ncert solutions for class 9 science chapter 3 atoms and
TRANSCRIPT
NCERT solutions for class 9 science chapter 3 Atoms and Molecules
Q 1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The
products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show
that these observations are in agreement with the law of conservation of mass.
sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water
Answer:
Given, the reaction
sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water
5.3g 6g 8.2g 2.2g 0.9g
Now,
Total Mass on the Left Hand Side = 5.3g + 6g = 11.3g
Total Mass on the Right Hand Side = 8.2g + 2.2g + 0.9g = 11.3g
As Mass on the LHS is equal to RHS, the observation is in agreement with the law of
conservation of mass.
Q 2. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What
mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Answer:
Given:
Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water.
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Now,
For every 1 g of Hydrogen, 8 g of oxygen is needed for the reaction to take place.
Therefore, for 3 g of Hydrogen, the mass of oxygen needed = 8 * 3 = 24 g.
Hence, 24 g of Oxygen is needed to complete a reaction with 3 g of Hydrogen.
Q 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of
mass?
Answer:
The postulate of Dalton’s atomic theory is the result of the law of conservation of mass
is
"Atoms can neither be created nor can be destroyed".
Q 4. Which postulate of Dalton’s atomic theory can explain the law of definite
proportions?
Answer:
The postulate of Dalton’s atomic theory can explain the law of definite proportions is
"The relative number and kinds of atoms are equal in given compounds."
CBSE NCERT solutions for class 9 science chapter 3 Atoms and Molecules
Topic 3.2 What is an atom?
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Q 1. Define the atomic mass unit.
Answer:
An atomic mass unit is a unit of mass used to express the weight subatomic particles,
where one unit is equal to exactly one-twelfth the mass of a carbon-12 atom.
Q 2. Why is it not possible to see an atom with naked eyes?
Answer:
We can't see the atom with the naked eye because they are minuscule in nature. they
are measured in the nanometres. Also, except for noble gases, all-atom do not exist
independently.they exist in the form of any compound.
NCERT free solutions for class 9 science chapter 3 Atoms and Molecules
Topic 3.4 Writing chemical formulae
Q 1. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide
Answer:
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The Formula of given compounds are :
(i) sodium oxide :
(ii) aluminium chloride:
(iii) sodium sulphide:
(iv) magnesium hydroxide:
Q 2. Write down the names of compounds represented by the following formulae:
(i)
(ii)
(iii)
(iv)
(v)
Answer:
The Names of the following compounds are :
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(i)
=Aluminium Sulphate
(ii)
=Calcium Chloride
(iii)
=Potessium Sulphate
(iv)
= Potessium Nitrate
(v)
= Calcium Carbonate.
Q 3. What is meant by the term chemical formula?
Answer:
The chemical formula of a compound is a symbolic representation of its composition.
For example, The chemical formula for common salt is NaCl as it is made up of Sodium
(Na) and Chlorine (Cl).
Q 4.(i) How many atoms are present in a
• molecule
Answer:
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molecule has 2 atoms of Hydrogen and 1 atom of Sulphur.
and hence,
Total of 3 atoms are present in molecule .
Q 4.(ii) How many atoms are present in a
• ion?
Answer:
ion has one atom of Phosphorus and 4 atoms of Oxygen.
And Hence,
A total of 5 atoms are present in ion.
Solutions for NCERT class 9 science chapter 3 Atoms and Molecules Excercise:
Topic 3.5 Molecular mass and mole concept
Q 1. Calculate the molecular masses
of
Answer:
The molecular mass of :
= 2 * Atomic mass of Hydrogen
= 2 * 1u
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=2u.
The molecular mass of :
= 2 * Atomic mass of Oxygen
= 2 * 16u
= 32u.
The molecular mass of :
= 2 * Atomic mass of Chlorine
= 2 * 35.5uu
= 71u.
The molecular mass of :
= Atomic mass of Carbon +2 * Atomic mass of Oxygen
= 12u + 2 * 16u
= 44u.
The molecular mass of :
= Atomic mass of Carbon +4 * Atomic mass of Hydrogen
= 12u + 4 * 1u
= 16u.
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The molecular mass of :
= 2 * Atomic mass of Carbon + 6 * Atomic mass of Hydrogen
= 2*12u+ 6 * 1u
=24u+ 6u
= 30u.
