Download Multiple-choice questions : 1. Which of the following solutions

Multiple-choice questions :
1.
Which of the following solutions contains the largest amount of chloride ions?
A. 100 cm3 of 1 M KCl(aq)
B. 70 cm3 of 2 M NaCl(aq)
C. 80 cm3 of 1 M CaCl2(aq)
D. 50 cm3 of 1 M AlCl3(aq)
□
2.
3
How many moles of ions are present in 200 cm of 1.5 M (NH4)2CO3?
A. 0.3 mole
B.
C.
D.
0.9 mole
1.2 mole
1.5 mole
□
3
3.
What volume of water has to be added to 200 cm of 0.7 M CaCl2 solution so as
to dilute it to 0.5 M?
A. 80 cm3
B. 100 cm3
C. 140 cm3
D. 280 cm3
□
4.
What is the molarity of the resultant MgCl2 solution formed by mixing 100 cm3
of 0.5 M MgCl2 solution and 150 cm3 of 0.9 M MgCl2 solution?
A. 0.70 M
B. 0.74 M
C. 1.23 M
D. 1.85 M
□
5.
What volume of 0.05 M Na2CO3 solution can be prepared from 2.65 g of
Na2CO3?
A. 0.05 dm3
B. 500 dm3
C. 0.5 dm3
D. 50 cm3
□
1
6.
Which of the following cases has different masses of solutes in the two
solutions?
Solution 1
Solution 2
500 cm3 of 0.50 M Na2SO4
100 cm3 of 2.50 M Na2SO4
B. 0.25 dm3 of 0.15 M Na2CO3
0.05 dm3 of 0.75 M Na2CO3
A.
C.
25 cm3 of 0.10 M NaOH
0.05 dm3 of 0.05 M NaOH
D.
100 cm3 of 0.05 M NaCl
0.025 dm3 of 0.02 M NaCl
□
7.
Which of the following information about the number of moles of ions present in
25 cm3 of 0.20 M ammonium sulphate solution is correct?
Ammonium ion
Sulphate ion
A.
5 × 103 mol
5 × 103 mol
B.
5 × 103 mol
1 × 102 mol
C.
1 × 102 mol
1 × 102 mol
D.
1 × 102 mol
5 × 103 mol
□
8.
What volume of water has to be added to 250 cm3 of 0.20 M K2CO3 solution to
dilute it to 0.05 M?
A. 250 cm3
B.
C.
D.
500 cm3
750 cm3
1000 cm3
□
9.
Which of the following information about a 0.025 M sodium carbonate solution
is correct?
Concentration of Na+ ions / M
Concentration of CO32 ions / M
A.
0.025
0.025
B.
0.025
0.05
C.
0.05
0.025
D.
0.05
0.05
□
2
10. Which of the following substances CANNOT be used to prepare a standard
solution directly?
(1) Concentrated H2SO4
(2) Solid NaOH
(3) Liquid NH3
A. (1) and (2) only
B. (1) and (3) only
C. (2) and (3) only
D. (1), (2) and (3)
□
11. During a titration experiment, which of the following pieces of apparatus should
be rinsed with the solution they would deliver (or hold)?
(1) Pipette
(2)
(3)
A.
B.
C.
D.
Conical flask
Burette
(1) and (2) only
(1) and (3) only
(2) and (3) only
(1), (2) and (3)
□
3
12. In a titration experiment, 25.0 cm of dilute sodium hydroxide solution is titrated
against a standard solution of sulphuric acid with phenolphthalein as an indicator.
Which of the following statements concerning this experiment is/ are correct?
(1) The colour of phenolphthalein changes from colourless to pink at the end
point.
(2) The colour of phenolphthalein changes from pink to colourless at the end
point.
(3) The volume of the dilute sodium hydroxide is measured by a pipette.
A. (1) only
B.
C.
D.
(2) only
(1) and (3) only
(2) and (3) only
□
3
13. Which of the following apparatus should be cleaned with the solution to be held
just before using it/ them?
(1) Burette
(2) Conical flask
(3) Pipette
A. (1) only
B. (2) only
C. (1) and (3) only
D. (2) and (3) only
□
14. In a titration experiment, 25.0 cm3 of ammonia solution is titrated against a
standard solution of hydrochloric acid with methyl orange as indicator. Which of
the following colour change is correct at the end point?
A.
B.
C.
D.
Yellow to orange
Red to orange
Red to colourless
Colourless to orange
□
15. What is the burette reading as shown in the diagram below?
3
cm
19
20
21
22
A.
