**Class 9**^{th }**Science ****Chapter 3 – ****Atoms and Molecules**

^{th }

**Exercise-3.1 (Page: 32)**

**In a reaction, 5.3g 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

5.3g + 6g —> 8.2g + 2.2g + 0.9g

The law of conservation of mass says that the sum of the total mass of the reactants is always equal to the sum of the total mass of the products.

Here, we get the same total on both sides, which means LHS=RHS=11.3g. These observations that have been made above are thus in conformation with the law of conservation of mass.

**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?**

**Given in the question, hydrogen and oxygen combine in the ratio of 1:8, which means, 1g hydrogen and 8g oxygen combine to form water.**

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**So, in order to react completely, 3 g of hydrogen gas would require **

**3 × 8= 24 g mass of oxygen. **

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

**“Atoms can neither be created nor be destroyed**” **is the postulate from Dalton’s atomic theory which is a result of the law of the conservation of mass.**

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

The postulate of Dalton’s atomic theory- **“The relative number and kinds of atoms are constant in a given compound” **explains the law of definite proportions. According to this postulate, the atoms in a compound will always combine in a specific ratio.

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**Exercise-3.2 (Page: 35)**

**Define the atomic mass unit?**

**Atomic mass unit or unified atomic mass is the unit of mass used in Chemistry and is defined as the 1/12 ^{th} of the mass of an atom of carbon-12.**

**Why is it not possible to see an atom with naked eyes?**

**Atoms are infinitely small in size and they are so small that even the light cannot interact with them. They are invisible to the light itself and thus cannot be seen even through the most powerful light-focussing microscopes.**

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**Exercise-3.3-3.4 (Page: 39)**

**Write down the formulae of**

**(i) sodium oxide**

**(ii) aluminium chloride**

**(iii) sodium sulphide**

**(iv) magnesium hydroxide**

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(i) The formula for sodium oxide is described as Na_{2}O.

(ii) The formula for aluminium chloride is described as AlCl_{3}

(iii) The formula for sodium sulphide is described as – Na_{2}S

(iv) The formula for magnesium hydroxide is described as Mg (OH)_{2}

**Write down the names of compounds represented by the following formulae:**

**(i) Al _{2}(SO_{4})_{3}**

**(ii) CaCl _{2}**

**(iii) K _{2}SO_{4}**

**(iv) KNO _{3}**

**(v) CaCO _{3}.**

**The names of the compounds with the above chemical formulae are—**

(i) Aluminium sulphate is the compound for the chemical formula Al_{2}(SO_{4})_{3.}

(ii) Calcium chloride is the compound for the chemical formula CaCl_{2. }

(iii) Potassium sulphate is the compound for the chemical formula K_{2}SO_{4}.

(iv) Potassium nitrate is the compound for the chemical formula KNO_{3}.

(v) Calcium carbonate is the compound for the chemical formula CaCO_{3}.

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**What is meant by the term chemical formula?**

**Chemical formula is a representation that makes use of symbols to offer information regarding the chemical compositions of a compound. **

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**In other words, it is the representation of the chemical proportion of atoms in a compound.**

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**How many atoms are present in a**

**(i) H _{2}S molecule and**

**(ii) PO _{4}^{3-} ion?**

The number of atoms present are mentioned as follows: –

(i) 2 atoms of hydrogen and 1 atom of sulphur i.e. a total of **3 atoms **are present in an H_{2}S molecule**.**

(ii) 1 atom of phosphorus and 4 atoms of oxygen i.e. a total of **5 atoms** in total are present in an PO_{4}^{3-} ion.

**Exercise-3.5.1-3.5.2 (Page: 40)**

**Calculate the molecular masses of H**_{2 }, O_{2}, Cl_{2}, CO_{2}, CH_{4}, C_{2}H_{6}, C_{2}H_{4}, NH_{3}, CH_{3}

The following are the molecular masses of the above molecules and compounds:—

The molecular mass of H_{2} – 2 x atoms atomic mass of H = 2 x 1u = 2u

The molecular mass of O_{2} – 2 x atoms atomic mass of O = 2 x 16u =32u

The molecular mass of Cl_{2} – 2 x atoms atomic mass of Cl = 2 x 35.5u =71u

The molecular mass of CO_{2} – atomic mass of C + 2 x atomic mass of O = 12 + (2×16)u = 44u

The molecular mass of CH_{4} – atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x 1)u = 16u

The molecular mass of C_{2}H_{6}– 2 x atomic mass of C + 6 x atomic mass of H = (2 x 12) + (6 x 1)u= 24+6=30u

The molecular mass of C_{2}H_{4}– 2 x atomic mass of C + 4 x atomic mass of H = (2x 12) + (4 x 1)u= 24+4=28u

The molecular mass of NH_{3}– atomic mass of N + 3 x atomic mass of H = (14 +3 x 1)u = 17u

The molecular mass of CH_{3}OH – atomic mass of C + 3x atomic mass of H + atomic mass of O + atomic mass of H

= (12 +3×1+16+1)u =(12+3+17)u = 32u

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**Calculate the formula unit masses of ZnO, Na**_{2}O, K_{2}CO_{3}, given atomic masses of Zn = 65u,

**Na = 23 u, K=39u, C = 12u, and O=16u.**

The given atomic mass of Na is 23u.