The molecular mass of :
= 2 * Atomic mass of Carbon + 64* Atomic mass of Hydrogen
= 2*12u+ 4* 1u
=24u+ 4u
= 28u.
The molecular mass of :
= Atomic mass of Nitrogen + 3 * Atomic mass of Hydrogen
= 14u+ 3 * 1u
= 17u.
The molecular mass of :
= Atomic mass of Carbon +4 * Atomic mass of Hydrogen + Atomic mass of Oxygen
= 12u+ 4 * 1u + 16u
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= 32u.
Q 2. Calculate the formula unit masses of ZnO, Na 2 O, K 2 CO 3 , given atomic masses
of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u
Answer:
Given,
the atomic mass of Zn = 65 u,
the atomic mass of Na = 23 u,
the atomic mass of K = 39 u,
the atomic mass of C = 12 u,
and the atomic mass of O = 16 u
Now,
Formula unit mass of ZnO = Atomic mass of Zinc + Atomic mass of O
= 65 u + 16 u
= 81 u.
Formula unit mass of Na 2 O = 2 * Atomic mass of Na+ Atomic mass of O
= 2 * 23 u + 16 u
= 62 u.
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Formula unit mass of K 2 CO 3 = 2 * Atomic mass of K+ Atomic mass of C + 3 *
Atomic mass of O
= 2 * 39 u + 12 u + 3 * 16 u
= 138 u.
NCERT solutions for class 9 science chapter 3 Atoms and Molecules
Topic 3.5.3 Mole concept
Q 1. If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1
atom of carbon?
Answer:
As we know,
Now,
the mass of 1 mole or = 12 g
The mass of 1 atom :
Hence the mass of 1 atom of Carbon is .
Q 2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron
(given, the atomic mass of Na = 23 u, Fe = 56 u)?
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Answer:
The number of moles of 100 g Na atoms :
The number of moles in 100 g of Fe atoms :
As we know, the one-mole atoms contain atoms.
So, more the number of moles, more the number of atoms. and hence 100 g of Na
atom has a greater number of atoms than 100 g Fe.
NCERT solutions for class 9 science chapter 3 Atoms and Molecules: Solved Exercise Questions
Q 1. A 0.24 g sample of a compound of oxygen and boron was found by analysis to
contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition
of the compound by weight.
Answer:
The total mass of the compound = 0.24 g
Mass of boron in the compound = 0.096 g
Mass of oxygen in the compound = 0.144 g
Now, As we know,
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The percentage of an element in the compound :
So,
The percentage of Boron in the compound by weight :
The percentage of Oxygen in the compound by weight :
Q 2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is
produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in
50.00 g of oxygen? Which law of chemical combination will govern your answer?
Answer:
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
So, by the law of definite proportions,
When 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will be used
to produce 11.00 gram of carbon dioxide. The remaining 42.00 g of oxygen will remain
unreacted. Law of constant proportion is Held.
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Q 3. What are polyatomic ions? Give examples.
Answer:
Polyatomic ions are ions that contain more than one atom. These atoms can be of the
same type or of a different type.
Some examples of polyatomic ions are NH 4 + , OH - , SO 4 2- , and SO 3 2- .
Q 4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:
The chemical formula of Given compounds are :
(a) Magnesium chloride
(b) Calcium oxide
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(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Q 5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer:
The names of the elements present in the given compounds are :
(a) Quick lime :
= Calcium and Oxygen.
(b) Hydrogen bromide:
= Hydrogen and Bromine.
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(c) Baking powder:
= Sodium, Hydrogen, Carbon, and Oxygen.
(d) Potassium sulphate:
= Potassium, Sulphur, and oxygen.
Q 6.(a) Calculate the molar mass of the following substances.
• Ethylene ,
Answer:
The molecular mass of ethyne, = 2 * Atomic Mass of C + 2 * Atomic Mass of H
= 2 * 12u + 2 * 1u
= 24u + 2u
= 26u
Q 6. (b) Calculate the molar mass of the following substances.
• Sulphur molecule , S 8
Answer:
The molecular mass of Sulphur molecule, = 8 * Atomic Mass of S
= 8 * 32u
= 256u
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Q 6.(c) Calculate the molar mass of the following substances
• Baking soda
Answer:
The molecular formula of Baking soda is NaHCO3
So the molecular mass will be some of the individual atoms involved in the formula.
