NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

NCERT Class 9 Science Chapter 3 Exercise Question Answer

Exercise-3.1 (Page: 32)

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.

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."

Exercise-3.2 (Page: 35)

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.

Exercise-3.3 and 3.4 (Page: 39)

Q 1. Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Answer:

The Formula of the given compounds are :

(i) sodium oxide :

$\Rightarrow N{a}_{2}O$

(ii) aluminium chloride:

$\Rightarrow AlC{l}_3$

(iii) sodium sulphide:

$\Rightarrow N{a}_{2}S$

(iv) magnesium hydroxide:

$\Rightarrow Mg(OH{)}_2$

Q 2. Write down the names of compounds represented by the following formulae:

(i) $A{l}_2(S{O}_4{)}_3$

(ii) $CaC{l}_2$

(iii) $K_{2}S{O}_4$

(iv) $KN{O}_3$

(v) $CaC{O}_3$

Answer:

The Names of the following compounds are :

(i) $A{l}_2(S{O}_4{)}_3$

=Aluminium Sulphate

(ii) $CaC{l}_2$

=Calcium Chloride

(iii) $K_{2}S{O}_4$

=Potessium Sulphate

(iv) $KN{O}_3$

= Potassium Nitrate

(v) $CaC{O}_3$

= 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

• $H_{2}S$ molecule

Answer:

$H_{2}S$ molecule has 2 atoms of Hydrogen and 1 atom of Sulphur.

and hence,

A total of 3 atoms are present in $H_{2}S$ molecule.

Q 4.(ii) How many atoms are present in a

• $P{O}_4^{3-}$ ion?

Answer:

$P{O}_4^{3-}$ ion has one atom of Phosphorus and 4 atoms of Oxygen.

And Hence,

A total of 5 atoms are present in $P{O}_4^{3-}$ ion.

Exercise-3.5.1-3.5.2 (Page: 40)

Q 1. Calculate the molecular masses of $H_2,O_2,C{l}_2,C{O}_2,C{H}_4,C_2{H}_6,C_2{H}_4,N{H}_3,C{H}_{3}OH$

Answer:

The molecular mass of $H_2$ :

= 2 * Atomic mass of Hydrogen

= 2 * 1u

=2u.

The molecular mass of $O_2$ :

= 2 * Atomic mass of Oxygen

= 2 * 16u

= 32u.

The molecular mass of $C{l}_2$ :

= 2 * Atomic mass of Chlorine

= 2 * 35.5uu

= 71u.

The molecular mass of $C{O}_2$ :

= Atomic mass of Carbon +2 * Atomic mass of Oxygen

= 12u + 2 * 16u

= 44u.

The molecular mass of $C{H}_4$ :

= Atomic mass of Carbon +4 * Atomic mass of Hydrogen

= 12u + 4 * 1u

= 16u.

The molecular mass of $C_2{H}_6$ :

= 2 * Atomic mass of Carbon + 6 * Atomic mass of Hydrogen

= 2*12u+ 6 * 1u

=24u+ 6u

= 30u.

The molecular mass of $C_2{H}_4$ :

= 2 * Atomic mass of Carbon + 64* Atomic mass of Hydrogen

= 2*12u+ 4* 1u

=24u+ 4u

= 28u.

The molecular mass of $N{H}_3$ :

= Atomic mass of Nitrogen + 3 * Atomic mass of Hydrogen

= 14u+ 3 * 1u

= 17u.

The molecular mass of $C{H}_{3}OH$ :

= Atomic mass of Carbon +4 * Atomic mass of Hydrogen + Atomic mass of Oxygen

= 12u+ 4 * 1u + 16u

= 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.

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.

Exercise-3.5.3 (Page: 42)

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,

$1mole=6.022\times {10}^{23}atoms$

Now,

the mass of 1 mole or $6.022\times {10}^{23}atoms$ = 12 g

The mass of 1 atom :

$=\frac{12}{6.022\times {10}^{23}}=1.99\times {10}^{-23}g$

Hence the mass of 1 atom of Carbon is $1.99\times {10}^{-23}g$ .

Q 2. Which has more atoms, 100 grams of sodium or 100 grams of iron (given, the atomic mass of Na = 23 u, Fe = 56 u)?

Answer:

The number of moles of 100 g Na atoms :

$=\frac{100}{23}=4.35moles$

The number of moles in 100 g of Fe atoms :

$=\frac{100}{56}=1.76moles$

As we know, the one-mole atoms contain $6.022\times {10}^{23}$ 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.

Exercise Page - 43

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,

The percentage of an element in the compound :

$=\frac{totalmassofelement}{totalmassofcompound}\times 100$

So,

The percentage of Boron in the compound by weight :

$=\frac{0.096}{0.24}\times 100$

$=40\mathrm{\%}$

The percentage of Oxygen in the compound by weight :

$=\frac{0.144}{0.24}\times 100$

$=60\mathrm{\%}$

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.

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

$=MgC{l}_2$

(b) Calcium oxide

$=CaO$

(c) Copper nitrate

$=Cu(N{O}_3{)}_2$

(d) Aluminium chloride

$=AlC{l}_3$

(e) Calcium carbonate

$=CaC{O}_3$

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 :

$Cao$ = Calcium and Oxygen.

(b) Hydrogen bromide:

$HBr$ = Hydrogen and Bromine.

(c) Baking powder:

$NaHC{O}_3$ = Sodium, Hydrogen, Carbon, and Oxygen.

(d) Potassium sulphate:

$K_{2}S{O}_4$ = Potassium, Sulphur, and oxygen.

Q 6.(a) Calculate the molar mass of the following substances.

• Ethylene , $C_2{H}_2$

Answer:

The molecular mass of ethyne, $C_2{H}_2$ = 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

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, HNO3

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

= 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 (Na2SO3)?

Answer:

The molecular mass of Sodium sulphite (Na2SO3) = 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 Na2SO3 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,

Since 32 g Oxygen = 1 mole

The number of moles in 12 g Oxygen:

$=\frac{1}{32}\times 12moles$

$=0.375moles$

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 :

$=\frac{givenmass}{molarmass}=\frac{20}{18}=1.11moles$

Q 8.(c) Convert into mole.

• 22 g of carbon dioxide

Answer:

Given Mass of the carbon dioxide = 22 g.

The molecular mass of Carbon dioxide in grams = mass of Carbon + 2 * mass of Oxygen

= 12u + 2 * 16u

= 44u

The number of mole :

$=\frac{givenmass}{molecularmass}=\frac{22}{44}=0.5moles$

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,

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 :

$=\frac{16}{256}$

$=0.0625moles$

Now

Number of molecules in 1 mole = $6.022\times {10}^{23}$ molecules

So,

The number of molecules in 0.0625 moles = $0.0625\times 6.022\times {10}^{23}$ molecules.

= $3.76375\times {10}^{22}$ Molecules.

Hence there are $3.76375\times {10}^{22}$ Sulphur molecules in 16 g of solid sulphur .

Q 11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

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 :

$\frac{0.051}{102}=0.0005moles$

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 = $6.022\times {10}^{23}$ ions.

The number of ions in 0.001 moles of Aluminium ion = $0.001\times 6.022\times {10}^{23}$ ions

= $6.022\times {10}^{20}$ ions.

Hence there are $6.022\times {10}^{20}$ Aluminium ions in 0.051 g of aluminium oxide.