The given atomic mass of Zn is 65u.

The given atomic mass of C is 12u

The given atomic mass of K is 39u.

The given atomic mass of O is 16u.

- The formula for unit mass of ZnO is described as

Atomic mass of Zn + Atomic mass of O

i.e. 65u+16u = 81u

The unit mass for ZnO is 81u.

- The formula for unit mass of
**K**_{2}CO_{3 }is described as

**= 2 x Atomic mass of K + atomic mass of C + 3 x atomic mass of O**

**= 2x39u + 12u + 3x16u **

**=78u+12u + 48u**

**=138u**

- The formula unit mass of Na
_{2}O = 2 x Atomic mass of Na + Atomic mass of O i.e. (2 x 23)u + 16u= 46u+16u= 62u

**Exercise-3.5.3 (Page: 42)**

**If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?**

As per the data provided in the above question,

The weight of 1 mole of carbon is 12g. This implies that weight of 1 mole of carbon atoms can be described as 6.022 X 10^{23}

Thus, the molecular mass of carbon atoms = 12g = an atom of carbon mass

Hence, the mass of 1 carbon atom is

= 12 / 6.022 X 10^{23}

=** 1.99 X 10 ^{-23}g**

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

Given that— the atomic mass of Na=23u,

And the atomic mass of Fe= 56u

To calculate the number of atoms in 100g of sodium:

23g of Na contains = 6.022 X 10^{23} atoms

1g of Na contains = 6.022 X 10^{23} atoms / 23

100g of Na contains = 6.022 X 10^{23} atoms X 100 / 23 = 2.6182 X 10^{24} atoms

To calculate the number of atoms in 100g of sodium:

56g of Fe contains = 6.022 X 10^{23} atoms

1g of Fe contains = 6.022 X 10^{23} atoms / 56

100g of Fe contains = 6.022 X 10^{23} atoms X 100 / 56 = 1.075 X 10^{24} atoms

Therefore, through comparison, it is clear that **100g of Na has more atoms.**

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**Exercise (Part 43)**

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**A 0.24g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.**- To calculate the percentage composition of the compound by weight, the formula to be used is:
**Mass of Boron/mass of the compound × 100**

- To calculate the percentage composition of the compound by weight, the formula to be used is:

The following information is given

Mass of boron = 0.096 g

Mass of oxygen -= 0.144g

Therefore, Percentage of Boron = 0.096g/0.24g × 100 = 40%

To calculate the percentage of oxygen = 100 – percentage of boron i.e,

100 – 40=60%

**When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?**- According to the information provided in the question, we know that to produce 11.00 g of carbon dioxide, we need to burn 3.0g of carbon in 8.00 g of oxygen.

Therefore, when 3.0 g of carbon is burnt in 50.00 g of the oxygen, we again get only 11 g of carbon dioxide since carbon and oxygen combine in the ratio of 3:8. So when 3 g of carbon is burnt in 50 g of oxygen, the 3g of carbon only reacts with 8 g of oxygen from the 50g available and produces 11g of carbon dioxide.

This is the law of constant or definite proportions where the given chemical compound contains its component elements in constant ratios, and therefore, the process of preparation doesn’t change the ratio.

**What are polyatomic ions? Give examples.**- A molecular ion which is made up of 2 or more atoms in a covalent bond and performs as a single unit is known as polyatomic ions.

Examples: Ammonium NH4+, Chlorite ClO_{2}^{–}

**Write the chemical formula of the following.**

**(a) Magnesium chloride**

**(b) Calcium oxide**

**(c) Copper nitrate**

**(d) Aluminium chloride**

**(e) Calcium carbonate**

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**The chemical formula for Magnesium chloride is described as MgCl**_{2}.**The chemical formula for****Calcium oxide****is described as****The chemical formula for Copper nitrate is described as Cu(NO**_{3})_{2}.**The chemical formula for****Aluminium chloride****is described as****AlCl**_{3}.**The chemical formula for****Calcium carbonate****is described as****CaCO**_{3}.

**Give the names of the elements present in the following compounds.**

**(a) Quick lime **

**(b) Hydrogen bromide **

**(c) Baking powder **

**(d) Potassium sulphate **

**The elements present in Quick lime are Calcium and oxygen i.e. CaO.****The elements present in Hydrogen Bromide are Hydrogen and bromine i.e. HBr****The elements present in Baking Powder are Sodium, Carbon, Hydrogen, and Oxygen i.e. NaHCO**_{3}.- The elements present in Potassium Sulphate are Sulphur, Oxygen, and Potassium i.e. K
_{2}SO_{4}.

