1. Choose the most correct option.
A. A sample of pure water, whatever the source always contains by mass of oxygen and 11.1 % by mass of hydrogen.
a. 88.9
b. 18
c. 80
d. 16
B. Which of the following compounds can NOT demonstrate the law of multiple proportions ?
a. NO, NO2
b. CO, CO2
c. H2O, H2O2
d. Na2S, NaF
C. Which of the following temperature will read the same value on celcius and Fahrenheit scales.
a. – 40
b. + 40
c. -80
d. -20
D. SI unit of the quantity electric current is
a. Volt
b. Ampere
c. Candela
d. Newton
E. In the reaction N2 + 3H2 → 2 NH3, the ratio by volume of N2, H2 and NH3 is 1 : 3 : 2 This illustrates the law of
a. definite proportion
b. reciprocal proportion
c. multiple proportion
d. gaseous volumes
F. Which of the following has maximum number of molecules ?
a. 7 g N2
b. 2 g H2
c. 8 g O2
d. 20 g NO2
G. How many g of H2O are present in 0.25 mol of it ?
a. 4.5
b. 18
c. 0.25
d. 5.4
H. The number of molecules in 22.4 dm3of nitrogen gas at STP is
a. 6.022 x 1020
b. 6.022 x 1023
c. 22.4 x 1020
d. 22.4 x 1023
I. Which of the following has the largest number of atoms ?
a. 1g Au(s)
b. 1g Na(s)
c. 1g Li(s)
d. 1g Cl2(g)
2. Answer the following questions.
A. State and explain Avogadro’s law.
Ans: (a). In the year 1811, Avogadro made a distinction between atoms and molecules and thereby proposed Avogadro’s law.
(b). Avogadro proposed that, ” Equal volumes o f all gases at the same temperature and pressure contain equal number of molecules ’’.
e.g. Hydrogen gas combines with oxygen gas to produce water vapour as follows:
According to Avogadro’s law, if 1 volume contains n molecules, then 2n molecules of hydrogen combine with n molecules of oxygen to give 2n molecules of water, i.e., 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2 molecules of water vapour as represented below:
B. Point out the difference between 12 g of carbon and 12 u of carbon
Ans: 12 g of carbon is the molar mass of carbon while 12 u of carbon is the mass of one carbon atom .
C How many grams does an atom of hydrogen weigh ?
Ans: The mass of a hydrogen atom is 1.6736 x 10-24 g.
D. Calculate the molecular mass of the following in u.
a. NH3 b. CH3COOH c. C2H5OH
Solution:
(a). Molecular mass of NH3 = (1 x Average atomic mass of N) + (3 x Average atomic mass of H)
= (1 x 14.0 u) + (3 x 1.0 u)
= 17 u
(b). Molecular mass of CH3COOH = (2 x Average atomic mass of C) + (4 x Average atomic mass of H) + (2 x Average atomic mass of O)
= (2 x 12.0 u) + (4 x 1.0 u) + (2 x 16.0 u)
= 60 u
(c). Molecular mass of C2H50H = (2 x Average atomic mass of C) + (6 x Average atomic mass of H) + (1 x Average atomic mass of O)
= (2 x 12.0 u) + (6 x 1.0 u) + (1 x 16.0 u)
= 46 u
Ans: (a). The molecular mass of NH3 = 17 u
(b). The molecular mass of CH3COOH = 60 u
(c). The molecular mass of C2H50H = 46 u
E. How many particles are present in 1 mole of a substance ?
Ans: A mole (mol) is the amount of a substance that contains 6.02 × 1023 representative particles of that substance.
F. What is the SI unit of amount of a substance ?
Ans: The SI unit for the amount of a substance is mole (mol).
G. What is meant by molar volume of a gas ?
Ans: The volume occupied by one mole of a gas at standard temperature (0° C) and pressure (1 atm) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is 22.4 dm3
H. State and explain the law of conservation of mass.
Ans: Law of conservation of mass:
(a). The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
(b). Antoine Lavoisier who is often referred to as the father of modern chemistry performed careful experimental studies for various combustion reactions, namely burning of phosphorus and mercury in the presence of air.
(c). Both his experiments resulted in increased weight of products.