Molecular Mass of baking soda = Molecular mass (Na + H + C + 3xO)
= 23 + 1 + 12 + 3x16
= 84 g/mol
Q 6.(d) Calculate the molar mass of the following substances.
• Hydrochloric acid, HCl
Answer:
The molecular mass of Hydrochloric acid, HCl = 1 * Atomic Mass of Cl + 1 * Atomic
Mass of H
= 35.5 u + 1u
= 36.5u
Q 6. (e) Calculate the molar mass of the following substances.
• Nitric acid, HNO 3
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Answer:
The molecular mass of Nitric acid, HNO 3 = 1 * Atomic Mass of N + 1 * Atomic Mass of
H + 3 * Atomic Mass of O
= 1 * 14u + 1 * 1u + 3 * 16u
= 14u + 1u + 48u
= 63u
Q 7.(a) What is the mass of—
• 1 mole of nitrogen atoms?
Answer:
Atomic Mass of Nitrogen atom = 14 u.
Mass of one mole of nitrogen atoms = molecular mass of nitrogen atoms in grams
= 14 g
Q 7.(b) What is the mass of -
• 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
Answer:
The atomic mass of Aluminium = 27 u.
Mass of 4 moles of aluminium atoms = 4 x Mass of 1 mole of Al atoms
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= 4 x molecular mass of aluminum atoms in grams
= 4 x 27
= 108 g
Q 7. (c) What is the mass of—
• 10 moles of sodium sulphite (Na 2 SO 3 )?
Answer:
The molecular mass of Sodium sulphite (Na 2 SO 3 ) = 2 * 23 + 32 + 3 * 16
= 126u
Mass of 10 moles of sodium sulphite = 10 x Mass of 1 mole of Na 2 SO 3
= 10 x molecular mass of Na 2 SO 3 in grams
= 10 x 126 g
= 1260 g.
Q 8.(a) Convert into a mole.
• 12 g of oxygen gas
Answer:
The molecular Mass of the Oxygen = 32 g
Now,
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Since 32 g Oxygen = 1 mole
The number of moles in 12 g Oxygen:
Q 8.(b) Convert into mole.
• 20 g of water
Answer:
Given, Mass of water = 20 g
The Molecular mass of the water = 2 * Mass of Hydrogen + Mass of Oxygen
= 2 * 1 + 16
= 18 g.
Now, Number of moles :
Q 8.(c) Convert into mole.
• 22 g of carbon dioxide
Answer:
Given Mass of the carbon dioxide = 22 g.
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The molecular mass of Carbon dioxide in grams = mass of Carbon + 2 * mass of
Oxygen
= 12u + 2 * 16u
= 44u
The number of mole :
Q 9 . (a) What is the mass of:
• 0.2 mole of oxygen atoms?
Answer:
The mass of 1 mole of oxygen atoms = 16 g
The mass of 0.2 mole of oxygen atoms = 0.2 x 16 g
= 3.2 g
Q 9.(b) What is the mass of:
0.5 mole of water molecules?
Answer:
The mass of 1 mole of water molecules = 18 g
Thus,
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the mass of 0.5 mole of water molecules = 18 x 0.5
= 9.0 g
Q 10. Calculate the number of molecules of sulphur (S 8 ) present in 16 g of solid
sulphur.
Answer:
Given the mass of Sulphur = 16 g
The molecular mass of the Sulphur molecule = 8 x 32 = 256 g.
The number of moles of Sulphur molecule :
Now
Number of molecules in 1 mole = molecules
So,
The number of molecules in 0.0625 moles = molecules.
= Molecules.
Hence there are Sulphur molecules in 16 g of solid sulphur .
Q 11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
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Answer:
Give, the mass of Aluminum Oxide = 0.051 g.
The molecular mass of Aluminium Oxide = 2 x mass of Aluminium + 3 x mass of
Oxygen.
= 2 x 27 + 3 x 16
= 102 g
Number of Moles of Aluminium Oxide :
Now, Since 1 mole of Aluminium Oxide contain 2 moles of Aluminium ion,
The Number of Moles of Aluminium ion = 2 x number of moles of Aluminium Oxide
= 2 x 0.0005
= 0.001 moles.
Now, As the number of ion in 1 mole = ions.
The number of ions in 0.001 moles of Aluminium ion = ions
= ions.
Hence there are Aluminium ions in 0.051 g of aluminium oxide. Aak
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