**Calculate the molar mass of the following substances.**

**–** To calculate the molar mass of the substances, add the standard atomic names of the constituent atoms.

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**Ethyne, C2H2 –**2 × Mass of C + 2×Mass of H

** = **2×12+2×1

** = **24+2 =26g

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**Sulphur molecule, S8 –**8×Mass of S =8×32=256g

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** (c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31) = ** Molar mass of Phosphorus molecule, P_{4} = 4 x Mass of P = 4 x 31 = 124g

**Hydrochloric acid, HCl =**Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g

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** ( e) Nitric acid, HNO3 =** Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+

3×16 = 63g

**What is the mass of –**

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**1 mole of nitrogen atoms?**- The atomic mass of Nitrogen atoms is 14u

Moreover atomic mass of nitrogen atoms is equal to the mass of 1 mole of nitrogen atoms.

Therefore, 14g will be the mass of 1 mole of a nitrogen atom.

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**4 moles of aluminium atoms (Atomic mass of aluminium =27)?**- The atomic mass of aluminium is 27u.

Thus, it can be inferred that 27g is the mass of 1 mole of aluminium atoms.

Now, from the above information it is implied that the mass of 1 mole of aluminium atoms is 27g,

Therefore, the mass of 4 moles of aluminium atoms can be described as

= 4 x 27g

= 108g.

Thus, the mass of 4 moles of aluminium atoms is 108g.

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** (c) 10 moles of sodium sulphite (Na _{2}SO_{3})?**

- It is pertinent to mention that the molecular mass of sodium sulphite is equal to the mass of 1 mole of sodium sulphite (Na
_{2}SO_{3}),

Therefore, the mass of 1 mole of sodium sulphite can be calculated as,

= 2 x Mass of Na + Mass of S + 3 x Mass of O

= (2 x 23) + 32 +(3x 16)

46+32+48 = 126g.

Therefore, the mass of 10 moles of sodium sulphite (Na_{2}SO_{3})= 10 x 126 = 1260g.

**Convert into mole.**

**12g of oxygen gas – **Mass of oxygen gas is 12g

Molar mass of oxygen gas = 2 Mass of Oxygen = 2 x 16 = 32g

Number of moles = Mass given / molar mass of oxygen gas = 12/32 = 0.375 moles

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**20g of water – **Mass of water is 20g

Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g

Number of moles = Mass given / molar mass of water = 20/18 = 1.11 moles

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** 22g of carbon dioxide – **Mass of carbon dioxide = 22g

Molar mass of carbon dioxide = Mass of C + 2 x Mass of Oxygen = 12 + 2x 16 = 12+32=44g.

Number of moles can be computed as Mass given / molar mass of carbon dioxide i.e.

= 22/44 = 0.5 moles.

**What is the mass of:****i) 0.2 mole of oxygen atoms?**

**It is known that the mass of 1 mole of oxygen atoms is equal to 16u, hence it weighs 16g**

Therefore, the mass of 0.2 moles of oxygen atoms will be calculated as follows,

= 0.2 x 16

= 3.2u

Therefore, the mass of 0.2 mole of oxygen atoms is 3.2u.

**ii) 0.5 mole of water molecules?**

It is known that the mass of 1 mole of water molecules is 18u, hence it weighs 18g.

Therefore, the mass of 0.5 moles of water molecules will be calculated as follows,

= 0.5 x 18

= 9u

Hence, the mass of 0.5 mole of water molecules will be 9u.

**Calculate the number of molecules of sulphur (S**_{8}) present in 16g of solid sulphur.

It is known that the molecular mass of Sulphur (S_{8}) can be calculated as follows: –

8 x Mass of Sulphur i.e.

= 8×32

= 256g.

Now, the mass given = 16g

Number of moles = mass given/ molar mass of sulphur i.e.

16/256 = 0.0625 moles

To calculate the number of molecules of sulphur in 16g of solid sulphur following formula will be used i.e.

Number of molecules = Number of moles x Avogadro number

= 0.0625 x 6.022 x 10²³ molecules

= 3.763 x 10^{22} molecules

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

**The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u**

1 mole of aluminium oxide = 6.022 x 10^{23} molecules of aluminium oxide

1 mole of aluminium oxide (Al_{2}O_{3}) = 2 x Mass of aluminium + 3 x Mass of Oxygen = (2x 27) + (3 x16) = 54 +48 = 102g

1 mole of aluminium oxide = 102g = 6.022 x 10^{23} molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has = 0.051 x 6.022 x 10^{23} / 102** **= 3.011 x 10^{20} molecules of aluminium oxide

Since, one molecule of aluminium oxide contains 2 aluminium ions, the number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 1020 molecules of aluminium oxide = 6.022 x 1020.