# Physical Sciences Grade 11 P2 KwaZulu Natal June 2019 Memo

> Grade 11 Physical Sciences Paper 2 memorandum from the KwaZulu Natal June 2019, 18 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/phy-g11-p2-jun-2019-and-memo/f102fd1e-4a1e-4d83-b1d8-7760f4784c21
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 11
- Document type: Memorandum
- Year: 2019
- Exam period: KwaZulu Natal June
- Paper: 2
- Curriculum: CAPS
- Pages: 18

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Physical Sciences · Grade 11 · KwaZulu Natal June · 2019. Memorandum, 18 pages. Read online or download the PDF.

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Physical Sciences P2

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INSTRUCTIONS AND INFORMATION TO CANDIDATES

ae

2.

©

While your name on the ANSWER BOOK

Answer ALL the questions in the answer book.

You may use a non-programmabie calculator.

You may use appropriate mathemalical instruments.

Number the answers correctly according to the numbering system
used in this question paper

YOU ARE ADVISED TO USE THE ATTACHED DATA SHEETS.

Give brief motivations, discussions, et cetera where required.
Show the formulae and substitutions in ALL calculations.

Round off answers to a minimum of TWO decimal places

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Physical Sciences P2 NSC June 2019 Common Test

1.5 2,50 mol of SO2 and 1 mol of O; are sealed in a 1 dm’ flask and allowed
to react completely at STP according to the following balanced equation.

2 SO2 (g)+ O2(g) -> 2 SOs (g)

The TOTAL number of moles of gas in the flask at the END of the reaction is:

A

8
c
D

2

3,50

2,50

0,50 (2)

1.6 The gas law that expresses the relationship between pressure and temperature
of a gas is known as:

A

8
c
D

Charle's Law
Gay- Lussac Law

Boyle's Law
Avogadro's Law (2)

1.7 The type of intermolecular force involved when CO; is added to water is :

A

B
Cc
D

Dipole- induced dipole forces

London forces

Covalent bonds

lon — induced dipole forces (2)

TOTAL : SECTION A [14]

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Physical Sciences P2 NSC June 2019 Common Test

QUESTION 3

Grade 11 learners investigated the effect of intermolecular force on capillarity. They
pour 100ml each of water; glycerine and nail polish remover in separate beakers.
A capillary tube is inserted into each liquid and after a while, the level of liquid in the

capillary tube is measured.
am Capillary tube

They recorded their results in a table as follows:

je Liquid és Height (mm) ]

| Water site (eat | aT

| Glycerine _ oes 5 |

[Nail polish remover 26 |
3.1 State the dependent variable in the above investigation. (1)
3.2 Which liquid displayed the greatest degree of capillarity? (1)
3.3. Explain the answer to question 3.2 above. (3)

3.4 __ Use the results in the table and arrange the liquids in order of INCREASING
STRENGTH of intermolecular force that is, from the weakest intermolecular force
to the strongest intermolecular force. (2)

3.5. Identify the liquid with the highest boiling point. (1)
[8]

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Physical Sciences P2 NSC June 2019 Common Test
QUESTION 5

A compound contains the elements carbon, hydrogen and oxygen only. It consists of
54.56% carbon and 36.36% hydrogen. The molar mass of the compound is 132 g. mol".

5.1 State the definition of empirical formula. (2)
5.2 Calculate the empirical formula of the compound. (6)
5.3 Determine the molecular formula of the compound. (2)

[10}
QUESTION 6

tron is recovered from iron ore (Fe2O3) in a blast furnace. The following reaction takes
place.

Fe203(s) + 3CO(g) — 2 Fe(s) + 3C02g)

In one such reaction, 160g of impure iron ore was reacted and 63 dm’ of CO2 was
produced at STP.

6.1 Write down the definition of the mole. (2)
6.2 Calculate the number of CO2 molecules that formed at STP (4)
6.3. Calculate the maximum no of moles of iron that will be formed in the above

reaction. (2)
6.4 Calculate the percentage purity of the iron ore sample used. ia
QUESTION 7

Industrially, vanadium metal, (V) which is used in steel alloys can be obtained by reacting
vanadium pentoxide( V2Os) with calcium at high temperatures. The balanced equation

for the reaction is:
5Ca+V20s5 +~5CaO +2V

During an industrial process 31850 g of V2Os reacts with 2 x 10* g of Ca.

