# NCERT Class 9 Science Notes & Solutions – Chapter 3 – Atoms and Molecules

## Exercises Questions & Answers - Atoms and Molecules - Chapter 3

#### In-Text Questions & Answers | NCERT Class 9 Science Chapter 3 | Atoms and Molecules

##### In-Text Questions & Answers | NCERT Class 9 Science | Atoms and Molecules | Page Nos. 32 – 33

1. In a reaction, 5.3 gram of sodium carbonate reacted with 6 gram of acetic acid. The products were 2.2 gram of carbon dioxide, 0.9 gram water &amp; 8.2 gram of sodium acetate. Prove that this follows the law of conservation of mass.

In this case –
sodium carbonate + acetic acid =  carbon dioxide + water +sodium acetate,
⇒ 5.3 gram + 6 gram = 2.2 gram + 0.9 gram + 8.2 gram, (after putting the quantities)
⇒ 11.3 gram = 11.3 gram,
Therefore, total mass before reaction = total mass after reaction.
So it follows the law of conservation of mass.

##### 2. Hydrogen & 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 gram of hydrogen gas?

Water ⇒ Hydrogen : Oxygen ⇒ 1 : 8. It means that 1 gram hydrogen reacts with 8 gram oxygen. So, 3 gram hydrogen will react with (3 x8) = 24 gram oxygen.

3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.

4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

The relative number & kind of atoms in a given compound remain constant.

##### In-Text Questions & Answers | NCERT Class 9 Science | Atoms and Molecules | Page Nos. 35

1. Define atomic mass unit.
One
atomic mass unit is equal to (1/12) times of weight of single atom of Carbon-12 isotope. One atomic mass is written as u. Therefore –

1 u = weight of single atom of carbon-12 isotope / 12

2. Why is it not possible to see an atom with naked eyes?
We can not see an atom with naked eyes because it is too small & does not exist independently.

##### In-Text Questions & Answers | NCERT Class 9 Science | Atoms and Molecules | Page Nos. 38-39

1. Write down the formulae of – i) sodium oxide, ii) aluminium chloride, iii) sodium sulphide, iv) magnesium hydroxide.
1) Sodium Oxide ⇒ Na2O, 2) Aluminium Chloride ⇒ AlCl3, 3) Sodium Sulphide ⇒ Na2S, 4) Magnesium Hydroxide ⇒ Mg(OH)2.

##### 2. Write down the names of compounds represented by the following formulae: i) Al2(SO4)3, ii) CaCl2, iii) K2SO4, iv) KNO3, v) CaCO3.

i) Al2(SO4)3 ⇒ Aluminium sulphate, ii) CaCl2 ⇒ Calcium chloride, iii) K2SO4 ⇒ Potassium sulphate, iv) KNO3, ⇒ Potassium nitrate, v) CaCO3 ⇒ Calcium carbonate.

3. What is meant by the term chemical formula?
It expresses the symbolic representation of the composition of a compound. We can easily know the number & kinds of atoms of different elements that a compound contains. For example – H2O (water). From the chemical formula H2O, we come to know that two hydrogen atom & one oxygen atom are chemically bonded to form one molecule of H2O i.e. water.

##### 4. How many atoms present in a – i) H2S molecule, & ii) PO43- ion?

i) H2S molecule = [02 nos. hydrogen atom + 01 no. sulphur atom] = 03 nos. atom.
ii) PO43- ion = [01 no. phosphorus + 04 nos. oxygen atom] = 05 nos. atom.

##### 1. Calculate the molecular masses of H2, O2, Cl2, CO2, C2H6, C2H4, NH3, CH3OH.

Atomic masses of H = 1 u, O = 16 u, Cl = 35.5 u, C = 12 u, N = 14.
H2 = (2×1) u = 2 u, O2 = (2×16) u = 32 u, Cl2 = (2×35.5) u = 71 u, CO2 = [(1×12) + (2×16)] u = 44 u,
C2H6 = [2×12  +  6×1] u = 30 u,
C2H4 = [2×12  +  4×1] u = 28 u, NH3 = [1×14  + 3×1] u = 17 u,
CH3OH = [(1×12) + (1×3) + (1×16) + (1×1)] u = 32 u.

##### 2. Calculate the formula unit masses of ZnO, Na2O, K2CO3. Given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, & O = 16 u.

Atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, & O = 16 u.
ZnO = [1×65  +  1×16] u = 81 u,
Na2O = [2×23  +  1×16] u = 62 u,
K2CO3 = [2×39  +  1×12  +  3×16] u = 138 u.

