UP Board Solutions For Class 9 Science Chapter 3 Atoms And Molecules

Class 9 Science Chapter 3 Atoms And Molecules Long Answer Type Questions

Question 1.

1. Explain why the number of atoms in one mole of hydrogen gas is double the number of atoms in one mole of helium gas.
2. Explain atomic mass unit.
3. How many atoms are present in
   1. MnO₂ molecule
   2. CO molecule?

Answer:

1. Hydrogen gas exists as a diatomic molecule, i.e. each hydrogen gas molecule (H₂) has two atoms. While helium gas exists as a monoatomic particle that is its atoms exist individually.
   1. Thus, one mole of hydrogen gas has a double number of atoms as compared to one mole of helium gas.
2. The unit of mass equivalent to the twelfth part of the mass of the C-12 isotope of carbon is called atomic mass unit (amu) or unified mass (u).
3. In the MnO₂ molecule, three atoms are present: one manganese atom and two oxygen atoms.
   1. In a CO molecule, two atoms are present: one carbon atom and one oxygen atom.

Question 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:

Mass of the compound = 0.24 g,

Mass of boron = 0.096 g

Mass of oxygen = 0.l44g

Percentage of boron = \(\frac{\text { Mass of boron }}{\text { Mass of compound }} \times 100\)

⇒ \(=\frac{0.096 \mathrm{~g}}{0.240 \mathrm{~g}} \times 100=40 \%\)

Percentage of oxygen =\(\frac{\text { Mass of oxygen }}{\text { Mass of compound }} \times 100\)

⇒ \(\frac{0.144 \mathrm{~g}}{0.240 \mathrm{~g}} \times 100=60 \%\)

Alternative method

Percentage of oxygen = 100 – a percentage of boron

= 100-40=60%

Read and Learn More Class 9 Science Solutions

Question 2. Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H_6, C₂H₄, NH₃, CH₃OH
Answer:

1. The molecular mass of H₂ (hydrogen)

= Atomic mass of hydrogen × 2 =1 × 2 = 2 u

2. The molecular mass of O₂ (oxygen)

= Atomic mass of oxygen × 2 = 16 × 2 = 32 u

3. The molecular mass of Cl₂ (chlorine)

Atomic mass of chlorine × 2 = 35.5 × 2 = 71 u

4. The molecular mass of CO₂ (carbon dioxide)

= (Atomic mass of carbon x 1) + (Atomic mass of oxygen × 2)

= 12 × 1 + (16 × 2) = 12 + 32 = 44 u

5. The molecular mass of CH₄ (methane)

= (Atomic mass of carbon × 1) +(Atomic mass of hydrogen × 4) =12 × 1 +(1 × 4) =12 +4 = 16 u

6. The molecular mass of C₂H₆ (ethane)

= (Atomic mass of carbon × 2) + (Atomic mass of hydrogen × 6)

= (12 × 2)+ (1 × 6) = 24 + 6 =30 u

7. The molecular mass of C₂H₄ (ethene)

= (Atomic mass of carbon × 2) + (Atomic mass of hydrogen × 4)

= (12 × 2) + (1 × 4) = 24 + 4 = 28 u

8. The molecular mass of NH₃ (ammonia)

= (Atomic mass of nitrogen × 1) + (Atomic mass of hydrogen × 3)

= (14 × 1) + (1 × 3)

= 14 + 3 =17u

9. Molecular mass of CH₃OH (methanol or methyl alcohol) = (Atomic mass of carbon × 1) + (Atomic mass ,of hydrogen × 3) + (Atomic mass of oxygen × 1 ) + (Atomic mass of hydrogen × 1)

= (12 × 1) + (1 × 3) + (16 × 1) + (1 × 1)

= 12 + 3 + 16 + 1

= 32 u

Question 3. Write the formulae for the following and calculate the molecular mass for each one of them.