(d). After several experiments, in burning of phosphorus, he found that the weight gained by the phosphorus was exactly the same as the weight lost by the air. Hence, total mass of reactants = total mass of products.
(e). When hydrogen gas bums and combines with oxygen to form water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed. Thus, this is in accordance with the law of conservation of mass.
I. State the law of multiple proportions.
Ans: The law states that, “ When two elements A and B form more than one compounds, the masses o f element B that combine with a given mass o f A are always in the ratio o f small whole numbers”.
3. Give one example of each
A. homogeneous mixture
Ans: Homogeneous mixture: Solution (An aqueous solution of sugar)
B. heterogeneous mixture
Ans: Heterogeneous mixture: Suspension (of sand in water)
C. element
Ans: Element: Gold
D. compound
Ans: Compound: Distilled water
4. Solve problems :
A. What is the ratio of molecules in 1 mole of NH3 and 1 mole of HNO3.
Solution:
One mole of any substance contains particles equal to 6.022 x 1023.
1 mole of NH3= 6.022 x 1023 molecules of NH3
1 mole of HNO3= 6.022 x 1023 molecules of HNO3
Ans: The ratio of molecules is = 1:1
B. Calculate number of moles of hydrogen in 0.448 litre of hydrogen gas at STP
Given: Volume of hydrogen at STP = 0.448 L
To find: Number of moles of hydrogen
Formula: 
Calculation: Molar volume of a gas = 22.4 dm3 mol-1 = 22.4 L at STP
Ans: Number of moles of hydrogen = 0.02 mol
C. The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen?
Solution:
Mass of 1 atom of hydrogen = 1.008 u
Mass of 18 atoms of hydrogen = 1 8 x 1.008 u = 18.144 u
Ans: The mass of 18 atoms of hydrogen = 18.144 u
D. Calculate the number of atom in each of the following (Given : Atomic mass of I = 127 u).
a. 254 u of iodine (I)
b. 254 g of iodine (I)
Solution:
(a). 254 u of iodine (I) = x atoms
Atomic mass of iodine (I) = 127 u
∴ Mass of one iodine atom = 127 u
∴ 
= 2 atoms
(b). 254 g of iodine (I)
Atomic mass of iodine = 127 u
∴ Molar mass of iodine = 127 g mol
Now,
= 2 mol
Now,
Number of atoms = Number of moles x Avogadro’s constant
= 2 mol x 6.022 x 1023 atoms/mol
= 12.044 x 1023 atoms
= 1.2044 x 1023 atoms
Ans: (a). Number of iodine atoms in 254 u = 2 atoms
(b). Number of iodine atoms in 254 g= 1.2044 x 1023 atoms
E. A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate.
a. The number of moles of carbon in his homework writing.
b. The number of carbon atoms in 12 mg of his homework writing.
(a). 5 mg carbon = 5 x 10-3 g carbon
Atomic mass of carbon = 12 u
Molar mass of carbon = 12 g mol-1
= 4.167 x 10-4 mol
(b)12 mg carbon = 12 x 10-3 g carbon
= 1 × 10-3 mol
Number of atoms = Number of moles x Avogadro’s constant
Number of atoms of carbon =1 x 10-3 mol x 6.022 x 1023 atoms/mol
= 6.022 x 1020 atoms
Ans: Number of moles of carbon in his homework writing = 4.167 x 104 mol
Number of atoms of carbon in 12 mg homework writing = 6.022 x 1020 atoms
F. Arjun purchased 250 g of glucose (C6H12O6) for Rs 40. Find the cost of glucose per mole.
Given: Mass of urea = 250 g, cost for 250 g glucose = Rs 40, molecular formula of glucose = C6H12O6
To find: Cost per mole of glucose
Calculation: Molecular formula of glucose is (C6H12O6).
Molecular mass of glucose = (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O)
= (6 × 12 u) + (12 × 1 u) + (6 x 16 u)
= 180 u
Molar mass of glucose = 180 g mol-1
mol
Ans: The cost of glucose per mole is Rs 28.8
G. The natural isotopic abundance of 10B is 19.60% and 11B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Solution:
Average atomic mass of Boron (B)
Ans: Average atomic mass of boron = 10.84 u
H. Convert the following degree Celsius temperature to degree Fahrenheit.
a. 40 0C b. 30 0C
Solution: (a). Given: Temperature in degree Celsius = 40 °C
To Find : Temperature in degree Fahrenheit
Calculation: Substituting 40 °C in the formula,
(b). Given: Temperature in degree Celsius = 30 °C
To Find: Temperature in degree Fahrenheit
Calculation: Substituting 30 °C in the formula 
Ans: (a). The temperature 40 °C corresponds to 104 °F.
