9th CBSE Science Chap 3 Atoms and Molecules Self study Questions

Solved problems:Class 9 _Atoms and Molecules 

1. Q. 5 g of calcium combine with 2 g of oxygen to form a compound. Find the molecular formula of the compound. (Atomic mass of Ca = 40 u; O = 16 u) 

Ans. Number of moles in 5g of calcium = mass / molar mass    =5/40    =  0.125 

Number of moles in 2g of oxygen = mass / molar mass         =  (2 / 16)  =  0.125


Now we will calculate the simplest ratio of the element by dividing the number of moles of each element by the smallest value. 

Since number of moles of each element is 0.125, therefore calcium and oxygen are present in a ratio of 1 : 1. Thus the empirical formula of the compound is CaO.


For calculating the molecular formula, we need the molecular mass of the compound. However, because a compound with the formula CaO is known,


Therefore the molecular formula of the compound is CaO.

. Q. (i) Name the body which approves the nomenclature of elements and compounds.
(ii) The symbol of sodium is written as Na and not as S. Give reason.
(iii) Name one element which form diatomic and one which form tetra atomic molecules. [CBSE 2011 ]

Ans. (i) IUPAC (International Union of Pure and Applied Chemistry)
(ii) Latin name of sodium is Natrium. The first two letters (Na) of this name represents the symbol of sodium.
(iii) Oxygen forms diatomic molecules and phosphorus forms tetra atomic molecules.

3. Q. Calcium and Oxygen are combined in the rates of 5:4 by mass to form calcium oxide.  What mass of Oxygen gas would be required to react with 2.5 g of calcium ? 

Ans:  Calcium and oxygen combine in the rates of 5:4 by mass to form calcium oxide
let x gram of oxygen is required to react with 2.5 g of calcium to form calcium oxide
Therefore 5/4 = 2.5 / x 
     5 x = 2.5 x  4 
       x = (2.5 x 4 ) / 5   
So, x = 2
Therefore 2 grams of oxygen is required to react with 2.5 grams of calcium to form calcium oxide .

Problems (based on mole concept)

4.Q. a. Calculate the number of moles in 81g of aluminium

Solution : Atomic mass of Al= 27gm
     27g  of  aluminium = 1 mole of aluminium  

     81g  of aluminium   = 1/27 x 81=3 moles of aluminium  

OR, Use formula , Number of moles = given mass/atomic mass

Self: ii) 4.6g sodium   iii) 5.1g of Ammonia  iv) 90g of water v) 2g of NaOH

5. Q. Calculate the mass of 0.5 mole of iron

Solution: Atomic mass of iron = 55.9 g

Mass of the 1 mole of iron   = 55.9 g

Mass of the 0.5 mole of iron   = 0.5 x 55.9 g = 27.95 g

Or, Using formula: mass = atomic mass x number of moles

FOLLOW UP: Find the mass of 2.5 mole of oxygen atoms

6.Q.Calculate the number of molecules in 11g of CO₂

Solution: gram molecular mass of CO₂ = 44g

 44g of CO₂  = 6.023 x 10²³  molecules

1 g of CO₂  = ( 6.023 x 10²³ ÷ 44 g ) molecules

11g of CO₂ = ( 6.023 x 10²³ ÷ 44 g ) x 11 = 1.51 x 10²³ molecules

FOLLOW UP: Calculate the number of molecules in 360g of glucose

7. Q. Calculate the mass of glucose in 2 x 10²⁴ molecules

Solution: Gram molecular mass of glucose = 180g

Mass of glucose  [180 x 2 x 10²⁴ ] / [6.023 x 10²³  ] = 597.7g

FOLLOW UP: Calculate the mass of 12.046 x 10²³ molecules in CaO.

8. Q. Calculate the number moles for a substance containing 3.0115 x 10^23 molecules in it.
Solution: Number of moles = Number of molecules/Avogadro Number

                       = [3.0115 x 10²³ ] / [6.023 x 10²³  ]      = 0.5 moles

9. Q. Calculate the mass of 18.069 x 10²³ molecules of SO₂

Solution:

Mass of a substance =  gram molecular mass x number of particles/ 6.023 x 10²³

Gram molecular mass SO2 = 64g

6.023 x 10²³ molecules of SO2     = 64 gm

1           molecules of SO2     = 64/(6.023 x 10²³ )  gm

18.069 x 10²³  molecules of SO2   = [ 64/(6.023 x 10²³ )  x  18.069 x 10²³  ]gm = 192 g

Other Download study links:  Chap 3 Atoms and Molecules class9

Class 9 Atoms and molecules NCERT Solutions-1
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Solved Problems (based on mole concept)
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Chapter based Multiple Choice Questions
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