Q/A, Clas -09, C-3, Atoms and molecules class 09(Science)

 

         Question’s Answer          

          Class 09 (Science)   

                         Chapter 3rd 

                 Atoms and molecules

 

Question’s Answer page no. 32 and 33 NCERT

Q.01- In a reaction, 5.3g of sodium carbonate reacted with 6g of acetic acid. The products were 2.2g of carbon dioxide, 0.9g water and 8.2g of sodium acetate. Show that these observation are in agreement with the law of conservation of mass.

sodium carbonate + acetic acid --->

 sodium acetate + carbon dioxide + water

Ans:- sodium carbonate + acetic acid--->

                 5.3g          +               6 g   ---> 

sodium acetate+carbon dioxide + water 

         8.2             +        2.2                +    0.9s

                11.3 g  (LHS)  =   11.3 g (R H S)

Mass of reactant  = Mass of product

As per law of conservation of mass, the total mass of reactants must be equal to the total mass of products.

   Hence the observation are in agreement with the law of conservation of mass.

 

Q.02- 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 3g of hydrogen gas?

Ans:- Ratio of H:O by mass in water is 1:8

then, the mass of oxygen gas required to react completely with 1gm of hydrogen gas = 8gm

So,the mass of oxygen gas required to react completely with 3gm of hydrogen gas = 

 8×3=24gm

 

Q.03- Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Ans:- The postulate of Dalton’s Atomic theory which is a result of the law of conservation of mass is,

“Atoms can neither be created nor destroyed”

 

Q.04- Which postulate of Dalton’s atomic theory can explain the law of definite proportion?

Ans:- The postulate of Dalton’s atomic theory that can  explain the law of definite proportions is,

“The relative number and kinds of atoms are equal in given compounds”

 

Question’s Answer page no.35 NCERT

Q.01- Define the atomic mass unit.

Ans:- An atomic mass unit is a unit of mass used to express weight of atoms and molecules where one atomic mass is equal to 1/12^th the mass of one carbon-12 atom.

 

Q.02-Why is it not possible to see an atom with naked eyes?

Ans:- Atom is too small to be seen with naked eyes. It is measured in nanometers.

1 m = 10⁹ nm (nanometer)

 

Question’s Answer page no.39 NCERT

Q.01- Write down the formulae of

 (1). Sodium oxide

Ans:- Na₂O

 

 (2). Aluminium chloride

Ans:- AlCl₃

 

(3). Sodium sulphide

Ans:- Na₂S

 

 (4). Magnesium hydroxide

Ans:- Mg(OH)₂

 

Q.02- Write down the names of compounds represented by the following formulae:

(1) Al₂(SO₄)₃

Ans:- Aluminium sulphate

 

(2) CaCl₂

Ans:- Calcium chloride

 

(3) K₂SO₄

Ans:- Potassium sulphate

 

(4) KNO₃

Ans:- Potassium nitrate

 

(5) CaCO₃

Ans:- Calcium carbonate

 

Q.03- What is meant by the term chemical formula?

Ans:- Chemical formula is the symbolic representation of a chemical compound.

Ex. The chemical formula of hydrochloric acid is HCl.

 

Q.04- How many atoms are present in a

(a). H₂S molecule and

(b). PO₄^---

Ans:- (a)- In an H₂S molecule, three atoms are present, two of hydrogen and one of sulphur.

Ans:- (b)- In a PO₄^--- ion, five atoms are present, one of phosphorus and four of oxygen.

 

Question’s Answer page no.40 NCERT

Q.01- Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Ans:-

*Molecular mass of H₂ = 2×Atomic mass of hydrogen

= 2 × 1 = 2u

*Molecular mass of O₂ = 2×Atomic mass of oxygen

= 2 × 16 = 32u

*Molecular mass of Cl₂ = 2 x Atomic mass of chlorine

= 2 × 35.5 = 71u

* Molecular mass of CO₂ = Atomic mass of C + 2 x Atomic mass of O

= 12+(2×16) = 12+32 = 44u

* Molecular mass of CH₄ = Atomic mass of C + 4 x Atomic mass of H

= 12+(4×1) = 12+4 = 16u

* Molecular mass of C₂H₆ =  2 x Atomic mass of C + 6 x Atomic mass of H

= (2 × 12) +(6 × 1) = 24+6 = 30u

* Molecular mass of NH₃ = Atomic mass of N + 3 x Atomic mass of H

= 14 + (3 × 1) = 14+3 = 17u

 

* Molecular mass of CH₃OH  = Atomic mass of C + 3 × Atomic mass of H + Atomic mass of O + Atomic mass of H

= 12+(3×1)+16+1=12+3+16+1= 32u

Q.02- Calculate the formula unit masses of ZnO, Na₂O, K₂CO_3, given atomic masses of Zn = 65u,   Na = 23u, K = 39u, and O = 16u.