##### In-Text Questions & Answers | NCERT Class 9 Science | Atoms and Molecules | Page Nos. 42

1. If one mole of carbon atoms weighs 12 grams, what is the mass in grams of 1 atom of carbon?
In this case –
one mole of carbon atoms = 12 grams carbon. And we know, Avogadro number = 6.022X1023 .
So, weight of 6.022X1023 nos. carbon atoms = 12 gram.
Then, weight of 01 no. carbon atom = [12/6.022X1023 ] gram = 1.9926 x 10-23 gram.

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

In this case –  atomic mass of Na = 23 u, Fe = 56 u. Avogadro number = 6.022X1023 .
Now, 23 gram Na contains 6.022X1023 nos. atom.
So, 100 gram Na contains = [(100/23) x 6.022X1023 ] = 2.6182X1024 nos. atom.
Similarly, 56 gram Fe contains 6.022X1023 nos. atom.
So, 100 gram Fe contains = [(100/56) x 6.022X1023 ] = 1.0753X1024 nos. atom.
Therefore, 100 gram Na contains more atoms than 100 gram Fe.

##### Exercises Questions & Answers | NCERT Class 9 Science | Atoms and Molecules | Page Nos. 43 - 44
###### 1. A 0.24 gram sample of compound of oxygen & boron was found by analysis to contain 0.096 gram of boron & 0.14 gram of oxygen. Calculate the percentage composition of the compound by weight.

In this case, total sample ( boron+oxygen) = 0.24 gram.
Boron = 0.096  gram. So % of  boron in sample = (0.096/0.24) x 100% = 40 %.
Oxygen = 0.14  gram. So % of  Oxygen in sample = (0.144/0.24) x 100% = 60 %.
Therefore, ratio by mass : Boron : Oxygen = 0.096 : 0.144 = 96 : 144  = 4 : 6

###### 2. When 3.0 gram of carbon is burnt in 8.00 gram of oxygen, 11.00 gram carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.0 gram of carbon is burnt in 50.00 gram of oxygen? Which law of chemical composition will govern your answer?

We apply laws of definite proportion in this case.
3 gram carbon + 8 gram oxygen = 11 gram carbon dioxide.
So, ratio by mass = carbon : oxygen = 3 : 8.
In this case, 3 gram carbon is burnt in 50 gram oxygen. But only 11 gram carbon dioxide is formed as there is only 3 gram carbon and as per laws of definite proportion, 3 gram carbon reacts with 8 gram oxygen.

3. What are polyatomic ions? Give examples.

Compound radical or poly-atom is a group of atom of different elements and it behaves like single atom in chemical reaction. It also turns into cation & anion by losing or adding electron respectively. For example – Ammonium – NH4+ ,Sulphite – SO32-, Sulphate – SO42-, Phosphate – PO43-, Nitrate – NO3-, Hydroxide – OH etc.

###### 4. Write chemical formulae of – a) Magnesium chloride , b) Calcium oxide, c) Copper nitrate, d) Aluminium chloride, e) Calcium carbonate.

a) Magnesium chloride [ MgCl2]
Symbol           Mg              Cl
Valency          2                  1
Formula ⇒  Mg1Cl2= MgCl2

b) Calcium oxide (CaO)
Symbol ⇒     Ca        O
Charge ⇒     2+        2-
Formula ⇒   Ca2O2=CaO

c) Copper Nitrate [ Cu(NO3)2]
Symbol          Cu              NO3
Valency          2                  1
Formula ⇒  (Cu)1(NO3)2= Cu(NO3)2

d) Ammonium Chloride [ AlCl3 ]
Symbol           Al                Cl
Valency           3                  1
Formula ⇒ (Al)1(Cl)3 = AlCl

e) Calcium Carbonate [ CaCO3 ]
Symbol           Ca                      CO3
Valency           2                          2
Formula ⇒ Ca2(CO3)2= CaCO3

###### 5. Write name of the elements present in the following compounds – a) Quick lime, b) Hydrogen bromide, c) Baking powder, d) Potassium sulphate.

Quick lime: Calcium (C) and Oxygen (O).
Hydrogen bromide: Hydrogen (H) and Bromine (Br).
Baking powder: Sodium (Na), Hydrogen (H), Carbon (C) and Oxygen (O).
Potassium sulphate: Potassium (K), Sulphur (S) and Oxygen (O).