B.
C.
D.
20.40
20.60
20.80
21.20
□
4
16. The following table shows the results of a titration. What is the average volume
of acid added in this titration?
Titration
Burette reading
Final reading (cm3)
Initial reading
(cm3)
Volume of acid added (cm3)
1
2
3
4
24.6
26.1
25.4
26.6
1.1
1.4
0.6
2.1
23.5
24.7
24.8
24.5
3
A.
B.
24.4 cm
24.7 cm3
C.
D.
24.8 cm3
24.5 cm3
□
17. Which of the following apparatus should be used to deliver 25.0 cm3 solution
accurately from a volumetric flask to a conical flask?
(1) Pipette
(2) Burette
(3) Measuring cylinder
A. (1) only
B. (3) only
C.
D.
(1) and (2) only
(2) and (3) only
□
18. The following were titration results for the reaction between 25.00 cm3 of a
sodium hydroxide solution and 0.100 M nitric acid:
Burette readings
1st
2nd
3rd
3
(cm )
Final reading
32.50 32.40 33.20
Initial reading
1.00
2.50
3.10
The molarity of the sodium hydroxide solution would be
A. 0.126 M.
B.
C.
D.
0.124 M.
0.122 M.
0.120 M.
□
5
19. Consider the following two solutions:
Solution P: 50 cm3 of 0.05 M HCl
Solution Q 25 cm3 of 0.1 M CH3COOH
Which of the following statements about solutions P and Q is correct?
A. P reacts with Mg while Q does not.
B. P and Q require the same volume of 0.1 M NaOH for neutralization.
C. P can turn blue litmus paper red while Q cannot.
D. P reacts with sodium hydrogencarbonate while Q does not.
□
20. The concentration of an aqueous solution of an acid is 2.0 M. 20.0 cm3 of this
acid solution requires 80.0 cm3 of 1.0 M sodium hydroxide solution for complete
neutralization. What is the basicity of the acid?
A. 1
B.
C.
D.
2
3
4
□
21. In an experiment, 2.0 M sodium hydroxide solution was added to 20.0 cm3 of 1.0
M sulphuric acid until the acid was completely neutralized. What is the
concentration of sodium sulphate (correct to two decimal places) in the resultant
solution?
A. 0.25 M
B. 0.33 M
C. 0.50 M
D. 1.00 M
□
22. The formula of a solid tribasic acid is H3X. 3.89 g of the acid is dissolved in
250.0cm3 of distilled water. 25.0 cm3 of the dilute solution requires 18.0 cm3 of
0.50 M sodium hydroxide solution for complete neutralization. What is the molar
mass of H3X? (correct to the nearest gram)
A. 100 g
B. 110 g
C. 120 g
D. 130 g
□
6
23. 0.57 g of a sample of hydrated sodium carbonate Na2CO3 ∙ nH2O required 20.0
cm3 of 0.20 M hydrochloric acid for complete neutralization. What is the number
of crystallization, n, in the formula?
A. 1
B. 5
C. 9
D. 10
□
24. 1.05 g of a mixture of anhydrous sodium carbonate and sodium chloride was
dissolved in 50 cm3 of deionized water. The resultant solution required 28.5 cm3
of 0.15 M sulphuric acid for complete reaction. What is the percentage purity of
the anhydrous sodium carbonate sample?
A. 21.4 %
B.
C.
D.
43.8 %
64.3 %
85.6 %
□
25. 25.0 cm3 of 0.20 M sulphuric acid is completely neutralized by 15.5 cm3 of
sodium hydroxide solution. What is the resultant concentration of the sodium
sulphate solution formed?
A.
B.
C.
D.
0.645 M
0.323 M
0.247 M
0.123 M
□
26. A 25.0 cm3 of 1.0 M ethanoic acid and a 25.0 cm3 of 1.0 M hydrochloric acid are
each titrated with a sodium hydroxide solution. Which of the following will be
the same for these two titrations?
(1) Initial pH
(2)
(3)
A.
B.
C.
D.
pH at the end point
Volume of sodium hydroxide solution required to reach the end point
(1) only
(3) only
(1) and (2) only
(2) and (3) only
□
7
Each question below consists of two separate statements. Decide whether each of
the two statements is true or false; if both are true, then decide whether or not
the second statement is a correct explanation of the first statement. Then select
one option from A to D according to the following table:
A. Both statements are true and the 2nd statement is a correct explanation
of the 1st statement.
B. Both statements are true and the 2nd statement is NOT a correct
explanation of the 1st statement.