1. Caustic potash
2. Baking soda
3. Caustic soda
4. Common salt

Answer:

1. KOH

Molecular mass of KOH =39 +16 + 1= 36 u

2. NaHCO₃

Molecular mass of NaHCO₃ = 23+ 1+ 12+ 3 × 16

= 23 + 1 + 12+ 48

=84 u

3. CaCO₃

The molecular mass of CaCO₃

= 40 + 12 + 3 × 16=100 u

4. NaOH

Molecular mass of NaOH =23+16 + 1

=40u

5. C₂H₅OH

Molecular mass of C₃H₅OH or C₂H₆O =2 ×12 + 6 × 1 + 16

=24 + 6 + 16

=46 u

6. NaCl,

Molecular mass of NaCl = 23 + 35.5

= 58.5 u

Question 4. A metal weighing 6 g formed diatomic oxide upon heating in the presence of air. The oxide thus formed weighed 10 g. Write the chemical name of the compound.
Answer:

Suppose the metal is M. The oxide being diatomic, should have the formula as MO.

The amount of oxygen present in the oxide

= 10 -6 = 4g

∵ 32 g O₂ =1 mol

∴ 4 g O₂ =0.125 mol

The reaction for the formation of oxide can be written as

M + O₂ → MO

Upon balancing we get,

M + O₂ → 2MO

1 mole O₂ reacts with 2 moles of metal

∴ 0.125 mole O₂ reacts with 0.25 mole of metal

Hence, we get

0.25 mole of metal = 6 g

or1 mole of metal =24 g

Thus, the metal is magnesium and the oxide is

magnesium oxide (MgO).

Question 5. Find the number of atoms in 120 g of calcium and 120 g of iron. Which one has more atoms and how much is the difference?

[Atomic mass of Ca =40 u, Fe =56 u]

Answer:

Gram molecular mass of Ca = 40 g

∵ 40 g of Ca contains, several atoms

= 6.022 × 10²³

120 g of Ca contains, several atoms

⇒ \(=\frac{6.022 \times 10^{23}}{40} \times 120\)

= 18.066 × 1023

= 1.8066 × 1024 atoms

Again, the gram molecular mass of Fe = 56 g

∵ 56 g of Fe contains several atoms

= 6.022 x 10²³

∴ 120 g of Fe contains several atoms

⇒ \(=\frac{6.022 \times 10^{25}}{56} \times 120\)

= 12.904 x 10²³ = 1.2904 x 1024 atoms

Therefore, 120g of calcium has more number of atoms.

Difference = (1.8066 -1.2904) x 10²⁴

= 5.162 x 10²³ atoms

Question 6. A sample of ethane (C₂H₆) gas has the same mass as 1.5 x 10²⁰ molecules of methane (CH₄ ). How many C₂H₆ molecules does the sample of gas contain?
Answer:

∵ 6.022 x 10²³ molecules of methane have mass

= Gram molecular mass of methane (CH₄)

=12+ 4 × 1 = 16 g

∴ 1.5 x 10²⁰ molecules of methane have mass

⇒ \(=\frac{16 \times 1.5 \times 10^{20}}{6.022 \times 10^{23}}=3.98 \times 10^{-3} \mathrm{~g}\)

This is also the mass of ethane (C₂H₆).

Gram molecular mass of ethane (C₂H₆)

= 2 × 12+ 6 × 1 = 24 + 6

= 30 g As 30 g of ethane has several molecules

=6.022 × 10²³

Question 7. If 3.42 g of sucrose is dissolved in 18 g of water in a beaker, then what will be the number of oxygen atoms in the solution?
Answer:

Step 1 Molar mass of sucrose,

C₁₂H₂₂O₁₁ =12×12+ 1 × 22+ 16 ×11 = 342g/mol

∵ 342 g of sucrose = 1 mole sucrose

∴ 3.42 g ofsucrose \(=\frac{1 \times 3.42}{342}=0.01 \text { mole sucrose }\)

∵ 1 mole of sucrose (C₁₂H₂₂O₁₁) contains O-atoms

= 11× 6.022 × 10²³

∴ 0.01 mole of sucrose will contain O-atoms = 0.01× 11× 6.023×10²³ = 6.6253 × 10²²

Step 2 18 g water (H₂O) = 1 mole O-atoms = 6.022 × 10²³ O-atoms

Step 3 By adding the number of O-atoms present in 3.42 g sucrose and 18 g water, we get 6.022 × 10²³ +6.6253 × 10²²

= 10²²(60.22+ 6.6253)

= 66.844 × 10²²

= 6.68 × 10²² atoms

Question 8. When 3.0 g 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 3.00 g of carbon is burnt in 50.0g of oxygen? Which law of chemical combination will govern your answer?
Answer:

First, we find the proportion of mass of carbon and oxygen in carbon dioxide.

