ANSWER SCHEME CHEMBASSADOR | CHECKPOINT CHAPTER 1

Chapter 1.3 : Mole Concept

Avogadro’s Number & Molar Mass
Calculate the number of moles in each of the following masses:
Q1 8 Marks
a) 64.1 g of aluminum
b)28.1 g of silicon
c) 0.255 g of zinc

n = mass/molar mass n = mass/molar mass

= 64.1 g/26.98 gmol-1 = 2.38 mol = 0.255 g/32.07 gmol-1= 7.95x10-3 mol
n = mass/molar mass n = mass/molar mass
Answer = 28.1 g/28.09 gmol-1 = 1.00 mol = 850.5 g/65.39gmol-1= 13.01 mol

Q. 2 Calculate the mass of each of the following amounts: 6 marks

(a) 1.22 mol sodium

(b) 14.5 mol copper
(c) 9.37 x10-3 mol magnesium

Mass = n x Molar mass (c)Mass = n x Molar mass

=1.22 mol x 22.99 gmol-1= 28.05 g = 9.37 x10-3 mol x 24.31 gmol-1
Answer Mass = n x Molar mass = 0.28 g
=14.5 mol x 63.55 gmol-1= 921.48 g

Q. 3 Calculate the amount in moles in each of the following quantities: 4 marks

(a) 3.01 x1023 atoms of rubidium

(b) 2.997x1025 atoms of vanadium

Answer (a) n = No of particles/ NA (b) n = No of particles/ NA

= 3.01 x10 23/6.02x1023 = 0.5 mol = 2.997 x1025/6.02x102 = 49.78 mol

Q. 4 A sample of oxygen gas has a volume of 8 dm 3 at STP. Count the number of 6 marks

(a) moles of the oxygen gas

(b) O2 molecule
(c) oxygen atoms in the sample

No moles of oxygen gas = V/ VM (c)1 O2 molecule contains 2 O atoms

= 8 dm3 /22.4 dm3= 0.357 mol 2.15 x 1023 O2 molecules contains:
Answer No of O2 = 0.357 mol x 6.02x1023 =2 x 2.15 x 1023 = 4.3 x 1023 O atoms
= 2.15 x 1023 molecules

Q. 5 In an experiment, 0.53 mol of NO2 is produced at the end of the reaction. Calculate the 2 marks
volume of produced NO2 at room condition.
Answer At STP, 1 mol gas ≡ 24.0 L
ANSWER SCHEME CHEMBASSADOR | CHECKPOINT CHAPTER 1

Volume of NO2 = n x Molar Volume

= 0.53 mol x 24.0 L
= 12.72 L

Empirical and Molecular Formula

Q. 6 Compound Q contains carbon, hydrogen and nitrogen. Combustion of 10

0.250 g of Q produces 0.344 g of water, H2O, and 0.558 g of carbon dioxide, CO 2. marks
Determine

(a) the empirical formula of Q,

(b) the molar mass of Q if the molecular formula is the same as the empirical formula.
(c) the number of hydrogen atoms present in the above sample Q.

Answer (a) Mass C=0.558 x 12 Mass H= 0.344 x 2 Mass N = 0.250 – (0.1522 + 0.0382)
44 18 = 0.0596 g
= 0.1522 g = 0.0382 g
Table:

Element Carbon Hydrogen Nitrogen

Mass 0.15218 0.03822 0.0596
0.1522 0.0382 0.0596
Number of moles 12.01 1.01 14.01
= 0.01267 = 0.0378 = 0.00425
0.01267 0.03878 0.00425
Smallest ratio 0.00425 0.00425 0.00425
= 2.98 = 9.12 =1
3 9 1
Empirical
formula of Q is C3H9N

(b)
Molar mass of Q = 3(12.01) + 9(1.01) + 1(14.01)
= 59.13 g mol-1
(c) Mol of Q = 0.25 g
59.13 gmol-1
= 4.23 x 10-3 mol
1 mol Q , C3H9N ≡ 9 mol of H
4.23 x 10-3 mol Q , C3H9N ≡ 9 x 4.23 x 10-3 mol H

No of H atom = (9 x 4.23 x 10-3) mol x 6.02 x 1023 atoms = 2.295 x 102

1 mol

Q. 7 An organic compound contains only carbon, hydrogen and oxygen with a mass composition of 41.40%
C, 3.47% H and 55.13% O. If 0.05 mole of this compound weighs 5.80 g, determine its
(a) empirical formula.