Ans:- The formula unit mass of

(a)- ZnO = 65u+16u = 81u

(b)- Na₂O =(23u×2)+16u=46u+16u =62u

(c)- K₂CO₃ = (39u × 2)+12u+(16u ×3) 

                   =  78u+12u+48u = 138u

 

Question’s Answer page no.42 NCERT

Q.01- If one mole of carbon atoms weight 12 gm, what is the mass (in gm) of 1 atom of carbon?

Ans:-

Mass of 1 mole of carbon atom  = 12gm

Then, mass of 6.022×10²³ number of carbon atom = 12gm

Therefore, mass of 1 atom of carbon   =12/6.022×10²³ = 1.9926 × 10^-23gm

Q.02- Which has more number of atoms, 100 gm of sodium or 100 gm of iron (given atomic mass of Na=23u, Fe=56u)?

Ans:- Atomic mass of Na = 23u,

Atomic mass of Fe = 56u

# To calculate the number of atoms in 100gm of sodium:

23gm of Na contains = 6.022×10²³atoms

1gm of Na contains = 6.022×10²³/23

100gm of Na contains

                = 6.022×10²³×100/23

                = 2.6182 ×10²⁴ atoms

# To calculate the number of atoms in 100gm of Iron:

56gm of Fe contain = 6.022×10²³atoms

1gm of Fe contains = 6.022×10²³/56

100gm of Fe contains 

                     = 6.022×10²³ × 100/56

                     = 1.075 ×10²⁴ atoms

->100gm of sodium contain more atoms

 

Question’s Answer Exercises

Q.01- A 0.24 gm Sample of compound of oxygen and boron was found by analysis to contain 0.096 gm of boron and 0.144 gm of oxygen. Calculate the percentage composition of the compound by weight.

Ans:- Boron and oxygen compound --> Boron + oxygen

0.24g ----> 0.096g + 0.144g

# Percentage composition of the compound

For Boron- 0.24 ---> 0.096g

                      100g ---> ?

= 100×0.096/ 0.24 = 40%

For oxygen-  0.24 ---> 0.144g

                        100g ---> ?

= 100×0.144/ 0.24 = 60%

 

Q.02- When 3.0 gm of carbon is burnt in 8.00 gm oxygen, 11.00gm of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00gm of carbon is burnt in 50.00 gm of oxygen? Which law of chemical combination will govern your answer?

Ans:- 3.0gm of carbon combine with 8.0gm of oxygen to give 11.0gm of CO_2.

   Carbon and oxygen are combined in the ratio 3:8 ti give carbon dioxide using up all the carbon and oxygen,

                Hence, for 3gm of carbon and 50gm of oxygen, 8gm of oxygen is used and 11gm of carbon is formed, the left oxygen is unused, 50-8=42gm of oxygen is unused.

     The answer governs the law of constant proportion.

Q.03- What are polyatomic ions? Give examples.

Ans:- The ions which contain more than one atoms (same kind or may be of different kind) and behave as a single unit are called polyatomic ions. Ex- OH^-, SO₄^--, CO₃^—etc

 

Q.04- Write the chemical formulae of the following.

(a)- Magnesium chloride

Ans:- MgCl₂

 

(b)- Calcium oxide

Ans:- CaO

 

(c)- Copper nitrate

Ans:- Cu(NO)₃

 

(d)- Aluminium chloride

Ans:- AlCl₃

 

(e)- Calcium carbonate

Ans:- CaCO₃

 

Q.05- Give the names of the elements present in the following compounds.

(a). Quick lime

Ans:- Quick lime ( Calcium oxide )

Elements -> Calcium and oxygen

 

(b). Hydrogen bromide

Ans:- Hydrogen bromide

Elements -> Hydrogen and bromide

 

(c). Baking powder

Ans:- Baking powder (Sodium hydrogen carbonate)

Elements -> Sodium, hydrogen, carbon and oxygen

 

(d). Potassium sulphate.