In CO₂, C: O =12: 32 or 3: 8

In other words, we can say that

12.00 g carbon reacts with oxygen =32.00 g

∴ 3.00 g carbon will react with oxygen = 8 g

Therefore, 3.00 g of carbon will always react with 8.0g of oxygen to form 11 g of carbon dioxide, even if a large amount (50.00 g) of oxygen is present. This means 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 g of carbon dioxide. The remaining 42.00 g of oxygen will remain as it is. This reaction will be governed by the law of constant proportions.

Question 9. Which has more number of atoms? 100 g of N₂ or 100 g of NH₃
Answer:

The molar mass of 1 mole of N₂ = 2 × 14= 28 g

∵ 28 g of N₂ has several molecules

=6.022 x 10²³

∴ 100 g of N₂ has several molecules

⇒ \(=\frac{6.022 \times 10^{23} \times 100}{28}=2.1 \times 10^{24}\)

Atoms in 100 g of N₂ = 2.1 x 10²⁴ × 2

= 4.2 × 10²⁴ atoms.

Similarly, the molar mass of 1 mole of

NH₃= 14 + 3X1 =17 g

∵ 17g NH₃ has number of molecules =6.022 × 10²³

∴ 100 g NH₃ has several molecules

⇒ \(=\frac{6.022 \times 10^{24} \times 100}{17}=3.54 \times 10^{24}\)

Atoms in 100 g of

NH₃ =3.54  × 10²⁴ × 4 = 1.416 x10²⁴

Thus, 100 g of NH₃ has more number of atoms.

Class 9 Science Chapter 3 Atoms And Molecules Short Answer Type Questions

Question 1.

1. State the law of constant proportion.
2. In a compound carbon and oxygen react in a ratio of 3:8 by mass to form carbon dioxide. What mass of oxygen is required to react completely with 9 g carbon?

Answer:

1. The law of constant proportion states that “a pure chemical compound always consists of the same elements that are combined in a fixed (or definite) proportion by mass”.
2. Carbon: oxygen (by mass) =3: 8 i.e. 3 g of carbon requires 8 g of oxygen to form carbon dioxide.

∴ 9 g of carbon requires (3 × 8) 24 g of oxygen to form carbon dioxide.

Question 2. State three points of differences between an atom and a molecule.
Answer:

Question 3. Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide. When 10 g of calcium carbonate is decomposed completely then 5.6 g of calcium oxide is formed. Calculate the mass of carbon dioxide formed. Which law of chemical combination will you use in solving this problem? State the law.
Answer:

The reaction occurs as follows:

According to the law of conservation of mass, Total mass of reactant(s) = Total mass of products(s)

⇒ 10 g = 5.6 g + Mass of CO₂

⇒ Mass of CO₂ =10 —5.6 =4.4g

This problem is solved using the law of conservation of mass according to which mass can neither be created nor be destroyed during a chemical reaction.

Question 4. Give the formulae of the compounds formed from the following sets of elements.

1. Calcium and fluorine 
2. Hydrogen and sulphur
3. Nitrogen and hydrogen
4. Carbon and chlorine

Answer:

Question 5. A vessel containing 28 g of carbon monoxide gas to which 16 g of oxygen gas is added and the mixture is left for a few hours. When checked there was only one kind of gas (carbon dioxide) in the vessel. What change would you observe in the overall mass of the content? Give a reason for your answer.
Answer:

Carbon monoxide (CO) reacts with oxygen (O₂) to form carbon dioxide (CO₂). There will not be any change in the overall mass of the content of the vessel as the mass of the product formed is equal to the mass of reactants.

i.e. Mass of CO₂ =Mass of CO + mass of O₂

= 28+16 = 44 g

The above observation is based on the law of conservation of mass.