C. The 1st statement is false but the 2nd statement is true.
D. Both statements are false.
27.
0.15 mole Na2CO3 contains 0.45 mole
ions.
1 formula unit of Na2CO3 gives 2
formula units of Na+ ions and 1
formula unit of CO32 ions.
□
28.
A 25.0 cm3 pipette is usually used to
deliver 22.5 cm3 solution in titration
experiment.
There is a graduation mark on a
pipette.
□
29.
Before doing a titration, distilled
water is used to wash a conical flask.
Water present in a conical flask will
not change the number of moles of
solute in the conical flask.
□
30.
In a titration, 1 mole of any acid always
neutralizes 1 mole of any alkali.
1 mole of hydrogen ions, H+(aq) reacts
with 1 mole of hydroxide ions, OH−(aq).
□
8
Conventional questions :
1.
Pauline has prepared 150 cm3 aqueous solution of 1 M MgSO4. She left the flask
unstoppered and noticed that there was a weight loss in the flask. The weight of
the flask changed as follows:
After 2 days
(a) What was the cause of the weight loss?
(b) Would the resultant solution become concentrated or diluted after two days?
Explain your answer.
(c) What was the molarity of the solution after two days? (Given: Density of
water = 1 g cm−3)
[5M]
9
2.
3.
Many oven cleaners contain a powder, but a few contain a liquid. One brand of
oven cleaner contains 350 cm3 of liquid (which is in fact sodium hydroxide
solution) per bottle. 25.0 cm3 of the liquid were diluted to 250 cm3. 25.0 cm3 of
the diluted solution required 28.5 cm3 of 0.100 M sulphuric acid for
neutralization. Calculate the mass of sodium hydroxide contained in one bottle of
the oven cleaner.
[7M]
(a) What volume of hydrogen chloride gas, measured at room conditions, is
required to make 100 cm3 of 0.25 M hydrochloric acid? (Hint: The volume
of 1 mole of any gas at room conditions is 24.0 dm3)
(b) What volume of water must be added to 25 cm3 of 1.0 M sodium hydroxide
solution to make it 0.20 M?
(c) Concentrated sulphuric acid of density 1.84 g cm−3 contains 98% by mass
of sulphuric acid. What volume of the concentrated acid should be diluted
to prepare 250 cm3 of 0.50 M solution?
10
4.
In an experiment to determine the concentration of ammonia solution, 25.0 cm3
of the ammonia solution was transferred into a conical flask and titrated against
0.1 M sulphuric acid. A few drops of indicator were added. The titration results
are listed in the table below:
1
2
3
4
Final reading (cm3)
15.90
16.70
18.40
18.50
Initial reading (cm3)
0.00
1.50
3.10
3.40
(a) Which indicator is suitable for this titration?
(b) What will be the colour change of the indicator at the end point?
(c) (i)
Calculate the reasonable average volume of sulphuric acid used.
(ii) Calculate the molarity of the ammonia solution.
(d) Name the salt formed and suggest ONE common use of the salt.
[8M]
11
In an experiment, 100 cm3 of 1.0 M hydrochloric acid was added to 0.1 g of
calcium carbonate granules. The graph below shows the results obtained in the
experiment.
Volume of carbon dioxide gas (cm3)
5.
Time of reaction (min)
(a) Write a chemical equation for the reaction between 1.0 M hydrochloric acid
and calcium carbonate granules.
(b) (i)
The experiment was repeated using the same amount of calcium
carbonate and 1.0 M ethanoic acid instead of 1.0 M hydrochloric acid.
Sketch, on the same graph, the results that would be obtained in the
repeated experiment.
(ii) Would the total volumes of carbon dioxide gas collected be the same
for both experiments? Explain your answer.
(c) 1.0 M hydrochloric acid and 1.0 M ethanoic acid have different electrical c
conductivities. State and explain which solution has higher electrical
conductivity.
[6M]
12
6.
25.0 cm3 of sodium carbonate solution was titrated against 0.50 M hydrochloric
acid. 22.0 cm3 of the hydrochloric acid was required for complete neutralization
of the carbonate. Find the concentration of the sodium carbonate solution in
(a) molarity,
(b) g dm−3.
[4M]
13
7.
4.15 g of a sample of hydrated sodium carbonate, of formula Na2CO3 • nH2O, is
dissolved in water and the solution made up to 250.0 cm3. With methyl orange as
indicator, 25.0 cm3 of this solution requires 29.0 cm3 of 0.05 M sulphuric acid
for neutralization. Calculate n, the number of molecules of water of
crystallization in a formula unit of the sodium carbonate hydrate sample.
