precision of 0.00 ±

0.01.

Student B

Weigh 1.00 g Y and add 100ml – 1.00%Transfer 1 ml of 1% to 999 ml = 0.001%

1%0.1% 0.01%0.001%

1 ml 1 ml 1 ml

1 ml

0.001%1%

Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/

Prepared by Lawrence Kok

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