Question 6. Write the Latin names of sodium and iron.
Answer:

Latin name ofsodium (Na) = Natrium

Latin name ofiron (Fe) =Ferrum.

Question 7. Which of the following symbols of elements are incorrect? Give their correct symbols.

1. Cobalt (CO)
2. Carbon (c)
3. Aluminium (AL)
4. Helium (He)

Answer:

1. CO is an incorrect symbol of cobalt. Its correct symbol is Co.
2. c is the incorrect symbol of carbon. Its correct symbol is C.
3. AL is an incorrect symbol of aluminium. Its correct symbol is Al.
4. For helium, He is the correct symbol.

Question 8.

1. What is the relative atomic mass of an element?
2. What is the average atomic mass of hydrogen?

Answer:

1. Relative atomic mass is defined as the number of times a given atom is heavier than 1 /12th of the mass of 1 atom of C-12.
2. The atomic mass of hydrogen is taken as lu whereas actually, it is 1.008 u.

Question 9. Define atomic mass unit. State how atoms exist.
Answer:

• The atomic mass unit (amu) is defined as the mass unit equal to exactly 1/12th of the mass of one atom of the C-12 isotope.
• Atoms may exist independently in a free state (For example., He, Ne and Ar) or in a combined state.
• The combined form of atoms may have similar kinds of atoms (H₂, N₂, O₂ etc.) or different kinds of atoms (H₂O, CO₂, H₂S etc).

Question 10. Classify each of the following based on their atomicity.

1. F2
2. NO₂
3. N₂O
4. C₂H₆
5. P₄
6. H₂O₂
7. P₄O₁₀
8. HCl
9. He
10. O₃
11. CH₄
12. Ag

Answer:

1. Monoatomic: Ag, He
2. Diatomic: HCl, F₂
3. Triatomic : NO₂, N₂O, O₃
4. Tetraatomic : P₄, H₂O₂
5. Polyatomic : C₂H₆, P₄O₁₀, CH₄

Question 11. How do molecules react?
Answer:

Molecules exist in the following two forms

1. Molecules of elements These are formed by the combination of two or more atoms of the same element, for example., O₂, H₂, P₄ etc.
2. Molecules of compounds These are formed by the combination of atoms of different elements. For example., NH₃, CO₂ etc.

Question 12. State the number of atoms present in each of the following chemical species.

1. CO^2-₃
2. PO^3-₄
3. P₂O₅
4. CO

Answer:

1. Number of atoms in CO^2-₃

= Number of C-atoms + Number of O-atoms

= 1 + 3 = 4

2. Number of atoms in PO^3-₄

= Number of P-atoms + Number of O-atoms

=1+4=5

3. Number of atoms in P₂O₅

= Number of P-atoms + Number of O-atoms

=2+5=7

4. Number of atoms in CO

= Number of C-atoms + Number of O-atoms =1+1=2

Question 13. Write the cations and anions present (if any) in the following compounds.

1. CH₃COONa
2. NaCl
3. H₂
4. NH₄NO₃

Answer:

Question 14. Give the names of any two elements present in the following compounds: Baking powder, common salt, sulphuric acid
Answer:

• Baking powder (NaHCO₃) Sodium, hydrogen, carbon and oxygen.
• Common salt (NaCl) Sodium and chlorine.
• Sulphuric acid (H₂SO₄) Sulphur, hydrogen and oxygen.

Question 15.

1. Give one word for the following:
   1. Positively charged ion
   2. A group of atoms carrying a charge
2. Mention any two important rules for writing a chemical formula.

Answer:

1. Cation
   1. Polyatomic ion
2. Two rules for writing a chemical formula are:
   1. When a compound consists of a metal and a non-metal, the symbol of the metal is written first and on the left whereas of non-metal on its right.
   2. The valencies or charges on the ions must be balanced in the chemical formula.

Question 16. Write the molecular formulae for the following compounds.

1. Copper (2) bromide
2. Aluminium (3) nitrate
3. Calcium (2) phosphate
4. Iron (3) sulphide

Answer:

Question 17.