[5M]
14
8.
9.
0.985 g of an anhydrous metal carbonate (formula XCO3) required 40.0 cm3 of
0.25 M hydrochloric acid for complete neutralization. Calculate the relative
atomic mass of the metal X.
[4M]
4.77 g of an impure sample of anhydrous sodium carbonate required 31.5 cm3 of
1 M sulphuric acid for complete neutralization. What is the percentage purity of
anhydrous sodium carbonate in the sample?
15
[5M]
10. Sodium chloride can be prepared in the school laboratory by reacting 0.5 M
sodium hydroxide solution with dilute hydrochloric acid. A student carried out a
titration to find out how much dilute hydrochloric acid was needed to react with
25.0 cm3 of 0.5 M sodium hydroxide.
(a) Write a full chemical equation for the reaction.
(b) Draw a labelled diagram for the set-up of the titration.
(c) Methyl orange can be used to determine the end point of the titration. State
the colour change at the end point.
(d) Suggest how the student can prepare large pieces of dry sodium chloride
crystals using the titration results.
[6M]
16
Answers :
Multiple-choice questions :
1
2
3
4
5
6
7
8
9
10
C
B
A
B
C
D
D
D
C
D
11
12
13
14
15
16
17
18
19
20
B
D
C
A
C
B
A
D
B
B
21
22
23
24
25
26
27
28
29
30
C
D
D
B
D
B
A
C
A
C
Conventional questions :
1.
(a) It was due to the evaporation of water. [1]
(b) The solution would become concentrated as the amount of MgSO4 remained the
same but there was less solvent. [2]
(c) Let X be the new molarity,
(158.3  135.4)
= 22.9 cm3 [1]
1
Since the number of moles of MgSO4 remained the same before and after
evaporation of water,
Volume of water evaporated =
1×
(150  22.9)
150
=X×
1000
1000
X = 1.18 [1]
The molarity of the solution after two days is 1.18 M.
2.
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l) [1]
From the equation, the mole ratio of H2SO4 to NaOH = 1:2 [1]
25.0 cm3 of the diluted solution required 28.5 cm3 of 0.100 M H2SO4 for
neutralization. [1]
25.0 cm3 of the original solution required 28.5 
H2SO4 for neutralization. [1]
17
250
cm3 = 285 cm3 of 0.100 M
25.0
350
cm3 = 3990 cm3 of 0.100 M
25.0
350 cm3 of the original solution required 285 
H2SO4 for neutralization. [1]
No. of moles of H2SO4 required to neutralize 350 cm3 of the original solution
= 0.100 
3990
mol = 0.399 mol [1]
1000
No. of moles of NaOH in a bottle of the oven cleaner = 0.399  2 mol = 0.798 mol
Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g mol1
Mass of NaOH = 0.798  40.0 = 31.9 g [1]
The oven cleaner contains 31.9 g of sodium hydroxide.
3.
(a) No. of moles of HCl = 0.25 
100
mol = 0.025 mol [1]
1000
Volume of HCl gas = 0.025  24 000 cm3 = 600 cm3 [1]
(b) No. of moles of NaOH remains unchanged on dilution.
(MV) before dilution = (MV) after dilution
1.0 
25
V
= 0.20 
[1]
1000
1000
Volume of 0.20 M NaOH solution = 1.0 
25
1000

cm3 = 125 cm3 [1]
1000
0.20
Volume of water added = (125 – 25) cm3 = 100 cm3 [1]
(c) No. of moles of H2SO4 = 0.50 
250
mol = 0.125 mol [1]
1000
Mass of H2SO4 = 0.125  (1.0  2 + 32.1 + 16.0  4) g = 12.3 g [1]
Mass of concentrated sulphuric acid = 12.3 
Volume of concentrated sulphuric acid =
100
g = 12.6 g [1]
98
12.6
cm3 = 6.85 cm3 [1]
1.84
4.
(a) Methyl orange [1]
(b) From yellow to orange [1]
(c) (i) Ignore the first titration data.
Average volume of sulphuric acid used =
15.20  15.30  15.10 cm3
3
= 15.20 cm3 [1]
(ii) 2NH3(aq) + H2SO4(aq)  (NH4)2SO4(aq)
Number of moles of sulphuric acid used = 0.1 ×
18
15.20
mol
1000
= 0.00152 mol [1]
From the equation, mole ratio of NH3 : H2SO4 = 2 : 1
Number of moles of ammonia in 25.0 cm3 ammonia solution = 0.00152  2
mol = 0.00304 mol [1]
Molarity of ammonia solution =
0.00304
M = 0.122 M [1]
25.0
1000
(d) Ammonium sulphate [1] It is used to make fertilizer. [1]
5.
