Download - STPM Trials 2009 Chemistry Answer (N Sembilan), Pra-2010

8/8/2019 STPM Trials 2009 Chemistry Answer (N Sembilan), Pra-2010

1/11

ANSWER TRIAL EXAM CHEMISTRY PAPER 1 2009

NO ANSWER NO ANSWER

1 C 26 A

2 B 27 D3 C 28 D

4 A 29 B

5 B 30 D

6 C 31 A

7 A 32 D

8 B 33 A

9 C 34 D

10 B 35 A

11 B 36 A

12 C 37 B

13 C 38 D

14 B 39 A

15 D 40 A

16 B 41 A

17 C 42 B

18 A 43 D

19 C 44 D

20 B 45 B

21 C 46 D

22 D 47 D

23 A 48 C

24 B 49 B

25 C 50 C


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Marking Scheme for Chemistry Paper 2 STPM Trial 2009

Q Answer Mark

1(a) (i) CH3CH2CH2OH CH3CHCH3

OH

CH3OCH2CH3 (either two)

1 + 1

2

(ii) Propan-1-ol / 1-propanol

The peak at m/e 31 indicates the presence of CH2OH group

1

1 2

1(b) (i)

( 1 mark)

frequencywavelength

1

1

2

(ii) Ultraviolet 1 1

(iii) f = cx RH ( 1/n12

- 1/n22)

= 3.0 x 108

x 1.097 x 107

m-1

(1/12

- 1/32)

= 2.93 x 1015 s-1

1

11

3

102(a)(i) Saturated potassium chloride solution/Saturated potassium

nitrate solution or any suitable saturated aqueous ionic salt

solution

1

(ii) Zn(s)/Zn2+(aq)//M+(aq)/M(s) 1

(iii) From zinc to metal M 1 3

(iv) E

cell= E

M- E

Zn

1.56= E

M-(-0.76)

E

M=0.80 V

1

1

2

(b)(i) Number of coulomb =I xt

= 0.50 x 20.0 x 60= 600 C ( mark for final answer with unit) 1

(ii) 0.197 g of Cu is deposited by 600 C

l.0 mol of Cu is deposited by

63.5 x 600 C = 193 000 C

0.197

0.672 g of X is deposited by 600 C.

1


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l.0 mol of X is deposited by

108 x 600 C = 96 400 C

0.672working shown in either above calculations award 1 mark

1

1

(iii) Charge is X+

/ n = 1 1 4

Total 103 (a) Element : Na Mg Al Si S Cl

Formula of oxide Na2O MgO Al2O3 SiO2 SO2 Cl2ONa2O2 SO3 Cl2O7

( 5, 6 correct ---- 3 marks

( 3, 4 correct ---- 2 marks

( 2 correct ----- 1 mark(0, 1 correct ----- 0 mark)

3

3(b)

(i)

Equation: Al2O3 + 2NaOH + 3H2O 2NaAl(OH)4 1

3(b)

(ii)

Equation: Al2O3 + 6HCl 2AlCl3 + 3H2O 1 2

3(c) (i) Equation: 2Sr(NO3)2 2SrO + 4NO2 + O2 1

3(c)(ii)

Magnesium nitrate 1

3(c)

(iii)

Mg2+ has a higher charge density

Mg2+ polarises / distorts the CO32- anion more strongly /

weakens the CO bond

1

1

4

3 (d) BeSO4 , MgSO4, CaSO4, SrSO4, BaSO4 1 1

Total 10

4 (a) H H H H

CCCC

Cl Cl

1 1

4(b)

Equation: CH=CHCl + H2 CH2CH2Cl

Conditions: catalyst: Ni / Pd / Pt and heat

1

1

2

4 (c)

(i)

1 +

1

2

4(c)(ii) Mechanism: electrophilic addition 1 3

A

H H

HCCCl

H Cl

B

H H

HCC=O

H


4/11

4 (d)

Br Br

C C

Cl Cl CH2BrBrH2CH H

mirror

1 +

1

2

4 (e)

CH3 CH3 CH3

+ 2 CH3CCl CCH3 +

O O

O=CCH3

( 1mark) (1 mark)

+ 2HCl (optional)

1 +

1

2

TOTAL 10

5 (a)

(i)

(a) (ii)

The standard enthalpy change of atomization of an element isthe heat absorbed when one mole of gaseous atoms is formedfrom the element in its standard state under standard conditions.

The standard heat of formation of a substance is defined as theheat evolved or absorbed when one mole of the substance is

formed from its elements under standard conditions.

