Sekolah Menengah Kebangsaan St. Michael
Chemistry Form 4 – Acids and Bases
Drilling Exercise – Set 2
Objective Questions
1.
When 2.0 g of sodium hydroxide are present in 500 cm3 of its solution,what is its molarity?
[ Relative atomic mass : Na, 23 ; O, 16; H , 1 ]
A 0.001 moldm-3
2.
0.1 moldm-3
D 1.0 mol dm-3
0.01
B
0.02
C
0.10
D 0.20
0.01
B
0.05
C
0.20
D 0.40
200 cm3 of 0.5 moldm-3 of hydrochloric acid was prepared from a standard solution of
2 moldm-3 hydrochloric acid. Calculate the volume of the standard solution needed to be
dilute with acid ?
A
5.
C
What is the concentration of hydroxide ions when 8.55 g of barium hydroxide is dissolved
in water to produce 250 cm3 of solution ?
[ Relative atomic mass : Ba, 137 ; O, 16 ; H, 1 ]
A
4.
0.01 moldm-3
Calculate the number of moles of sulphuric acid in 100 cm3 0.2 moldm-3 of its solution ?
[ Relative atomic mass : H,1 ; S, 32 ; O, 16 ]
A
3.
B
50 cm3
B
75 cm3
C
100 cm3
D 150 cm3
What is the mass of potassium hydroxide powder used to produce 100 cm3 of 0.5 moldm-3
potassium hydroxide solution ?
[ Relative atomic mass : O, 16; K, 39, H,1 ]
6
A 1.4
B 2.8
C 4.2
D 5.6
Calculate the volume of water needed to be added to 40 cm3 of 2.0 moldm-3 sulphuric acid
to obtain 80 cm3 of 1 moldm-3 sulphuric acid .
A 20 cm3
B 40 cm3
C
60 cm3
D 80 cm3
Questions 7 and 8 are based on the information below :
Zinc powder reacts with hydrochloric acid to produce zinc chloride and hydrogen gas,
according to the equation :
Zn(s) + 2 HCl(aq)
7.
 ZnCl2 (aq) + H2 (g)
Calculate the mass of zinc chloride that is formed when 1.3 g of zinc powder reacts with
excess hydrochloric acid.
[ Relative atomic mass : Zn, 65; Cl, 35.5 ]
A 1.36
B 2.72
C 3.48
1
D
6.72
8
Calculate the volume of the hydrogen gas released at room conditions when 1.3 g of zinc
powder reacts with excess hydrochloric acid.
[ Molar volume : 24 dm3mol-1 at room conditions ]
A
9.
120 cm3
B
C 360 cm3
D 480 dm3
2 HNO3 + Ba(OH)2  Ba(NO3)2 + 2 H2O
25 cm3 of 0.5 mol dm-3 nitric acid reacts with 25 cm3 of barium hydroxide solution. What is
the concentration of barium hydroxide solution ?
A 0.25 moldm-3
10.
240 cm3
B 0.5 moldm-3
C
1.0 moldm-3
D
2.0 moldm-3
H2SO4 + 2 LiOH  Li2SO4 + 2 H2O
50 cm3 of 0.1 moldm-3 sulphuric acid is neutralized by 100 cm3 of lithium hydroxide solution.
What is the concentration of lithium hydroxide solution used in the reaction ?
A 0.005 moldm-3
11.
B 0.1 moldm-3
Na2CO3 + 2 HCl
C 0.2 moldm-3
D 0.025 moldm-3
 2 NaCl + CO2 + H2O
Based on the equation above, what is the volume of gas released at room condition if
25 cm3 of 0.5 moldm-3 hydrochloric acid reacts with sodium carbonate ?
[ Relative atomic mass : H, 1; C, 12; O, 16; Na,23; Cl 35.5; 1 mol of gas occupies 24dm3 at
room condition ]
A 0.15
12.
B
0.30
C 0.24
D 0.50
The equation below represents the reaction between hydrogen sulphide and nitric acid.