###### 6. Calculate the molar mass of the – a) Ethyne (C2H2), b) Sulphur molecule (S8), c) Phosphorus molecule (P4 atomic mass of Phosphorus is 31), d) Hydrochloric acide (HCl), e) Nitric Acid ( HNO3).

See Q7 especially to know atomic mass of various elements.
a) Ethyne (C2H2) : [ 12×2 +  1×2  ] gram = 26 gram.
b) Sulphur molecule (S8) : [ 32×8 ] gram = 256 gram.
c) Phosphorus molecule (P4) : [ 31×4 ] gram = 124 gram.
d) Hydrochloric acide (HCl) : [ 1×1 + 35.5×1  ] gram = 36.5 gram.
e) Nitric acid ( HNO3) : [ 1×1 +  14×1  + 16×3  ] gram = 63 gram.

###### Basically one mole of any substance is the quantity in grams of it’s molecular mass. And molecular mass of a molecule of an element or compound is the summation of atomic masses of all atoms it contains. Therefore -

a) 1 mole of Nitrogen atoms : [ 1×14 ] gram = 14 gram.
b) 4 moles of Aluminium atoms : [ 4 x 27 ] gram = 108 gram.
c) 10 moles of Sodium sulphate ( Na2 SO3) : 10 x [ 2×23 + 1×32 +  3×16 ] gram = 1260 gram.
[For a and b : 1 mole of atom is mentioned. So we have taken atomic mass. If the question is like – what is the mass of 1 mole of Nitrogen gas ? Nitrogen gas is N
2. It contains two Nitrogen atom. So the answer would be – [ 2×14 ] gram i.e. 28 gram.

###### 8. Convert into mole – a) 12 gram of oxygen gas, b) 20 gram of water, c) 22 gram of carbon dioxide.

a) 1 mole of oxygen gas = 1 mole O2 = (16×2) gram oxygen = 32 gram oxygen.
So 12 gram of oxygen gas = (12/32) mole of oxygen gas = 0.375 mole of oxygen gas.
b) 1 mole of water = 1 mole H2O = [1X2   + 16×1) gram water = 18 gram water.
So 20 gram of water = (20/18) mole of water = 1.11 mole of water.
c) 1 mole of carbon dioxide = 1 mole CO2 = (12X1 + 16×2) gram CO2 = 44 gram CO2 .
So 22 gram of carbon dioxide = (22/44) mole of carbon dioxide = 0.50 mole of carbon dioxide.

###### 9. What is the mass of – a) 0.2 mole of oxygen atoms, & b) 0.5 mole of water molecules?

a) 1 mole of oxygen atom = 1 mole O = 16 gram oxygen atom.
So 0.2 mole of oxygen atom = (16×0.2) gram oxygen atom  = 3.2 gram oxygen atom.
b) 1 mole of water molecule = 1 mole H2O = [1X2   + 16×1) gram water = 18 gram water.
So 0.5 mole of water molecule = (0.5×18) gram water molecule = 9 gram water molecule.

###### 10. Calculate the number of molecules of sulphur(S8) present in 16 gram of solid sulphur.

Basically one mole of any substance contains 6.022 X 1023 nos. molecule.
In this case, 1 mole sulphur (S8)  = (32×8) gram sulphur (S8)  = 256 gram sulphur (S8).
So one mole of sulphur (S8) i.e. 256 gram sulphur (S8) contains 6.022 X 1023 nos. molecule.
So 16 gram sulphur (S8)  contains = (16/256) x 6.022 X 1023 nos. atom = 3.76 X 1022 nos. molecule.

###### 11. Calculate the number of aluminium ions present in 0.051 gram of aluminium oxide (Atomic mass of Al = 27 u).

In this case, 01 mole  Al2O3 = [27×2 + 16×3] gram Al2O3 = 102 gram Al2O3. So 102 gram Al2O3 contains 6.022 X 1023 nos. molecule. So 0.051 gram Al2O3 contains [(0.051/102) x 6.022 X 1023] = 3.011 x 1020  nos. molecule .
Aluminium oxide (Al2O3
Symbol    Al          O
Charge    3+         2-

Therefore, we get that one Al2O3 molecule contains 2 nos. positive Al ion and 3 nos. negative O ion. 2 nos. positive Al ion = (2×3)+ = 6+ charge and 3 nos. negative O ion = (3×2)- = 6- charge. Thus both nullify each other. So 0.051 gram Al2O3  i.e.  3.011 X 1020  nos. molecule contains  [2 x (3 X 1020)] = 6.022 x 1020  nos. Al ion.

Atoms and Molecules – To Be Continued – Atoms and Molecules