1

1

2


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5 (b)

(i)

Born-Haber Cycle

For each enthalpy change correctly shown

in diagram above ------------ x 8 = 4 marksEnthalpy diagram with E labeled -------- 1 mark

8 x

+ 1

5

By Hesss Law

Hf(CuO) = Hat (Cu) + HIE1st(Cu) + HIE2nd (Cu+) +Hat(O) + HEA1st(O) + HEA2nd (O-) +Hlat (CuO)

-155.2 = (+339.3) + (+750.0) + (+2 000) + (+249) +(-

140.0) + (+786.0) + Hlat (CuO)

Lattice energy CuO = - 4139.5 kJmol-1

1

12

c(i) Molten- AlCl3 exists as simple covalent molecules.Aqueous- AlCl3 dissociates to form Al

3+ / [Al(H2O)6]3+ and

Cl-ions

11 2

Cu2+

(g)+ O(g) +2e

Cu2+

(g) + O2(g) +2e

Cu+

(g) + O2(g) +e

Cu(g) + O2(g)

Cu(s) + O2(g)

CuO(s)

Cu2+

(g) + O-(g) +e

Cu2+

(g) + O2-

(g)

Hat(O)

HIE2nd

Cu+

(g)

HIE1st Cu(g)

HatCu(s)

HfCuO(s)

HEA 1

st(O) HEA 2

nd(O-)

H1stCuO

E (kJ)


6/11

c(ii) AlCl3 is a covalent molecule.

AlF3 is ionic compound

Cl-ion is much larger than F

-ion

Al3+ polarises Cl- ion easily / Al3+ cannot polarise F- easily

1

1

11

4

6(a)(i) pH is negative logarithm to the base 10 of the concentration ofH+(aq) ions in mol dm

-3or

pH = -lg [H+]

1

(ii) Buffer solution- a solution which resists changes in pH when

small amounts of an acid or alkali are added.

1 2

(b)(i) Ka= [H+][CH3COO

-]

[CH3COOH][H+]= [CH3COO

-]

1.75 x 10-5

= [H+]2 or [H+] = Ka x c

0.100[H+] = 1.32 x 10-3

pH= -log [H

+

]= -log (1.32 x 10

-3)

= 2.88

1

1

1

(ii) pH = pKa + lg [CH3COONa][CH3COOH]

= - lg (1.75 x 10-5

) + lg [ (16.4/82.0) ]

[ 250/1000 x 0.100 ]= 4.76 + lg ( 0.20 / 0.025)

= 4.76 + 0.903

= 5.66

1

1

1 6

(c)(i) The partition law states that a solute, Q will distribute itself

between the two immiscible solvents, ether and water untilequilibrium is achieved. At equilibrium the ratio of the

concentrations in the two solvents is constant orConcentration of Q in ether = 12.50

Concentration of Q in water

1

(ii) The partition law holds true when the temperature is constant

the solute is in the same molecular condition in both solvents / (no dissociation or association)

low concentration ( any 2)

11

(iii) Let x= amount of Q extracted by one portion of 25.0 cm3

of

ether

x / 25.0____________ = 12.50

8.00-x/25.0

x= 7.41 g

Amount of Q left in aqueous layer = 8.00 7.41 = 0.59 g

1

1


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1st extraction with 12.5 cm3 ether:

Let y = amount of Q extracted by first 12.5 cm3

of ether

y/ 12.5____________ = 12.50

8.00-y/25.0Therefore amount extracted is 6.90 g

The amount of Q left in aqueous layer after first extraction is

8.00g 6.90g = 1.10 g

2nd

extraction with 12.5 cm3

of ether.

Let z= amount of Q extracted by the 2nd portion of 12.5 cm3 of

ether.

z / 12.5____________ = 12.50

1.10 z / 25.0

z = 0.95 g

Hence amount left in aqueous layer =1.10 0.95 = 0.15 g

1

1

1

max

4

marks

Total 15

Q7 (a) Mr = ( m x R x T ) / pV

= 0.500 x 8.31 x 473

1.00 x 105

x 73.6 x 10-6

= 267 covalent bond anyone labelled

correct shape ---- 1Cl Cl: Cl 2 types of bonds

Al Al labelled ----- 1

Cl :Cl Cl

coordinate bonding / dative bonding (both labelled)

1

1

1+ 1

4

7 (b) Aqueous solution of aluminium sulphate contains [Al(H2O)6]3+

/

hexaaqua aluminium ion

This ion undergoes hydrolysis to produce H

+

/ H3O

+

ions whichmakes the solution acidic with a pH of 4.8.

