Given below are the Class 9 Science CBSE Test Paper for Atoms and Molecules
a. Concepts questions
b. Calculation problems
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Concepts questions
Question 1
State the law of mass conservation. Answer
Law of mass conservation states that, mass can neither be created nor destroyed in a chemical reaction or a physical transformation.
Question 2
What do you understand by the law of constant proportion? Answer
According to the Law of constant proportion, a chemical compound always has its compounding elements in definite proportion by mass, irrespective of the source and the type of chemical reaction.
Question 3
Who was responsible for coining the term atom? Answer
In modern science, John Dalton widely used it and he can be said to be responsible for coining it, but a more factually correct answer would be a Greek philosopher who goes by the name of Lucretius.
A molecule is a group of atoms bonded together which represents the most fundamental unit of a chemical compound capable of taking part in a chemical reaction.
An ion is an atom or a molecule with an electric charge (+ve or -ve), caused by gain or loss of a single or multiple electrons.
Question 9
Give two drawbacks of Dalton’s atomic theory? Question 10
Define Avogadro’s constant. Answer
Avogadro’s Constant (6.022 x 1023) is the number of atoms of an element present in one mole of that element.
Question 11
(a)Explain how atoms exist.
(b)What do you understand by atomicity?
(c) Explain polyatomic ions. Answer
(a) Atoms of most elements don’t exist independently they exist as molecules, for molecules are more stable. However, atoms of inert gasses are chemically unreactive and they exist independently. E.g. helium.
(b) Atomicity is the number of atoms in a molecule. E.g. The atomicity of H2O = 3.
(c) A polyatomic ion is an ion composed of multiple atoms acting as a single charged unit.
Question 12
Explain atomic mass and gram atomic mass. Why does mass have different expressions viz, ‘u’ and ‘gm’? Answer
Atomic mass is the unit in which the mass of an atom is expressed, where one atomic mass unit is 1/12ththe mass of a carbon-12 atom.
Gram atomic mass is the atomic mass of an element expressed in grams.
The mass of an atom or a molecule is expressed in ‘u’, whereas, the molar mass is expressed in ‘gm’.
Question 13
Define a mole. Give the importance of the mole. Answer
One mole of atoms, molecules, or particles is that amount of the particle(atoms, molecules and ions) whose mass is equal to that particle's atomic or molecular mass in grams. 1 mole = 6.022 x 1023 particles of that substance.
Importance of a mole: -
Atoms and molecules are very small. So, it gets bothersome weighing them in grams and trying to count them. Mole concept, however, allows us to count atoms and molecules by weighing macroscopic amounts of materials.
It gives us a universally accepted standard of mass.
It provides a standard for reaction stoichiometry.
Question 14.
H2O is the formula for water. What information do you get from this formula? Answer
H2O represents water
H2O is a single molecule of water
H2O is a single mole of water. Thus, it contains 6.022 x 1023molecules of water.
H2O contains 2 atoms of hydrogen and 1 atom of oxygen.
Question 16
Why are Dalton's symbol not used in chemistry? Answer
Dalton was the first scientist to use the symbol for the name of the elements a specific sense but it was difficult to memorize and in uses so Dalton's symbol are not used in chemistry
Question 17
What is the difference between 2O, O2and O3. Answer
2O represents 2 atoms of oxygen, and it is not possible for it to exist independently.
O2represents an oxygen molecule which has two constituent oxygen atoms.
O3represents a single ozone molecule and it does exist independently.
Calculation Problems
Question 18
How many molecules are present in 1 ml of water? Answer
we know that density of water is 1gm/ml.
Hence, 1 gm water will = 1 ml water.
Now, we have molecular mass of water $H_2O$ = 1x2 + 16 = 18 gm
18 gm of water contain $6.022 \times 10^{23}$ molecules
1 gm of water will contain = $\frac {6.022 \times 10^{23}}{18} = 0.33 \times 10^{23}$ molecules
So, the no. of molecules of water in 1ml of water = $3.3 \times 10^{22}$
Question 19.
Calculate the molecular mass of glucose, C6H12O6. Answer
Molecular mass of C6H12O6= (12 x 6) + 12 + (16 x 6)
= 180u.
Question 20
What is the formula unit mass of CaCl2and NaCl.
(Na = 23, Cl=35.5, Ca=40) Answer
Formula Unit Mass of NaCl = 23 + 35.5 = 58.5u
Formula Unit Mass of CaCl2= 40+(2 x 35.5)= 111u.
Question 21
10.0 g CaCO3 on heating gave 4.4 g of CO2 and 5.6 g of CaO. Show that these observations are in agreement with law of conservation of mass. Answer
Mass of reactants: 10gm
Mass of product = 4.4 + 5.6 = 10g
Since Mass of reactants is equal to Mass of products , therefore these observations are in agreement with law of conservation of mass.
Question 22
Find out
(1) the mass of a single oxygen atom
(2) the mass of a single oxygen molecule
(3) the mass of a mole of oxygen gas
(4) the mass of an oxygen ion
(5) the number of atoms in a mole of an oxygen molecule. Answer
(1) Mass of a single oxygen atom
1 mole of oxygen atom = 16gm =6.022 x 1023atoms
Therefore, Mass of one oxygen atom = 16/6.022 x 1023= 2.65 x 10-23gm
(2) Mass of a single oxygen atom
1 molecule oxygen = O2=2 x 16 = 32u
(3) Mass of a mole of oxygen gas
1 mole of oxygen = O2= 2 x 16 =32u
(4) Mass of an oxygen ion = mass of an oxygen atom (since electrons have negligible mass)
(5) Number of atoms in a mole of oxygen molecule
We know, 1 mole of oxygen molecule, O2= 6.022 x 1023molecules.
1 molecule of O2= 2 atoms
Therefore in a mole of O2, there are =6.022 x 1023x 2 atoms
= 1.022 x 1024atoms.
Question 23
What is the atomicity of oxygen and phosphorous? Answer
Atomicity of oxygen, O2= 2 and atomicity of phosphorous, P4= 4.
Question 24
How many moles of $SO_2$ have same mass as 3 moles of oxygen? Answer
mass of 3 moles of oxygen = 3 x 16 = 48g
Now, mass of $SO_2$ = 32 + 2 x 16 = 64g
As 64g of $SO_2$ = 1 mole
then 48 g of $SO_2$ = (1/64) x 48 =0.75 mole
Summary
This Class 9 Test Paper for Atoms and Molecules with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.
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