CHEMISTRY FORM THREE QUESTIONS AND ANSWERS
FORM III TOPIC 4. MOLE CONCEPT AND RELATED CALCULATIONS: BANK OF QUESTIONS WITH SOLUTIONSQUESTIONS
1. Multiple choice.
Choose the most correct answer from the given alternatives.
(i)The mass of one molecule of a compound compared with the mass of one atom of carbon 12 is called _______
A: molar mass
B: relative atomic mass
C: moles
D: molarity
(ii) Molar mass of 2H2O (water) ...............
A: 18 B: 36 C: 17 D: 54
(iii) .............. is a ratio of mass or volume of the substance to its molar mass or molar volume.
A: mole
B: molarity
C: concentration
D: relative atomic mass
(iv) In order to obtain ................ , Avogadro's constant is multiplied with number of moles of the substance.
A: molar mass
B: molar solution
C: electrons
D: number of particles.
(v) .................. is any known amount of solute dissolved in water to make one litre.
A: molar solution
B: concentration
C: molarity
D: number of particles
(vi) The relationship of quantities between reactants and products in a chemical reaction is known as...............
A: stoichiometry
B: quantitative ratio
C: Avogadro's number
D: mole ratio
(vii) The mass of one mole of the substance is known as
A: molarity
B: mass
C: molar mass
D: molar solution
(viii) One mole of solute forms ................ when dissolved to make one litre of the solution.
A: molarity
B: molar solution
C: molar volume
D: moles
(ix) A number of atoms in exactly 12g of carbon 12 isotope
A: number of particles
B: Avogadro's number
C: number of moles
D: number of molecules
(x) At standard temperature and pressure equal volume of all gases contain same number of particles in ..................
A: 22.4 L
B: 22.4cm3
C: 1 L
D. 1000cm3
2. Matching items.
Match items in list A with responses in list B
List A
(i) concentration in mole per litre
(ii) number of particles of 20g of Ca
(iii) volume occupied by one mole of gas in 22.4dm3 at s.t.p
(iv)molarity of 10g of NaOH in 500cm3
(v) Avogadro's constant
List B
A: molar volume
B: 6.02 x 10^23
C: molarity
D: 3.01 x 10^-23
E: molarity
F: 0.5
3. Define the terms
(a) mole
(b) molar mass
(c) molarity
(d) concentration
(e) molar volume
(f) molar solution.
4. (a) What volume of hydrogen gas is produced at STP when 0.65g of zinc reacts completely with excess hydrochloric acid?
(b) What mass of sulphuric acid is found in 400 cm^3 of its 0.1M aqueous solution?
(c) What is the molarity of :-
(i)solution containing 5.3g of sodium carbonate in 100ml of the solution?
(ii) solution containing 80g of sodium hydroxide in 2 litre?
5. Calculate amount of substance in the following:-
(a) 0.69g of sodium
(b) 180g of carbon
(c) 0.98g of sulphuric acid
6.(i) Find the mass of the following substances
(a) 0.78 mol of Ca(CN)2
(b) 7 mol of hydrogen peroxide
(c) 3.01 x 10^-23 particles of Ca.
(ii) Find the number of particles in the following substances
(a) 16.8 litres of chlorine gas at STP.
(b) 0.8 mol of nitric acid.
(c) 160g sodium carbonate
7. A gas occupies 1.95dm^3 at STP. Calculate the number of :-
(i) moles in the gas
(ii) molecules in the gas.
8.(a) Ammonium chloride reacts with calcium hydroxide as shown in the following equation
2NH4Cl (aq) + Ca(OH)2 (aq) ----->
2NH3 (g) + CaCl2 (aq) + 2H2O(l)
Calculate the mass of ammonium chloride that will react completely with 14.8g of calcium hydroxide.
(b) Sodium sulphate reacts with barium nitrate as shown in the equation below.
Na2So4 (aq) + Ba(NO3)2 --------->
BaSo4 (s) + 2NaNO3 (aq)
How many moles of barium sulphate are produced when 80.5g of sodium sulphate react completely with barium nitrate?