1. An element X has a valency of 2. Write the chemical formula for
   1. Bromide of the element
   2. Oxide of the element.
2. Define the formula unit mass of a substance.

Answer:

1. Valency of X = 2

Valency of bromine =1

∴ The formula of compound

= XBr₂

2. Valency of X =2

Valency of oxygen = 2

∴ The formula of compound

= X₂O₂ or XO

Formula unit mass is the sum of atomic masses of all atoms present in a formula unit of a compound. It is calculated by adding the atomic masses of all the atoms present in one formula unit.

Question 18. Write the chemical formulae of the following compound, using the criss-cross method.

1. Magnesium bicarbonate
2. Barium nitrate
3. Answer:

Question 19. Find the ratio by mass of the combining elements in the following compounds.

1. CaCO₃
2. MgCl₂
3. H₂SO₄
4. C₂H₅OH

Answer:

CaCO₃ → Ca : C : O = 40 : 12 : 48

= 10: 3:12

MgCl₂ → Mg : Cl = 24 : 2 × 35.3 = 24 : 71

H₂SO₄ H : S : O = 2 × 1 : 32 : 4 × 16

= 2:32:64 = 1:16:32

C₂H₅OH → C : H : O = 2x 12 : 6x 1 : 16

= 24:6:16 =12:3:8

Question 20. Nitrogen and hydrogen atoms combine in a ratio of 14 : 3 by mass to form an ammonia molecule. Find the formula of the ammonia molecule by calculating the molar ratio. [Given atomic mass of N = 14 u and H = 1 u]
Answer:

Number of nitrogen atoms present in the molecule

⇒ \(=\frac{\text { Proportion by mass }}{\text { Atomic mass }}=\frac{14}{14}=1\)

Number of hydrogen atoms present in the molecule _ Proportion by mass Atomic mass

⇒ \(=\frac{\text { Proportion by mass }}{\text { Atomic mass }}\)

⇒ \(=\frac{3}{1}=3\)

This means many nitrogen and hydrogen atoms combined in ratio =1:3

Thus, the formula of a molecule of ammonia is NH3.

Question 21. Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.

1. Ammonia
2. Carbon monoxide
3. Hydrogen chloride
4. Aluminium fluoride

Answer:

Question 22. An element Y has a valency of 4, write the formula for its

1. Chloride Sol.
2. Nitrate
3. Chloride
4. Nitrate

Answer:

Question 23. Out of the following which has more atoms?

1. 50 g of Cu,
2. 50 g of Fe

[Atomic mass of Cu = 63.5 u, Fe =56 u]

Answer:

Number of atoms in 63.5 g Cu = 6.022 × 1023

Number of atoms in 1 g \(\mathrm{Cu}=\frac{6.022 \times 10^{23}}{63.5}\)

Number of atoms in 50g Cu

⇒ \(=\frac{6.022 \times 10^{23}}{63.5} \times 50\)

⇒ \(4.74 \times 10^{23} \text { atoms }\)

Number of atoms in 56 g Fe = 6.022 x 1023

Number of atoms in 1 g Fe \(=\frac{6.022 \times 10^{23}}{56}\)

Number of atoms in 50 g Fe \(=\frac{6.022 \times 10^{23}}{56} \times 50\)

⇒ \(=5.37 \times 10^{23} \text { atoms }\)

∴ 50 g of Fe has more number of atoms.

Question 24. A compound XH is formed by the combination of an element X with hydrogen. Find the valency of the element. State the formula of the compound formed by the combination of X with nitrogen and X with oxygen.
Answer:

In the compound XH, the valency of elements X is 1.

1. The valency of nitrogen is 3.

The valency of element X is 1.

∴ The formula of the compound = NX₃

2. The valency of oxygen is 2.

The valency of element X is 1.

∴ The formula of the compound = OX₂.

Question 25. During photosynthesis, a six-carbon atom molecule, glucose (C₆H₁₂O₆), is formed. Calculate its molecular mass.