(b). The temperature 30 °C corresponds to 86 °F.
I. Calculate the number of moles and molecules of acetic acid present in 22 g of it.
Given: Mass of acetic acid = 22 g
To find: The number of moles and molecules of acetic acid
Formulae: (a). 
(b) Number of molecules = Number of moles x Avogadro’s constant
Calculation:
Mass of acetic acid = 22 g
Molecular mass of acetic acid, CH3COOH
= (2 x Average atomic mass of C) + (4 x Average atomic mass of H) + (2 x Average atomic mass of O)
= (2 x 12 u) + (4 x 1 u) + (2 x 16 U) = 60 u
Molar mass of acetic acid = 60 g mol-1
= 0.36 mol
Now,
Number of molecules of acetic acid = Number of moles x Avogadro’s constant
= 0.367 mol x 6.022 x 1023 molecules/mol
= 2.210 x 1023 molecules
Ans: Number of moles = 0.367 mol
Number of molecules of acetic acid = 2.210 x 1023 molecules
J. 24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Solution:
Given: Mass of carbon (reactant) = 24 g, mass of carbon dioxide (product) = 88 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 32 g oxygen to form 44 g of carbon dioxide as follows:
Carbon + Oxygen → Carbon dioxide
12 g 32 g 44 g
Hence, (2 x 12 = 24 g) of carbon will combine with (2 x 32 — 64 g) of oxygen to give (2 x 44 = 88 g) carbon dioxide.
Ans: Mass of oxygen used = 64 g
K. Calculate number of atoms is each of the following. (Average atomic mass : N = 14 u, S = 32 u)
(a). 0.4 mole of nitrogen
(b). 1.6 g of sulfur
Solution:
(a). 0.4 mole of nitrogen (N)
Number of atoms of N = Number of moles x Avogadro’s constant
= 0.4 mol x 6.022 x 1023 atoms/mol
= 2.4088 x 1023 atoms of N
(b). 1. 1.6 g of Sulphur (S)
Molar mass of sulphur – 32 g mol-1
= 0.05 mol
Number of atoms of S = Number of moles x Avogadro’s constant
= 0.05 mol x 6.022 x 1023 atoms/mol
= 0.3011 x 1023 atoms
= 3.011 x 1023 atoms of S
Ans: (a). Number of nitrogen atoms in 0.4 mole = 2.4088 x 1023 atoms of N
(b). Number of sulfur atoms in 1.6 g = 3.011 x 1023 atoms of S
L. 2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 of its oxide. Which law is verified by these data ?
Solution:
Here, metal oxide is obtained by two different methods; reactions of metal with oxygen and reaction of metal with water vapour (steam).
In first reaction (reaction with oxygen),
The mass of oxygen in metal oxide = 3.2 – 2.0 = 1.2 g
In second reaction (reaction with steam),
The mass of oxygen in metal oxide = 2.27 – 1.42 = 0.85 g 
Therefore, irrespective of the source, the given compound contains same elements in the same proportion.
The law of definite proportions states that “A given compound always contains exactly the same proportion of elements by weight ”.
Hence, the law of definite proportions is verified by these data.
Ans: The law of definite proportions is verified by given data.
M. In two moles of acetaldehyde (CH3CHO) calculate the following
a. Number of moles of carbon
b. Number of moles of hydrogen
c. Number of moles of oxygen
d. Number of molecules of acetaldehyde
Solution:
Molecular formula of acetaldehyde: C2H4O
Moles of acetaldehyde = 2 mol
(a). Number of moles of carbon atoms = Moles of acetaldehyde x Number of carbon atoms
= 2 x 2
= 4 moles of carbon atoms
(b). Number of moles of hydrogen atoms = Moles of acetaldehyde x Number of hydrogen atoms
= 2 x 4
= 8 moles of hydrogen atoms
(c). Number of moles of oxygen atoms = Moles of acetaldehyde x Number of oxygen atoms
= 2 x 1
= 2 moles of oxygen atoms
(d). Number of molecules of acetaldehyde = Moles of acetaldehyde x Avogadro number
= 2 mol x 6.022 x 1023 molecules/mol
= 12.144 x 1023 molecules of acetaldehyde
Ans: (a). Number of moles of carbon, hydrogen and oxygen are 4 ,8, 2 respectively.