9. (a) What is the relative molecular mass of gas Y if the mass of 560 cm^3 of gas Y measured at STP is 1.1g? (Y is not the actual symbol of the gas)
(b) 4.9g of copper (II) oxide yield 3.92g of copper when reduced by hydrogen gas. Calculate the mass of off copper (II) oxide required to yield 4g of copper. (Cu=64, O=16)
10. (a) 20cm^3 of NaOH solution containing 8gdm^3 was required for complete neutralization of 0.18g of dibasic acid. Calculate the relative molecular mass of the acid(Na=23, O=16, H=1).
(b) What mass of CO2 (g) will occupy 224cm^3 at STP (C=12, O=16
11.🏾What is the concentration of resulting solution of HCl, when 100cm3 of 0.1M HCl mixed with 100cm3 of 1M HCl?
SOLUTIONS.
1. (i) C (ii) A (iii) A (iv) D (v) C
(vi) A (vii) C (viii) B (ix) B (x) A
2. (i) C (ii) D (iii) A (iv) F (v) B
3.(a) Mole of substance is the amount of a substance which contains the Avogadro’s number of particles.
(b) Molar mass:- is a mass of one mole of the compound.
For example. 2H2SO4 (sulphuric acid)
-its molar mass = Mole of substance is the amount of
a substance which contains the Avogadro’s
number of particles.
-its composition by mass
= 2((1x2)+32+(16x4))
= 196g.
(c) Molarity :- is a number of moles of moles of solute in one litre of the solution.
(d)molar volume:- is number of moles of gas in 22.4L at standard temperature and pressure.
(e) molar solution :- is a solution which contains one mole of solute dissolved in a solvent to make one litre(1dm^3) of the solution.
4. (a) given data.
Mass = 0.65g
Molar mass = 98g/mol.
Volume = needed
Firstly calculate number of moles then convert into volume
n= mass
Molar mass
= 0.65
65.5
= 0.0099mol.
From the eqn.
Zn + 2HCl ------> ZnCl2 + H2
This means
1mol.(Zn) -------> 1mol. (H)
0.0099 ----------> ?
? = 0.0099 x 1
1
= 0.0099mol(H)
Convert the obtained moles of hydrogen into volume.
From.
n = volume
Molar gas volume
Volume = n x molar gas volume
= 0.0099 x 22.4
= 0.222dm^3.
(b) given data.
Molarity = 0.1M
Volume = 400cm^3
Mass = needed
Molar mass = 98.
Firstly find the number of moles in 400cm^3 then convert moles into mass.
Remember that Molarity means the number of moles of solute that dissolved to make one litre of the solution.
0.1mol. -------> 1000cm^3
? <----------- 400="" cm="" p="">
?= 0.1 x 400
1000
0.04mol(in 400cm^3)
Convert 0.04mol. Into mass.
n = mass.
Molar mass
Mass = n x molar mass
= 0.04 x 98
= 3.92g.
(c) (i) data given.
Volume = 100cm^3
Mass = 5.3g
Molarity = required
Molar mass = 106g/mol.
But molarity = number of mole in one litre of the solution.
Also = concentration(g/dm^3)
Molar mass
Convert 5.3g in 100cm^3 into g/L(litre)
5.3g ---------> 100cm^3
? <----------- 1000cm="" p="">
? = 5.3 x 1000
100
= 53g/dm^3
Molarity = concentration
Molar mass
= 53
106
= 0.5M
(ii)
data given.
Volume = 2L (2000cm^3)
Mass = 80g
Molarity = required
Molar mass = 40g/mol.
But molarity = number of mole in one litre of the solution.
Also = concentration(g/dm^3)
Molar mass
Convert 80g in 2000cm^3 into g/L(litre) or dm^3
80g ---------> 2000cm^3
? <----------- 1000cm="" p="">
? = 80 x 1000
2000
= 40g/dm^3
Molarity = concentration
Molar mass
= 40
40
= 1M
5. (a) Given data.