Equation: Al(H2O)63+

+ H2O [Al(H2O)5OH]2+

+ H3O+

1

1

1

3

7 (c) Carbon dioxide - simple molecular structure with weak van

der waals forces between moleculesSiO2 has a macromolecular structure/ has a giant covalent

structure with strong Si-O covalent bonds

1

1


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CO2 dissolves slightly in water to produce carbonic acid or

CO2 + H2O H2CO3Both CCl4 and SiCl4 are simple non-polar moleculesCCl4 and SiCl4 are liquids, CO2 is gas because:-

the intermolecular forces in CCl4 or SiCl4 are stronger

than in CO2

due to larger molecular sizeSiCl4 can be hydrolysed by waterbut CCl4 cannotSi has empty valence 3d orbitals to accept lone pairs of

electrons to form dative bonds with water molecules

Carbon has no 3d orbitals / no empty valence orbitals

1

1

1

11

1

8

8 (a)

(i)

All the three halides react with aqueous silver nitrate to form

precipitates / insoluble silver salts orAg+ + X- AgX

AgCl is white, AgBr is cream/light yellow and AgI is yellow

With aqueous ammonia, AgCl dissolvesAg+ + 2NH3 Ag(NH3)2+

AgBr and AgI are insoluble in aqueous ammonia

1

1

11

1 5

(a) (ii) X-+ H2SO4 HX + HSO4-

For Cl-, white fumes of HCl gas is evolved

For Br-

white fumes of HBrandbrown fumes of Br2 evolved

For I-, purple (violet)

vapour of I2 are evolved

(equation for oxidation of HBr or HI to Br2 or I2)

1

1

1

11 max

4

(b) is an element that forms at least a stable ion that has partially

filled d-orbitals

Any two properties below (must give correct examples)

green colour of aqueous Fe2+ / yellow colour of aqueous Fe3+

catalytic activity of Fe2+

/ Fe3+

in the reaction between I-and

S2O82-

/ Fe is a catalyst in the Haber processvariable oxidation states +2 , +3, +6 (any 2)

complex formation example [Fe(CN)6]3-

/ FeO42-

1

1 + 1

3

(c) Blue precipitate of Cu(OH)2 is formed or

Cu2+

+ 2OH- Cu(OH)2(s)

blue precipitate dissolves in excess ammonia to form a darkblue solution (complex) - [Cu(NH3)4]2+

or

Cu(OH)2(s) + 4NH3 (aq) [Cu(NH3)4]2+(aq) + 2OH- (aq)

When EDTA is now added, the dark blue solution becomes

lighter blue due to the formation of [Cu(EDTA)]2-

complexor

1

1


9/11

[Cu(NH3)4]2+

+ [EDTA)]4-

[Cu(EDTA)]2- + 4NH3 1 3

15

9 (a)(i)

4-methylphenol dissolves in NaOH(aq) to form a soluble salt

CH3 OH + NaOH CH3 ONa + H2O

Phenylmethanol does not dissolve in NaOH(aq)

1

1

1 3

(a) (ii) 4-methylphenol decolourises bromine

and forms a white precipitateBr

CH3 OH + 2Br2 CH3 OH+ 2 HBr

Br

Phenylmethanol colour of Br2 remains unchanged and there isno white precipitate

1

1

13

(b) (i)

C = O

CH3 C

O

1

1 2

(b) (ii)

CH3 C CH3 CH3 CH2 C- H

O O

1+ 1 2

(b)

(iii)

Both R and S are heated with Tollens reagent / Fehlings

reagent

S - metallic silver (silver mirror) is formed / brick redprecipitate of Cu2O obtained and

R - no silver mirror or metallic silver / no brick red precipitate

1

1

2

(b)

(iv) Cl

CH3 C CH3 (P)

Cl

1


10/11

Cl

CH3 CH2 C H (Q)

Cl

1

2

(b) (v) CH3C(Cl2)CH3 + 2KOH CH3COCH3 + 2KCl + H2O

CH3CH2CHCl2 + 2KOH CH3CH2CHO + 2KCl + H2O

1

1 2max 9

15

10 (a)

(i)

B contains a chiral carbon because optically active

A fumes in moist air, thus A has acyl functional group / -C-Cl

Structure B Hence, Structure A O

Reaction of A with NaOH produces

(C)

On acidification, C is obtained.

1

1

1+1

1 + 1

(a) (ii)

+ _

+ H + HCl

( 1 mark for organic product)(1 mark for equation)

1 + 1 8

CH3

Cl-C-COOH

H

CH3

HO-C-COONa

H

CH3

HO-C-COOH

H

CH3

Cl-C-COCl

H

CH3

Cl-C-COCl

H

OH

CH3-CH-CH3

CH3

CH3

Cl-CH-COOCH-CH3


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(b)(i) Geometrical isomerism (cis-trans isomerism) for L

(cis isomer)

(trans-isomer )

1

1

1

(ii) Reaction of L with an aqueous solution of hot sodium

hydroxide

and

CH3(CH2)3 CH2OH

1

1

(c) Compound: 1- pentanolConditions: concentrated sulphuric acid , reflux / heat

11 7

Total 15

COOCH2(CH2)3CH3

H H

COOCH2(CH2)3CH3

C=CCH3O

H

H

C=C

CH3O

CH3O

O-Na

+

O

CH=CH C

Download - STPM Trials 2009 Chemistry Answer (N Sembilan), Pra-2010

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