[Atomic mass C = 12.0uH = 1.0u and O = 16.Ou]

Answer:

The molecular mass of glucose

= 6 × atomic mass of C+ 12 × atomic mass of H + 6 atomic mass of O

= 6 × 12 +12 × 1+6 × 16

= 72 + 12 + 96

=180 u

Question 26. 6 g of coke consisting of 100% carbon, is burnt in the air. Find the number of oxygen atoms consisting of the carbon dioxide gas thus formed.
Answer:

Each carbon atom combines with two oxygen atoms on combustion as follows:

6 g of coke consist \(\frac{6}{12} \times N_A\)

Hence, the number of oxygen atoms

⇒ \(=2 \times \frac{6}{12} \times N_A=N_A\)

⇒ \(\left(N_A=6.022 \times 10^{23} \text { atoms }\right)\)

Question 27. The mass of a single atom of M is 3.05x 10^-22g. What is its atomic mass? What would this element be? Check the periodic table for possible answers.
Answer:

1 mole = atomic mass of an element

= 6.022 x 10²³ atoms

∵ Mass of 1 atom = 3.05 x 10^-22 g

∴ Mass of 6.022 x 10²³ atoms

= 3.05 x 1(T22 x 6.022 x 10²³ =183.671g

This is nearly the atomic mass of tungsten (W^183.8).

Question 28. What are polyatomic ions? Give examples.
Answer:

A group of atoms carrying a charge and behaving like one entity is known as a polyatomic ion. For example., an oxygen atom and hydrogen atom combine to form a hydroxide ion (OH^–) and one C-atom and three O-atoms combine to form a carbonate ion (CO^2-₃).

Question 29. Give the names of the elements present in the following compounds.

1. Quicklime
2. Hydrogen bromide
3. Baking powder
4. Potassium sulphate

Answer:

1. Quicklime Calcium oxide — CaO Elements Calcium and oxygen
2. Hydrogen bromide — HBr Elements Hydrogen and bromine
3. Baking powder Sodium hydrogen carbonate — NaHCO₃ Elements Sodium, hydrogen, carbon and oxygen
4. Potassium sulphate — K₂SO₄ Elements Potassium, sulphur and oxygen.

Question 30. Write the molecular formulae of all the compounds that can be formed by the combination of the following ions.

Cu²⁺, Na⁺, Fe³⁺, Cl^–, SO^2-₄ , PO^3-₄

Answer:

Question 31. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.

Sodium carbonate + Ethanoic acid → Sodium ethanoate + Carbon dioxide + Water

Answer:

Total mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid = 5.3 + 6.0 = 11.3 g

Total mass of products =Mass of sodium ethanoate + Mass of carbon dioxide + Mass of water = 8.2 + 2.2 + 0.9 = 11.3 g

Since the sum of masses of reactants is equal to the sum of masses of products, therefore, the observation made is in agreement with the law of conservation of mass.

Question 32. Hydrogen and oxygen combine in a 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:

Since H and O combine in the ratio of 1: 8 by mass.

Therefore, \(\frac{\text { Mass of } \mathrm{H}}{\text { Mass of } \mathrm{O}}=\frac{1}{8}\)

Let the mass of oxygen required to react completely with 3 g of hydrogen gas be x.

⇒ \(\frac{3}{x}=\frac{1}{8}\)

or x = 24 g

Therefore, 24 g of oxygen is required to react with 3 g of hydrogen to form water.

Question 33. Why is it not possible to see an atom with the naked eye?
Answer:

• Atoms are very small, they are smaller than anything we can imagine. More than millions of atoms when stacked would make a layer barely as thick as a sheet of paper.
• These are very small in radii and measured in terms of nanometers (1nm =10^-9m). Hence, it is not possible to see an atom with the naked eye.

Question 34. Write down the formulae of

1. Sodium oxide
2. Aluminium chloride
3. Sodium sulphide
4. Magnesium hydroxide

Answer:

Sodium oxide

Question 35. What is meant by the term chemical formula?
Answer:

It is the shortest way to represent a compound with the help of symbols and valency (charge) of elements, for example., Element

Question 36. How many atoms are present in a

1. H₂S molecule and
2. PO^3-₄ ion?

Answer:

1. In H₂S molecule, three atoms
   1. [i.e. 2 atoms of H + 1 atom of S] are present.
2. InPO^3-₄ ion, five atoms
   1. [i.e. 1 atom of P + 4 atoms of O] are present.