Volume of carbon dioxide gas (cm3)
(a) 2HCl(aq) + CaCO3(s)  CaCl2(aq) + CO2(g) + H2O(l) [1]
(b) (i)
Time of reaction (min)
[1]
(ii) Yes. This is because calcium carbonate is the limiting reagent. [1]
(c) 1.0 M hydrochloric acid should have higher electrical conductivity. [1]
1.0 M hydrochloric acid is a strong acid which completely ionizes in water, [1]
while 1.0 M ethanoic acid is a weak acid which slightly ionizes in water. [1]
6.
(a) 2HCl(aq) + Na2CO3(aq)  2NaCl(aq) + CO2(g) + H2O(l)
Number of moles of HCl in 22.0 cm3 of 0.50 M HCl
= 0.50 ×
22.0
mol
1000
= 0.011 mol [1]
From the equation, mole ratio of HCl : Na2CO3 = 2 : 1
Number of moles of Na2CO3 in 25.0 cm3 of solution
19
=
0.011
mol
2
= 5.5 × 10−3 mol [1]
Concentration of Na2CO3
5.5  10 3
=
M
25.0
1000
= 0.22 M [1]
(b) Molar mass of Na2CO3
= (23.0 × 2 + 12.0 + 16.0 × 3) = 106.0 g mol−1
From (a), molarity of Na2CO3 is 0.22 mol dm−3,
Concentration of Na2CO3 in g dm−3
= 0.22 × 106.0 g dm−3
= 23.3 g dm−3 [1]
7.
Na2CO3 • nH2O(s) + H2SO4(aq)  Na2SO4(aq) + CO2(g) + (n + 1) H2O(l) [1]
Molar mass of Na2CO3 • nH2O = (106.0 + 18.0n) g mol−1
Number of moles of H2SO4 in 29.0 cm3 of 0.05 M H2SO4
= 0.05 ×
29.0
mol
1000
= 1.45 × 10−3 mol [1]
From the equation, mole ratio of Na2CO3 • nH2O : H2SO4 = 1 : 1
Number of moles of Na2CO3 • nH2O in 25.0 cm3 solution
= 1.45 × 10−3 mol [1]
Number of moles of Na2CO3 • nH2O in 250.0 cm3 solution
250.0
mol
25.0
= 1.45 × 10−3 ×
= 0.0145 mol [1]
∴
4.15
= 0.0145
106.0  18.0n
n = 10 [1]
8.
XCO3 + 2HCl  XCl2 + CO2 + H2O [1]
Number of moles of HCl in 40.0 cm3 of 0.25 M HCl
= 0.25 ×
40.0
mol = 0.01 mol [1]
1000
From the equation, mole ratio of XCO3 : HCl = 1 : 2
Number of moles of XCO3 reacted with HCl
20
=
0.01
mol
2
= 5 × 10−3 mol [1]
Let the relative atomic mass of X be y,
0.985
= 5 × 10−3
y  12.0  16.0  3
y = 137.0 [1]
9.
Na2CO3(s) + H2SO4(aq)  Na2SO4(aq) + CO2(g) + H2O(l) [1]
Number of moles of H2SO4 in 31.5 cm3 of 1 M H2SO4
=1×
31.5
mol
1000
= 0.0315 mol [1]
From the equation, mole ratio of H2SO4 : Na2CO3 = 1 : 1
Number of moles of Na2CO3 in the sample
= 0.0315 mol [1]
Molar mass of Na2CO3
= 106.0 g mol−1
Mass of Na2CO3 in the sample
= 0.0315 × 106.0 g
= 3.339 g [1]
Percentage purity of the Na2CO3 sample
=
3.339
× 100%
4.77
= 70 % [1]
21
10.
(a) NaOH(aq) + HCI(aq)  NaCl(aq) + H2O(l) [1]
(b)
burette
dilute hydrochloric acid
0.5 M sodium hydroxide solution
(+ methyl orange indicator)
white tile
For correct drawing [1]
For correct labeling [1]
(c) From yellow to orange [1]
(d) Repeat the titration, but do not add indicator. Boil the solution to concentrate it.
Then cool the hot concentrated sodium chloride solution under room conditions.
Large pieces of sodium chloride crystals will separate out. Filter the crystals and
dry in an oven. [2]
22