7.1 State the definition of a limiting reagent. (2)

7.2 Calculate the theoretical yield of vanadium. (6)

7.3 Calculate the percentage yield if 8,67 x 10°g of vanadium is obtained. (2)
[10}

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J0AO WN] OSE%qg

TABLE 3: THE PERIODIC TABLE OF ELEMENTS

5 6 7

Atomic number
Atoomgetal

KEY/SLEUTEL

4

9 10 © 120:«13—C—“‘“ HK 61 18
ay (yy) (vil) (vill)

Electronegativity
Elektronegatiwiteit_y|

29
2 Cu

=

63,5

Symbol
|_—Simbool

Approximate

Benaderde relatiewe atoommassa

atomic mass

241 25
2 Crj2Mn
52| 55

26
= Fe
56

27] 28
2Co/2 Ni
59 | 59

42| 43
2 Mo|? Te
96

44
= Ru
101

45| 46
a Rhit Pd
103| 106

74 75

W)| Re
184| 186

76

Os
190

7 78

Ir| Pt
192| 195

Pr | Nd
141 | 144

61
Pm

Sm
150

91 92

Pa | U
238

93
Np

94
Pu

QQUENS feoeAUg

S01 VOWWOD 6102 OUNr

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Physical Sciences P2             NSC-Marking Guideline          Common Test June 2019
                                              1

                       PHYSICAL SCIENCES P2 (CHEMISTRY)

                                    COMMON TEST

                                       JUNE 2019

                               MARKING GUIDELINE

                                    NATIONAL
                               SENIOR CERTIFICATE

                                        GRADE 11

TIME: 2 hours

MARKS: 100

                         This marking guideline consists of 7 pages.

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Physical Sciences P2     NSC-Marking Guideline                         Common Test June 2019
                                                   2

QUESTION 1

1.1.       C   
1.2.       B   
1.3.       A   
1.4.       D   
1.5.       C   
1.6.       B   
1.7.       A                                                                  7x2 = (14)

QUESTION 2

2.1.1 A and I                                                                                (2)

2.1.2. (B and F)   OR (D and F)           (2 or 0)                                           (2)

2.1.3. G                                                                                      (1)

2.1.4. G                                                                                      (1)

2.1.5. C                                                                                      (1)

2.1.6. E                                                                                      (1)

2.2 H2O has hydrogen bonding and H2S has dipole- dipole forces.  The intermolecular

       forces in water are stronger Therefore more energy is required to break the IMF

           in water.                                                                          (4)

2.3.                     H
                         x
                         ●
                    H x● C ●x H
                         ●
                         x
                         H                                                                   (2)

2.4.1 NON- POLAR
       -
           CO2 is a symmetrical molecule with even distribution of electrons

           There is no net dipole moment/dipoles cancel out/. There is no distinct opposite

           positive and negative ends.                                                         (3)

2.4.2 London forces                                                                           (1)

                                                                                              [18]

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Physical Sciences P2                    NSC-Marking Guideline           Common Test June 2019
                                                     3
QUESTION 3

3.1. Capillarity / height                                                                    (1)

3.2.   Nail polish remover                                                                   (1)

3.3.       It has the weakest intermolecular forces /cohesive forces  molecules
           separate easily.  The adhesive forces are stronger than the cohesive
           forces/IMF                                                                        (3)

3.4.   Nail polish remover; water ; glycerine                                               (2)

3.5    Glycerine                                                                             (1)

                                                                                                  [8]

QUESTION 4

4.1.1 Charle’s Law -- The volume of an enclosed gas is directly proportional to its
      Kelvin temperature provided the pressure is kept constant.                            (3)

4.1.2. Amount/mass of gas                                                                    (1)

4.1.3. volume is directly proportional to Kelvin temperature/ Vα T / volume is linearly

           proportional to Celsius temperature

       ACCEPT As temperature increases, volume increases.                                     (1)

4.1.4. P3                                                                                    (1)
       -
4.1.5. From pV = nRT , gradient which is V/T is equal to nR/P. Since R and n are constant, 

           Gradient is inversely proportional to the pressure/ higher pressure represents smaller

           gradient.                                                                         (3)

OR

From Boyles law --- at a constant temperature,  the gas with the highest volume will have
the lowest pressure pV = k

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Physical Sciences P2    NSC-Marking Guideline                             Common Test June 2019
                                                   4
4.2.
                                                              OPTION 2
          OPTION 1