Mass = 0.69g
Molar mass = 23g/mol.
Number of moles(n) = needed
n = mass
Molar mass
= 0.69
23
= 0.03mol.
(b) given data.
Mass = 180g
Molar mass = 12g/mol.
Number of moles(n)=required
n = Mass/ Molar mass
= 180
12
= 15mol.
(c) Mass = 0.98g
Molar mass = 98g/mol.
Number of moles(n)=required
n = mass /Molar mass
= 0.98/ 98
=o.1mol.
6. (i) a) given data.
Number of mol. = 0.78mol.
Molar mass(Ca(CN)2) = 92g/mol
Mass of (Ca(CN)2) = needed
From.
n= mass
Molar mass
Mass = n x molar mass
= 0.78 x 92
= 71.76g
(b) given data.
n (H2O2) = 7mol.
Molar mass (H2O2) = 34g/mol
mass (H2O2) = needed
Mass = n x molar mass
= 7 x 34
= 34g/mol.
(c) given data.
Number of particles(N)
=3.01 x 10^23
LA• = 6.02 x 10^23
Mass= needed
Firstly find the number of moles (n)
From.
N = n x LA•
n= N
LA•
= 3.01 x 10^23
6.02 x 10^23
= 0.5
Mass = n x molar mass
= 0.5 x 40
= 20g/mol.
(ii) (a) given data.
Volume = 16.8L
Molar gas volume = 22.4L
Number of particles(N) = needed
Firstly calculate number of moles.
From.
n = volume
Molar volume
= 16.8
22.4
= 0.75mol.
Where by
N= n x LA•
= 0.75 x 6.02 x 10^23
=4.515 particles
(b) given data.
n = 0.8mol.
LA• = 6.02 x 10^23
N = needed
From.
N= n x LA•
= 0.8 x 6.02 x 10^23
= 4.816 particles
(c) given data.
Mass = 160g
Molar mass = 106g/mol.
N= needed
LA• = 6.02.x 10^23
N= n x LA•
Whereby
n = mass
Molar mass
= 160
106
= 1.509mol.
N= 1.509 x 6.02 x 10^23
= 9.084 particles
5.
7. (i) given data.
Volume of gas = 1.95dm^3
Molar gas volume = 22.4dm^3
Number of moles = needed
From.
n = Volume
Molar gas volume
= 1.95
22.4
= 0.087mol.
(ii) Available data.
Number of moles(n) = 0.087mol.
Avogadro's constant (LA•) = 6.02 x 10^23
Number of moles = needed
Number o particles = n x LA•
= 0.087 x 6.02 x 10^23
= 0.524 x 10^23 particles.
8. (a) Given data.
Mass(NH4Cl) = needed
Mass((Ca(OH)2) = 14.8
From the eqn.
2NH4Cl + Ca(OH)2 ---------> 2NH3 + CaCl2 + 2H2O
Mass(NH4) = 2(14 + (1 x 4))
= 36g
Mass(Ca(OH)2) = (40 + 2(16 + 1)
= 74g
36g(NH4) --------> 74g (Ca(OH)2)
?<----------- 14.8="" p=""> ?= 36 x 14.8
74
=7.2g
7.2g of NH4 were needed
Alternative solution.
Number of moles can be used then converted into mass.
Firstly convert 14.8g of Ca(OH)2 into moles.
n = mass
Molar mass
= 14.8
74
0.2mol.
Equation shows that:-
2mol.(NH4) ----------> 1mol.(Ca(OH)4)
? <------------- 0.2mol="" nbsp="" p=""> ? = 2 x 0.2
1
=0.4 mol(NH4)
Convert it into mass.
Mass = n x molar mass
Molar mass of ammonium = 18
Mass of ammonium needed
0.4 x 18
= 7.2g
(b) data given.
Mass(Na2SO4) =80.5g
From the eqn.