Question 37. Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃. [Given, atomic mass of Zn = 65 u, Na = 23u, K = 39u, C = 12u andO = 16u]
Answer:

Formula unit mass of ZnO (zinc oxide)

= 65 +16 =81 u

Formula unit mass of Na₂O(sodium oxide)

= (23 x 2) + (16 x 1) = 46 + 16 = 62 u

Formula unit mass of K₂CO₃ (potassium carbonate) = (39 x 2) + (12 x 1) + (16 x 3)

= 78+12 + 48 =138 u

Class 9 Science Chapter 3 Atoms And Molecules Very Short Answer Type Questions

Question 1. How did Berzelius assign symbols to the elements?
Answer:

Berzelius assigned symbols to the elements by taking the first one or two letters of the element’s name in English and in some cases the symbols have been taken from the names of elements in different languages such as Latin, German, Greek etc.

Question 2. Give an example of a triatomic molecule of an element.
Answer: Ozone (O₂)

Question 3. Which is octatonic, carbon or sulphur?
Answer: Sulphur (S₈)

Question 4. Write one example of each.

1. Tetra-atomic molecule
2. Diatomic molecule

Answer: Phosphorus (P₄)

Question 5. Is argon monoatomic or diatomic?
Answer: Argon is monoatomic because its atom can exist independently.

Question 6. Give the difference between a cation and an anion.
Answer:

Cation It is a positively charged ion.

For example., Na⁺, K⁺,Ca²⁺,Mg²⁺ etc.

Anion is a negatively charged ion.

For example., Cr^–, Br^–, F^–,O^2-, N^3- etc.

Question 7. What is the difference between a sodium atom and a sodium ion?
Answer:

Question 8. Define a polyatomic ion. Give one example of it.
Answer:

A group of atoms carrying charge in a single entity is called a polyatomic ion. For example., sulphate ion (SO^2-₄) has five atoms.

Question 9. Choose an ionic compound among S₈ , Cu(NO₃)₂ , P₄ , H₂ and O₂?
Answer: Cu(NO₃ )₂ is an ionic compound because it has Cu²⁺ and NO^–₃ ions.

Question 10. What is the molecular formula of aluminium hydroxide?
Answer:

Question 11. What is the chemical formula of ammonium phosphate?
Answer:

Question 12. Choose the correct formulae of sodium sulphide and sodium sulphite: NaS, Na₂SO₄, Na₂S or Na₂SO₃.
Answer: Sodium sulphide = Na₂S Sodium sulphite = Na₂SO₃

Question 13. An element Z has a valency of 3. What is the formula of the oxide of Z?
Answer:

Question 14. If an element X has its valency equal to 3, what will be its formula with carbonate ion?
Answer:

Question 15. Calculate the formula unit mass of Na₂CO₃. [Atomic mass of Na = 23 u, C = 12 u, 0= 16 u]
Answer: Na₂CO₃ =2 × 23+ 12 + 3 × 16=106 u

Question 16. Calculate the formula unit mass of NaHCO₃. [Atomic mass of Na =23 u, H =1 u, C =12 u, O = 16 u]
Answer:

NaHCO₃ – (Atomic mass of Na) + (Atomic mass of H) + (Atomic mass of C) + (3 × Atomic mass of O)

= (23 + 1 + 12 + 3 × 16)

= 84 u

Question 17. What is the mass of sodium in 58.5 g of NaCl?
Answer:

23g [∵ Mass of sodium = 58.5— mass of Cl =58.5-35.5 = 23]

Question 18. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:

‘The relative number and kinds of atoms are constant in a given compound This postulate explains the law of definite proportions.

Question 19. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:

The postulate which is the result of the law of conservation of mass is “atoms are indivisible particles, which can neither be created nor be destroyed in a chemical reaction”.

Question 20. Define atomic mass unit.
Answer:

One atomic mass unit (u) is the mass unit equal to exactly 1/12th of the mass of one atom of the C-12 isotope.