(b) molecular formula.

Answer (a) Assume mass of the sample is 100 g

Mass of C = 41.4 g
ANSWER SCHEME CHEMBASSADOR | CHECKPOINT CHAPTER 1

Mass of H = 3.47 g
Mass of O = 55.13 g
Element C H O
Mass 41.4 3.47 55.13
n 41.4 3.47 55.13
12.01 1.01 16.00
= 3.45 =3.44 = 3.45
Smallest ratio 3.45 3.44 3.45
3.44 3.44 3.44
= 1 = 1 = 1
Empirical formula is CHO
3.3 Concentration
Q. 8 Saline solution is prepared by dissolving 9.0 gram of NaCl in deionized water in 500 mL 3 marks
volumetric flask. Calculate the molarity of the solution.
nsolute = 9.0g / 58.5 gmol-1 Molarity = nsolute / Vsolution ( L )
Answer = 0.1538 mol = 0.1538 mol / 0.5 L
= 0.31 M

Q. 9 (a) 120.0 grams of a salt solution contains 20.0 grams of salt. What is the concentration 4 marks
(w/w) of this solution?
(b) What is the percent by mass of a solution that contains 26.5 g of glucose in 500 g
of solution?
Answer (a) % w/w = 20 g x 100% (b) % w/w = 26.5 g x 100%
120 g 500 g
= 16.67 % = 0.28 %

Q. 10 Calculate the mass of Na2Cr2O7 needed to prepare a 0.025 M solution in a 50 mL 3 marks

volumetric flask.
Answer
Mol of Na2Cr2O7 = 50.0 x 10-3 L x 0.025 mol L-1
= 1.25 x 10 -3 mol

Mass Na2Cr2O7 = 1.25 x 10-3 mol x 262.0 g

= 0.33 g
Q. 11 Sodium carbonate, Na2CO3, dissolves in ethanol to give 2.5 M solution. Calculate the % by 5 marks
mass of the solution if the density of the solution is 1.430 g mL-1.
Answer
Assume: Assume V of solution = 1 L,
Molarity solution= 2.5M Density solution= 1.430 g mL-1
Formula molarity: nsolute/volsoln Formula density: mass soln/vol soln
no mol solute, Na2CO3= 2.5 mol L-1x1L =2.5mol Mass solution= density x vol soln = 1.43gml-1x1000ml
= 1430g
Mass of solute, Na2CO3 = 2.5 mol x 106g mol─1
= 265 g
mass of solute
%w/w =
mass of solution x 100%
ANSWER SCHEME CHEMBASSADOR | CHECKPOINT CHAPTER 1

265
¿
1430 x 100% = 18.53 %
Q. 12 A reagent bottle contains a stock solution of 0.90% by mass of sodium chloride, NaCl. The density of
the solution is 1.00g cm-3. Calculate
(a) The molarity of NaCl solution. 5 marks
(b) The volume of the stock solution required to prepare 100 mL of 0.01 M NaCl 1 mark
solution.
Assume mass solution = 100 g
Answer (a)
x 100% Molarity =
mass solute NaCl = 0.9 g =
n solute NaCl = 0.9g/58.44gmol-1 = 0.154 M
= 0.0154 mol

Volume solution = mass solution/ density

= 100 g/1.00g cm-3
= 100 cm3

(b) M1V1 = M2V2

V1 =
( 0.01 M ) ( 1000
100
)L
( 0.1538 )
= 6.50 mL
Q. 13 Calcium acetate, Ca(C2H3O2)2 solution is the substance used for reducing phosphate level in 5 marks
late-stage kidney failure. In an experiment, 250 mL of 0.25 M Ca(C 2H3O2)2 solution was
prepared. Determine the percentage by mass of the solution with a density of 1.509 g mL -1.

Answer Assume volume of solution = 1L

Mole of Ca(C2H3O2)2 = 0.25 mol

Mass of solute = 0.25 mol x 158.1 g mol-1

= 39.525 g
Mass of solution = ρ x volume
= 1.509 g mL -1 x 1000 mL
= 1509 g

mass of solute
% w/w = mass of solution x 100%
39.525 g
= 1509 g x 100 %
= 2.62 %