8/13/2019 2009 CT Ans

1/7

1

2009 JC1 H2 Chemistry Common Test (Suggested Solutions)

Section ACCBCC CCBDB DBADA

Section BB1 (a) Amount of H2SO4reacted = (15.00/1000) x 0.0250 = 3.75 x 10

4mol

Amount of OHions in 25.0 cm3diluted soln = 3.75 x 104 x 2 = 7.50 x 104mol

Amount of OH

ions in 250 cm3

diluted soln = 7.50 x 104

x 250/25.0 = 7.50 x 103

mol

Concentration of NaOH in FA1= 7.50 x 103/ (20.0/1000) = 0.375 mol dm3

(b) (i) 5H2O2(aq) + 2MnO4(aq) + 6H+(aq)5O2(g) + 2Mn

2+(aq) + 8H2O(l)

(ii) 10Cl-(aq) + 2MnO4(aq) + 16H+(aq)5Cl2(aq) + 2Mn

2+(aq) + 8H2O(l)

(c) (i) Amount of H2O2reacted = (15.00/1000) x 0.0800 = 1.20 x 103mol

Amount of MnO4that reacts with H2O2= (2/5) x 1.20 x 10

3= 4.80 x 104mol

(ii) Total amount of MnO4= (20.00/1000) x 0.0500 = 1.00 x 103mol

Amount of MnO4that reacts with Clions = 1.00 x 103 4.80 x 104

= 5.20 x 104mol

(d) Total amount of Clions in 25.0 cm3diluted solution = 5 x 5.20 x 104= 2.60 x 103mol

Amount of Clions from NaCl in 250 cm3diluted solution = 250/25.0 x 2.60 x 103= 0.0260 mol

Concentration of NaClin FA1= 0.0185/ (20.0/1000) = 1.30 mol dm3

(e) Concentration of Na+ions in FA1= 0.375 + 1.30 = 1.675 mol dm3

B2 (a) XeF5+(square pyramidal), XeF2(linear)

XeF3 (T-shaped), XeF4(square planar)

(b) In decreasing bond angle from left to right: XeF2> XeF4> XeF3

(c) The availability of vacant low-lying orbitals allow the electrons to be unpaired and used itfor varied number of covalent bonds formation. In addition, the valence shell is far awayfrom the nucleus which thus required minimal amount of energy for the electrons to bepromoted to the vacant orbitals.

8/13/2019 2009 CT Ans

2/7

2

(d) PtF6 has a simple molecular/covalent structure. It consists of discrete molecules heldtogether by weak van der Waals forces. The six highly electronegative fluorine atomscaused the Pt atom to be highly electron deficient, thus making it likely to accept electronand undergo reduction.

B3 (a) (i) pT(20.0 x 103) = (0.200+0.400)(8.31)(25+273)

pT= 74300 Pa (to 3 s.f)

OR

pT(20.0) = (0.200+0.400)(0.082)(25+273)pT= 0.733 atm (to 3 s.f)

(ii) Partial pressure of radon =0.2

743000.6

= 24800 Pa (or 0.245 atm) (to 3 s.f)

Partial Pressure of fluorine = 74300 24800 = 49500 Pa (or 0.489 atm)(or use partial pressure to find)

Comment: Students need to correspond the correct R value to the T and V valuesused.

(iii) Rn + F2 RnF2initial mol 0.2 0.4 0final mol 0 0.4-0.2=0.2 0.2

Total amount of gases = 0.200+0.200 = 0.400 molpf(20 x 10

3) = 0.400(8.31)(127+273)pf= 66500 Pa (to 3 s.f)

OR

pf(20 x 103) = 0.400(0.082)(127+273)

pf= 0.656 atm (to 3 s.f)

*Note the units of V with the different values of R used! Refer to Lecture Notes 4 Pg 5!*Remember to leave your answers in the appropriate units!

B3 (b) (i)

(ii) RnF2will deviate more from ideal behaviour.RnF2has a greater number of electrons (or larger electron cloud) resulting in itselectron cloud being more easily polarised, hence possessing strongerinstantaneous dipole-induced interactions between molecules. The

V / m3

T / 0C273

Constant p

V vs T

8/13/2019 2009 CT Ans

3/7

3

intermolecular forces of attraction between RnF2molecules are more significant,thus it deviates more from ideality.

OR

RnF2 molecules are bigger and the volume occupied is more significantcompared to the volume of the container. Hence, it deviates more from ideality.

B4 (a) Amount of CO2evolved = 50.0 24000 = 2.08 103mol

Na2CO3(s) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l)

Amount of Na2CO3present in weighed sample of FA2 = Amount of CO2evolved

= 2.08 103mol

Molar mass of Na2CO3= (23.0 2) + 12.0 + (16.0 3) = 106.0 g mol1

Mass of Na2CO3in weighed sample of FA2 = 2.08 103 106.0 = 0.221 g

Assume FA2 contains 100 % by mass of Na2CO3(s).

Then mass of FA2 to be used = 0.221 g.

A suitable mass of FA2 to be used is0.220 g

(b) Assume FA2 contains 100 % by mass of Na2CO3(s).