Na2SO4 + Ba(NO3)2 -------> NaNO3 + BaSO4
It shows that:-
Mass(Na2SO4) = 142g
Mass (Ba(NO3)2 = 261.3g
142g(Na2SO4)-->261.3g(Ba(NO3)2
80.5g -----------> ?
? = 80.5 x 261.3
142
= 148.131g Ba(NO3)2 required
Convert nto moles.
n = mass
Molar mass
= 148.131
261.3
= 0.567mol.
Alternative solution.
Firstly convert 80.5g(Na2SO4) into moles.
n = 80.5
142
= 0.567mol.
Eqn. Shows that
1mol (Na2SO4 ---->1mol(Ba(NO3)2
0.567 ----------> ?
= 0.567mol.(Ba(NO3)2
9. (a) given data.
Volume of gas Y at STP= 560 cm^3
Mass of gas Y at STP = 1.1g
Calculate number of moles then use number of moles obtained to find the molar mass of the gas since its mass is given.
Fro n= volume (dm^3)
Molar gas volume(dm^3)
(b) data given.
Mass of copper oxide = 4.9g
Mass if copper = 3.9
Mass of copper oxide = needed, when 4g of copper produced.
CuO + H2 -----> Cu + H2O
4.9g(CuO) ---------> 3.92(Cu)
? <----------- 4g="" p=""> ? = 4.9 x 4
3.92
=5g
5g of CuO were needed
10. Given data
Volume of NaOH used=20
Concentration NaOH = 8g/dm^3
Mass of the acid used = 0.18g
Molar mass of acid = needed.
From balanced chemical equation
2NaOH + H2X -----> Na2X + 2H2O
Base acid mole ratio is 2:1
This means base volume used is twice of acid used.
There fore in order to get volume of acid, volume of base must be divided by 2
Volume of acid = 20cm^3
2
=10 cm^3
Now find the molarity of acid base then molarity if acid.
From
Molarity(NaOH) = conc.
Molar mass
8
40
0.2M
From
MaVa = na
MbVb nb
Whereby Mb = 0.2M
Ma =needed
Vb = 20cm^3
Va = 10cm^3
na = 1
nb = 2
Ma x 10 = 1
0.2 x 20 2
Ma = 0.2
Find concentration (g/dm^3)
Concentration = mass(g)
Volume(dm^3)
.convert 10cm^3 into dm^3 but the mass of acid is 0.18g.
Concentration = 0.18g
0.01dm^3
=18g/dm^3
Since Molarity = concentration
Molar mass
Molar mass = concentration
Molarity
= 18
0.2
=90
Relative mass of the acid is 90g/mol
Alternative solution.
Given data.
Vb=20cm^3 (0.02dm^3)
Concentration of base(b) = 8g/dm^3
Mass of acid = 0.18g
Mb=concentration
Molar mass
= 8
40
= 0.2M
From.
MaVa = na
MbVb nb
Mava = MbVbna
nb
= 0.2 x 20 x 1
2
MaVa = 0.002
MaVa = na
There fore number of moles of acid=0.002
From n= mass
Molar mass
Molar mass = mass
n(number of moles)
= 0.18
0.002
= 90g/mol.
11. Data given:
Molarities of HCl = 0.1M and 1M
Volumes=both have 100cm3
Calculate the number of moles of solutions in 100cm3 per each solution because 0.1M and 1M are number of moles in 1 litre in each solution.
* number of moles of 0.1M HCl in 100cm3
0.1mol ---------> 1000cm3
? <----------- 100cm3="" p=""> = 0.01mol.
*number of moles of 1M HCl in 100cm3
1mol ---------> 1000cm3
? <----------- 100cm3="" p=""> = 0.1mol.
Sum up the number of moles obtained then find the concentration by using total number of moles and total volume.
Total number of mol=0.01+ 0.1
=0.11mol.
Total volume= 100cm^2 + 100cm^3
= 200cm^3 = 0.2dm^3
From.
Concentration = Number of moles
Volume(dm3)
= 0.11
0.2
= 0.55M
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