(b). Number of molecules of acetaldehyde = 12.044 x 1023
N. Calculate the number of moles of magnesium oxide, MgO in
(a).8 0 g and (b) .10 g of the compound. (Average atomic masses of Mg = 24 and O = 16).
Solution:
Given: (a). Mass of MgO = 80 g
(b). Mass of MgO = 10 g
To find: Number of moles of MgO
Formulae:
Calculation:
(a). Molecular mass of MgO = (1 x Average atomic mass of Mg) + (1 x Average atomic mass of O)
= (1 x24u) + (l x 16 u)
= 40 u
∴ Molar mass of MgO = 40 g mol-1
Mass of MgO = 80 g
= 2 mol
(b). Mass of MgO = 10 g, Molar mass of MgO = 40 g mol-1
= 0.25 mol
Ans: (a). The number of moles in 80 g of magnesium oxide, MgO = 2 mol
(b). The number of moles in 10 g of magnesium oxide, MgO = 0.25 mol
O. What is volume of carbon dioxide, CO2 occupying by
i. 5 moles and ii. 0.5 mole of CO2 gas measured at STP.
Given: i. Number of moles of CO2 = 5 mol
ii. Number of moles of CO2 = 0.5 mol
To find: Volume at STP
Formula:
Calculation: Molar volume of a gas = 22.4 dm3 mol at STP.
i. Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 5 mol × 22.4 dm3 mol-1 = 112 dm3
ii. Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 0.5 mol × 22.4 dm3 mol-1 = 11.2 dm3
Ans: i. Volume of 5 mol of C 02= 112 dm3
ii. Volume of 0.5 mol of C 02= 11.2 dm3
P. Calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP. Molar mass of KClO3 is 122.5 g mol-1
Solution:
The molecular formula of potassium chlorate is KCIO3
Required chemical equation:
2KClO3 → 2KCl + 3O2 ↑
[2 moles] [3 moles]
2 moles of KCIO3= 2 x 122.5 = 245 g
3 moles of O2 at STP occupy = (3 x 22.4 dm3) = 67.2 dm3
thus, 245 g of potassium chlorate will liberate 67.2 dm3 of oxygen gas
Let ‘x’ gas of KCLO3 liberate 6.72 dm3 of oxygen gas at S.T.P.
Ans: Mass of potassium chlorate required = 24.5 g
Q. Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NH2)2CO. Also calculate the number of atoms of N, C and O.
Given: Mass of urea = 5.6 g
To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen
Calculation: Molecular formula of urea: (NH2)2CO
Molar mass of urea = 60 g mol-1
0.0933 mol
Moles of urea = 0.0933 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O.
Number of H atoms in 5.6 g of urea = (4 × 0.0933) mol × 6.022 × 1023 atoms/mol
= 2.247 × 1023 atoms of hydrogen
Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 1023 atoms/mol
= 1.124 × 1023 atoms of nitrogen
Number of C atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 1023 atoms/mol
=0.562 × 1023 atoms of carbon
Number of O atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 1023 atoms/mol
= 0.562 × 1023 atoms of oxygen
Ans: 5.6 g of urea contain 2.247 × 1023 atoms of H, 1.124 × 1023 atoms of N, 0.562 × 1023 atoms of C and 0.562 × 1023 atoms of O.
R. Calculate the mass of sulfur dioxide produced by burning 16 g of sulfur in excess of oxygen in contact process. (Average atomic mass : S = 32 u, O = 16 u)
Solution:
Given: Mass of sulfur (reactant) = 16 g
To find: Mass of sulfur dioxide (product)
Calculation: 32 g of sulfur combine with 32 g oxygen to form 64 g of sulfur dioxide as follows:
Sulfur + Oxygen → Sulfur dioxide
32 g 32 g 64 g
Hence, (0.5 x 32 – 16 g) of sulfur will combine with (0.5 x 32 — 16 g) of oxygen to give (0.5 x 64 = 32 g) sulfur dioxide.