Ans:- Potassium sulphate

Elements -> Potassium, sulphur and oxygen

 

Q.06- Calculate the molar mass of the following substances.

(a). Ethyne, C₂H₂

Ans:- C₂H₂ = (2×12)+(2×1) = 24+2= 26gm

 

(b). Sulphur molecule, S₈

Ans:- S₈ = 8 × 32 = 256gm

 

(c). Phosphorus molecule, P₄ (Atomic mass of phosphorus =31)

Ans:- P₄  = 4 ×31 = 124gm

 

(d). Hydrochloric acid, HCl

Ans:- HCl = (1×1)+(1×35.5)

                  = 1+35.5=36.5gm

 

(e). Nitric acid, HNO₃

Ans:- HNO₃ = (1×1)+(1×14)+(3 ×16)

                      = 1+14+48 = 63gm

 

Q.07- What is the mass of –

(a). 1 mole of nitrogen atoms?

Ans:- The mass of 1 mole of nitrogen atoms is = 14gm

(Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen atoms)

 

(b). 4 mole of aluminium atoms (atomic mass of aluminium = 27) ?

Ans:- The mass of 4 moles of aluminium atoms

  = 4×27= 108gm

 

(c). 10 moles of sodium sulphite(Na₂SO₃)?

Ans:-The mass of 10 moles of sodium sulphite

Na₂SO₃ = 10×[2×23+32+3×16]  

              =10×[46+32+48]

              = 10 ×126 = 1260gm

 

Q.08- Convert into mole.

(a).  12 gm of oxygen gas

Ans:- Give: Mass of oxygen gas = 12gm

      Molar mass of oxygen gas = 2 mass of oxygen

                                      = 2 ×16=32gm

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

                                     = 0.375 moles

 

(b). 20 gm of water

Ans:- Given: Mass of water = 20gm

Molar mass of water = 2 × Mass of hydrogen + Mass of oxygen 

                = (2 ×1)+16=2+16=18gm

Number of moles = Mass given/ molar mass of water

                           = 20/18 = 1.11moles

 

(c). 22 gm of carbon dioxide.

Ans:- Given: Mass of CO₂ = 22gm

Molar mass of CO₂ = Mass of C + 2 ×Mass of oxygen

= 12+(2 ×16) = 12+32 = 44gm

Number of moles = Mass given/ molar mass od CO₂

                                  = 22/44  

                                  = 0.5 moles

 

Q.09- What is the mass of –

(a). 0.2 mole of oxygen atoms?

Ans:- Mass of 1 mole of oxygen atoms = 16u,

Then, mass of 0.2 moles of oxygen atoms = 0.2×16=3.2u

 

(b) 0.5 mole of water molecules?

Ans:- Mass of 1 mole of water molecules = 18u,

Then, mass of 0.5 moles of water molecules

= 0.5×18=9u

 

Q.10-Calculate the number of molecule of sulphur (S₈) present in 16 gm of solid sulphur.

Ans:- Molecular mass of sulphur (S₈) = 8×32=256gm

Mass given = 16gm

 

Number of moles = mass given/ molar mass of sulphur = 16/256 = 0.0625

 

:- Number of molecules = Number of moles × Avogadro number

= 0.0625 × 6.022 × 10²³ molecules

= 3.763 ×10²² molecules

 

Q.11- Calculate the number of aluminium ions present in 0.051 gm of aluminium oxide. (The mass of an ion is the same as that of an atom of the same element. Atomic mass of [aluminium=27)

Ans:-1 mole of aluminium oxide =6.022 × 10²³ molecules of aluminium oxide

 

1 mole of aluminium oxide(Al₂O₃)

= 2 × Mass of aluminium+3 × Mass of oxygen

= (2 ×27)+(3 ×16)=54+48=102gm

 

1 mole of aluminium oxide

 =102gm=6.022 × 10²³ molecules of aluminium oxide

Therefor, 0.051gm of Al₂O₃ has

= 0.051 ×6.022 ×10²³/102

=3.011 ×10²⁰ molecules of Al₂O₃

 

Therefor, one molecules of aluminium oxide has 2 aluminum ions(Al⁺⁺⁺), hence number of aluminium ions  present in 0.051gm of aluminium oxide

 = 2 ×3.011 ×10²⁰ = 6.022 ×10²⁰