3 H2 S
+ 2 HNO3  3 S + 2 NO
+ 4 H 2O
What is the mass of nitrogen monoxide gas released if 240 cm3 of hydrogen sulphide gas
completely reacts with nitric acid at room condition ?
[ Relative atomic mass : N, 14; O, 16; 1 mol of gas occupies 24 dm3 at room condition ]
A 0.2
13.
B 0. 23
C 0.3
D 0.34
In the reaction below, 50 cm3 of hydrochloric acid completely reacts with magnesium ribbon
to produce 120 cm3 of hydrogen gas at room condition.
Mg +
2 HCl  Mg Cl2 + H2
What is the concentration of the hydrochloric acid used ?
[ 1 mol of gas occupies 24 dm3 at room condition ]
A
14.
0.005
B
0.01
C 0.1
D 0.2
H2SO4 + Pb(NO3)2  PbSO4 + 2 HNO3
Calculate the volume of 0.2 moldm-3 sulphuric acid, H2SO4 that is needed to react
completely with 10 cm3 of 0.5 moldm-3 of lead(II) nitrate, Pb(NO3)2 solution.
A
5 cm3
B
10 cm3
C 15 cm3
2
D 25 cm3
15.
Calcium carbonate reacts with hydrochloric acid according to the chemical equation:
CaCO3

+ 2 HCl
CaCl2
+ CO2
+ H2 O
What is the mass of calcium carbonate that is required to produce 600 cm3 of carbon
dioxide gas at room conditions when it reacts with excess hydrochloric acid ?
[ Relative atomic mass : Ca, 40; C, 12 , O, 16. Molar volume : 24 dm3mol-1 at room
conditions ]
A 2.5 g
16.
B
3.5 g
C 6.5 g
D 25 g
5 cm3 of lead (II) nitrate solution requires 20 cm3 of 0.3 moldm-3 potassium iodide solution
to react completely according to the ionic equation below :
Pb2+ (aq)
+ 2 I- (aq)
 PbI2 (s)
What is the concentration of the lead(II) nitrate solution in moldm-3 ?
A
17.
2.4 g
0.6
C
0.9
D 1.2
B 4.8 g
C 6.0 g
D 14.4 g
AgNO3 + KI  AgI + KNO3
Calculate the mass of silver iodide, AgI produced when 25 cm3 of 1.5 moldm-3 silver nitrate,
AgNO3 is mixed with 50 cm3 of 0.5 moldm-3 potassium iodide, KI solution.
[ Relative atomic mass : Ag ,108; I, 127 ]
A 2.5 g
19.
B
CuO + 2 HNO3  Cu(NO3)2 + H2O
Copper (II) oxide powder reacts with excess nitric acid to produre copper(II) nitrate and
water .If 11.28 g of copper(II) nitrate is formed, calculate the mass of copper (II) oxide
used in the reaction.
[ Relative atomic mass; Cu, 64; N, 14; O, 16 ]
A
18.
0.3
B
3.75 g
C
5.875 g
D 8.813 g
Nitric acid reacts with metal X to produce a salt according to the chemical equation below:
X + 2 HNO3 
X ( NO3)2 + H2
Calculate the volume of 1.5 moldm-3 nitric acid that is needed to completely react with
0.54 g of metal X.
[ Relative atomic mass : X, 24 ]
A
20.
15 cm3
B
20 cm3
C
30 cm3
D
45 cm3
10 cm3 of 0.25 moldm-3 T(OH)x solution is neutralized by 5 cm3 of 1.0 moldm-3 HxU solution.
The salt produced from the reaction is
A
TU
B
T 2U
C TU2
3
D
T3U2
Sekolah Menengah Kebangsaan St. Michael
Chemistry Form 4 – Acids and Bases
Drilling Exercise – Set 2
Answer :
1. C
2. B
3. D
4. A
5. B
6. B
7. B
8. D
9. A
10. B
11. A
12. A
13. D
14. D
15. A
16. B
17. B
18. C
19. C
20. B
4