Ans: Mass of sulfur dioxide produced = 32 g
5. Explain
A. The need of the term average atomic mass.
Ans: (a) Several naturally occurring elements exist as a mixture of two or more isotopes.
(b) Isotopes have different atomic masses.
(c) The atomic mass of such an element is the average of atomic masses of its isotopes.
(d) For this purpose, the atomic masses of isotopes and their relative percentage abundances are considered.
Hence, the term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes.
B. Molar mass.
Ans: (a) The mass o f one mole o f a substance (element/compound) in grams is called its molar mass.
n. The molar mass of any element in grams is numerically equal to atomic mass of that element in u.
eg.
in. Similarly, molar mass of polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u.
eg.
C. Mole concept.
Ans: (a) Even a small amount of any substance contains very large number of atoms or molecules. Therefore, a quantitative adjective ‘mole’ is used to express the large number of sub microscopic entities like atoms, molecules, ions, electrons, etc. present in a substance.
(b) Thus, one mole is the amount o f a substance that contains as many entities or particles as there are atoms in exactly 12 g (or 0.012 kg) o f the carbon -12 isotope.
(c) One mole is the amount of substance which contains 6.0221367 x 1023 particles/entities
D. Formula mass with an example.
Ans: (a) The formula mass o f a substance is the sum o f atomic masses o f the atoms present in the formula.
(b) In substances such as sodium chloride, positive (sodium) and negative (chloride) entities are arranged in a three-dimensional structure in a way that one sodium (Na+) ion is surrounded by six chloride (Cl–) ions, all at the same distance from it and vice versa. Thus, sodium chloride do not contain discrete molecules as the constituent units.
(c) Therefore, NaCl is just the formula which is used to represent sodium chloride though it is not a molecule.
(d) In such compounds, the formula (i.e., NaCl) is used to calculate the formula mass instead of molecular mass,
e.g. Formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine = 23.0 u + 35.5 u = 58.5 u
E. Molar volume of gas.
Ans: (a) It is more convenient to measure the volume rather than mass of the gas.
(b) It is found from Avogadro law that one mole of any gas occupies a volume of 22.4 dm3 at standard temperature (0 °C) and pressure (1 atm) (STP).
(c) The volume of 22.4 dm3 at STP is known as molar volume of a gas.
(d) The relationship between number of moles and molar volume can be expressed as follows:
[Note: IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions the molar volume of a gas is 22.71 L mol-1 ]
F. Types of matter (on the basis of chemicalcomposition).
Ans: Matter on the basis of chemical composition can be classified as follows:
(1) Pure substances: They always have a definite chemical composition. They always have the same properties regardless of their origin.
e.g. Pure metal, distilled water, etc.
They are of two types:
(A) Elements: They are pure substances, which cannot be broken down into simpler substances by ordinary chemical changes.
Elements are further classified into three types:
(a). Metals:
i. They have a lustre (a shiny appearance).
ii. They conduct heat and electricity.
iii. They can be drawn into wire (ductile).
iv. They can be hammered into thin sheets (malleable).
eg. Gold, silver, copper, iron. Mercury is a liquid metal at room temperature.
(b). Nonmetals:
i. They have no lustre. (except diamond, iodine)
ii. They are poor conductors of heat and electricity, (except graphite)
iii. They cannot be hammered into sheets or drawn into wire, because they are brittle,
e.g. Iodine
(c). Metalloids: Some elements have properties that are intermediate between metals and nonmetals and are called metalloids or semimetals.
e.g. Arsenic, silicon and germanium.
(B) Compounds: They are the pure substances which are made up of two or more elements in fixed proportion,
e.g. Water, ammonia, methane, etc.
(2) Mixtures: They have no definite chemical composition and hence no definite properties. They can be separated by physical methods.
e.g. Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.
Mixtures are of two types:
(a). Homogeneous mixture: In homogeneous mixture, constituents remain uniformly mixed throughout its bulk.
e.g. Solution, in which solute and solvent molecules are uniformly mixed throughout its bulk.
(b) Heterogeneous mixture: In heterogeneous mixture, constituents are not uniformly mixed throughout its bulk.
e.g. Suspension, which contains insoluble solid in a liquid.