Amount of Na2CO3present = 2.08 103mol

Minimum amount of HCl(aq) used = 2 x 2.08 103mol = 4.16 x 103mol

Minimum volume of HCl(aq) used = 4.16 x 10-3 0.200 = 0.0208 dm3= 20.8 cm3

HCl

(aq) should be used in excess, thus we could use 30 cm

3

of 0.200 mol dm

3

HCl

(aq).(Any answer above 20.8 cm3is accepted)

(c)

Weigh accurately about 0.220 g of FA2 and place it in a small test tube tied to a string.

Introduce 30 cm3of HCl(aq) using a measuring cylinder into a clean and dry 250 cm3conical flask.

Set up the apparatus as shown above in the diagram.

Diagram showing small testtube with FA 2 in conicalflask + showing gassyringe[1m]

Small tube containing

FA2 tied to a string

Stopper

HCl(aq)

Retort stand

Graduated gas

syringe

8/13/2019 2009 CT Ans

4/7

4

11 12 13 14 15 16 17 18 19

Proton number

Na

Ar

S

Al

Cl

P

Si

Mg

K

Lower the filled tube into the conical flask taking care that the reagents do not mix.Stopper the conical flask.

Check that the initial reading of the 100.0 cm3graduated gas syringe is set at the zeromark.

At a suitable time, loosen the stopper slightly to release the string. Stopper the conicalflask immediately. (Or other forms of starting the reaction)

Swirl the conical flask to ensure that the reagents are well mixed.

Allow the reaction to progress until it has ceased as indicated by a constant volumereading of the syringe.

Record the final volume, V1 cm3, on the graduated gas syringe.

Repeat the experiment to get consistent results.

Section C

C1 (a) Cl(g)Cl+(g) + e 1s22s22p63s23p4

B (b)

Important points:Correct axesGeneral increaseExceptionally low for Al and S

(compared to previous member)

8/13/2019 2009 CT Ans

5/7

5

(c) (i) First ionisation energy generally increases across the period from Nato Ar.

From Na to Ar, Nuclear charge increases

Shielding effect remains approximately constant

Effective nuclear charge increases

Electrostatic attraction on the electrons by the nucleusincreases

Hence, more difficult to remove electron, IE increases.

(ii) Mg: 1s22s22p63s2

Al: 1s22s22p63s23p1

Less energy is needed to remove the 3p electron of Alwhich is at ahigher energy level than the 3s electron of Mg. Hence first ionisation

energy of Al is lower than that of Mg.

P: 1s22s22p63s23p3 S: 1s22s22p63s23p4

Extra inter-electron repulsion is predominant between the 3p electronsin the doubly filled 3p orbital of S.Hence less energy is required to remove the paired 3p electron from Scompared to the unpaired 3p electron from P.

(iii) First Ionisation energy of Na is higher than that of K.

Down a group from Na to K,

Nuclear charge increases. Shielding effect experienced by the

valence electrons increases due to greater number of innerfilled principal quantum shells.

Increase in shielding effect outweighs increase in nuclearcharge.

Distance of valence electrons from nucleus increases due togreater number of inner filled principal quantum shells.

Electrostatic attraction on the electrons by the nucleusdecreases.

Hence, easier to remove electron, IE decreases.

C2 (a)

It should be valence electrons of K which experiences stronger shielding effect thanthe valence electrons of Na. NOTK experiences stronger shielding effectPlease write complete sentences, your poor phasing of explanations made it very

difficult for examiner to understand you!

For discontinuities, please include electronicconfigurations.

Quite a number of students are still writing thewrong quantum number (3p electrons NOT 2pelectrons of Mg) or electronic configurations

8/13/2019 2009 CT Ans

6/7

6

(b) NO is a free radical or has an unpaired electron (or mention NO beingelectron deficient). This result in the NO molecule being more reactive thanthe CO molecule.

(c) (i) sp2

(ii)

(d) Increasing order of boiling point from left to right: N2< NO < (NH4)2SO4

(NH4)2SO4 is an ionic compound with strongest electrostatic attractionbetween the ions, thus it has the highest boiling point.

NO consists of polar molecules hence there are permanent dipole-permanentdipole interactions as well as instantaneous dipole-induced dipole interactionsbetween the NO molecules.

N2 consists of non-polar molecules hence there are only instantaneousdipole-induced dipole interactions between the N2 molecules.

NO has stronger intermolecular interactions due to the additional pd-pdinteractions compared to N2.

C3 (a) Na2O (basic) Al2O3(amphoteric) SO3(acidic)

Al2O3(s) + 6H+(aq) 2Al3+(aq) + 3H2O(l)

Al2O3(s) + 2OH-(aq)+ 3H2O(l) 2[Al(OH)4]

(aq)

(b) Na2O dissolves in water to give an alkaline solution (pH = 10 14)

Na2O(s) + H2O(l)2NaOH(aq)

Al2O3is insoluble in water (pH = 7)

SO3dissolves in water to give an acidic solution (pH = 0 4)SO3(l)+ H2O(l) H2SO4(aq) (or SO2(g) + H2O(l) H2SO3(aq))

(c) CO2

AlCl3undergoes appreciable hydrolysis to give an acidic solution. The acidicsolution formed will react with Na2CO3to form CO2.

AlCl3(s) + 6H2O(l) [Al(H2O)6]3+(aq) + 3Cl(aq)

[Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5(OH)]

2+(aq) + H3O+(aq)

(d) (i) SeO2+ 2NaOHNa2SeO3+ H2O

8/13/2019 2009 CT Ans

7/7

7

(ii) SeO2is acidic thus it does not react with HCl(aq)