          Change in pressure = 100-55 = 45kPa                pV = nRT

                                                              100 x 103 x 10 x 10-3 = n x 8.31 x 298
          pV = nRT
                                                              n= 0.404mol ( original no. of
          45 x 103 x 10 x 10-3 = n x 8.31 x 298
                                                              moles of oxygen in vessel)
          n = 0.182mol
                                                              pV = nRT
          n = m/M
                     [20]                                     55 x 103 x 10 x 10-3 = n x 8.31 x 298
          0.182 = m/32
                                                              n= 0.222mol ( no of mol in vessel
          m = 5.824g  of O2 is lost                          after leak is repaired)

                                                              no of moles of gas leaked
NB: ACCEPT p values in kPa with V values in dm3               =0.404 – 0.222
                                                              = 0.182mol

                                                              n = m/M

                                                              0.182 = m/32

                                                              m = 5.824g  of O2 is lost

                                                                                                       (8)

                                                                                                       [17]

QUESTION 5

5.1 Simplest whole number ratio in which elements in a compound combine                          (2)

5.2.

    Element     Mass per 100 g       n=m/M(mol)                   Simplest ratio
    C           54.56                54.56/12 = 4.547            4.547/2.2725 = 2
    H           9.08                9.08/1 = 9.08               9.08/2.2725 = 4   
    O           36.36                36.36/16 = 2.2725           2.2725/2.2725 = 1

       Empirical formula is C2H4O                                                                    (6)

5.3 n = True Mr/ Empirical Mr

       = 132/44

       = 3

Molecular formula is C6 H12 O3 (award both marks if answer correct without calculation)           (2)
                                                                                                  [10]

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Physical Sciences P2                  NSC-Marking Guideline           Common Test June 2019
                                                   5
QUESTION 6

6.1.   The amount of substance having the same number of particles as there
       are atoms in 12g of C-12. / Amount of substance having 6,02 x 1023
       elementary particles.                                                               (2)

6.2.
        n = V/Vm                         N = 2.8125 x 6.02 x 1023

          = 63/22.4                        = 1.693 x 10 24 molecules of CO2

          = 2.8125 mol
                                                                                               (4)

6.3
        POSITIVE MARKING FROM 6.2

        Fe : CO2

              2:3       

             nFe = 2.8125 x2/3
                                                                                               (4)
                     = 1.875mol
                                                                                                 (2)

6.4.
            Positive marking form 6.2/6.3

            CO2 : Fe2O3                   OR           Fe : Fe2O3
             3 :1                                       2 :1
            nFe2O3 = 2.8125/3                   nFe2O3 = 1.875/2
                    = 0.9375 mol                       = 0.9375 mol

            mFe2O3 = nM
                    = 0.9375 x 160
                                                                                (4)
                    = 150 g
                                                                                           [12]
            % purity = 150/160 x 100

                         = 93,75% 

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Physical Sciences P2     NSC-Marking Guideline                    Common Test June 2019
                                              6
QUESTION 7

7.1 Substance that is used up completely in a reaction.                                 (2)

7.2

       Ratio           5         1                5           2
       Moles           Ca        V2O5             CaO         V

       Initial         500      175             0           0

       Change          500      100              500         200

       End             0         75               500         200

Ca                                        V2O5

n = m/M                                   n = m/M

  = 20 000/40                             = 31 850/182

      = 500 mol                           = 175mol

nV formed = 200 mol

m = nxM

  = 200 x 51

      = 10 200 g                                                                   (6)

7.3 % yield = 8670/10 200 x 100

                 = 85 %                                                            (2)

                                                                                   [10]

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Physical Sciences P2              NSC-Marking Guideline                Common Test June 2019
                                               7

QUESTION 8

8.1 Amount of solute per litre of solution.                                               (2)

8.2. n = m/M

  = 4.14/69 

      = 0.06mol                                                                            (2)

8.3 Positive marking from 8.2.

              c = n/V

         0.05 =0.06/V

       V = 1,20 dm3                                                                        (3)

8.4

         Positive marking from 8.3
                                            Positive marking from 8.2. and 8.3
            c1V1 = c2V2 
                                            c = n/V
      0.05 x 1.20 = c2 x 1.45
                                                  =0.06/1.45
       C2 = 0.04 mol.dm-3
                                                   c= 0.04 mol.dm-3

                                                                                           (4)

                                                                                         [11]

                                                                       TOTAL MARKS : 100

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