Atoms and Molecules – Class 9 Chemistry Important Questions

1. Who gave the law of constant or definite proportion? Also state the law.

Ans.  Joseph.L. Proust gave the law of constant or definite proportion.

The law of constant proportion states that ‘in a chemical compound the elements always combine in a definite proportion by mass irrespective of the source from which it is obtained’.

2. What are the postulates of Dalton’s atomic theory?

Ans.  The postulates of Dalton’s atomic theory are as follows:-

a)All matter is made of very tiny particles called atoms.

b) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.

c) Atoms of a given element are identical in mass and chemical properties.

d) Atoms of different elements have different masses and chemical properties.

e) Atoms combine in the ratio of small whole numbers to form compounds.

f) The relative number and kind of atoms are constant in a given compound.

3.  a) An element Z forms an oxide with formula X {Z}_{2}}{{O}_{3} . What is it’s valency?

b) The valency of an element A is 4. Write the formula of its oxide.

c)An element X has valency 3, while the element Y has valency 2. Write the formula of the compound formed between X and Y.

Ans.: a) Valency of Z is 3.

b) The formula of the oxide of element A is {A}_{2}}{{O}_{4}  or A{{O}_{2} .

c) The formula of the compound is {{X}_{2}}{{Y}_{3}}.

4.  a) Define mole.

b) How many particles are present in one mole of a substance?

c) What is Avogadro’s number?

d) What is molar mass?

Ans: a) A mole is defined as the amount of substance that contains as many particles (atoms, molecules or ions) of that substance as the number of atoms present in exactly 12g of Carbon–12.

b) 6.022\times {{10}^{23}} particles are present in one mole of a substance.

c) The number of particles present in one mole of a substance i.e. 6.022\times {{10}^{23}}  is known as Avogadro’s number/Constant.

d) The mass of one mole substance is known as its molar mass. It is the atomic or molecular mass of that substance expressed in grams.

5. Before selecting C-12 atom for calculating relative atomic masses, oxygen was taken as reference. Why? What was the reason for changing the reference to C-12?

Ans:  Oxygen was selected because of the following two reasons:

a) Oxygen combines with most of the elements to form compounds.

b) By comparing with mass of oxygen taken as 16, the relative atomic masses of most of the elements were found to be whole numbers (not fractions).

Now oxygen is not taken as reference because it was found that naturally occurring oxygen is a mixture of atoms of slightly different masses (called isotopes).

6.  Calculate the number of particles in each of the following:

(i) 46 g of Na atoms (number from mass)

(ii) 8 g   molecules (number of molecules from mass)

(iii) 0.1 mole of carbon atoms (number of atoms from given moles).

Ans: (i) \text{The number of atoms }\,\,\,\,\,\,\,\,\,\,=~\frac{\text{given mass}}{\text{molarmass}}\text{ }\times \text{ Avogadro number}

\Rightarrow \,\,N=\frac{m}{M}\times {{N}_{o}}

\Rightarrow \,N=\frac{46}{23}\,\times 6.022\times {{10}^{23}}

\Rightarrow \,\,N=12.044\times {{10}^{23}}

ii) \text{The number of molecules }=~\frac{\text{given mass}}{\text{molarmass}}\text{ }\times \text{ Avogadro number}

\Rightarrow \,\,N=\frac{m}{M}\times {{N}_{o}}

atomic mass of oxygen = 16 u

\therefore  molar mass of  {{O}_{2} molecules =16\times 2=32\,\,g

\Rightarrow \,N=\frac{8}{32}\times 6.022\times {{10}^{23}}

\Rightarrow \,\,N=1.5055\times {{10}^{23}}

(iii) The number of particles (atom) = number of moles of particles × Avogadro number

N=n\times {{N}_{o}}

=0.1\times 6.022\times {{10}^{23}}=6.022\times {{10}^{22}}

7. Define (a.m.u.) atomic mass unit.

Ans: Atomic mass unit or a.m.u. is defined as 1/12th of the mass of one atom of Carbon-12 atom.

1 amu = \frac{1}{12}\text{th of mass of C-12 atom}\text{.}

8. What mass of N{{H}_{3}} will contain same number of molecules as that in 36 grams of {{H}_{2}}O ? 

Ans: Molar mass of {{H}_{2}}O=18g

No. of molecules in 18\,\,g\,\,{{H}_{2}}O=6.022\times {{10}^{23}}

Therefore number of molecules in 36 grams of H2O = \frac{6.022\times {{10}^{23}}}{18}\times 36

12.044 \times 10^{23}

Molar mass of N{{H}_{3}}=17g 

6.022\times {{10}^{23}} molecules of  N{{H}_{3}}  have mass = 17 g

Therefore 12.044\times {{10}^{23}}  molecules of NH3 have mass =\frac{17}{6.022\times {{10}^{23}}}\times 12.044\times {{10}^{23}}

=17\times 2=34\,g

9. Calculate number of atoms in the following :-

(atomic mas of He = 4u)

i) 16 moles of He   (ii) 16 u of He    (iii) 16 g of He

Ans.:(i) Number of atoms in 1mole of He =6.022\times {{10}^{23}}atoms

Number of atoms in 16 moles of He = 6.022\times {{10}^{23}}\times 16=96.352\times {{10}^{23}}atoms

ii) Number of atoms in 4u of He =1atom

Number of atoms in 16u of He = \frac{1}{4}\times 16=4\,\,atoms of He

iii) 16g of He No.of\,Moles=\frac{16}{4}=4\,moles

Number of atoms in 1 mole of He =6.022\times {{10}^{23}}atoms

Number of atoms in 4 moles of He =6.022\times {{10}^{23}}\times 4=24.088\times {{10}^{23}}atoms

10. Mass of {{10}^{22}}  an element ‘X’ is found to be 930 mg. Calculate the molar mass of X.

Ans.:    930 mg = 0.930 g

Mass of {{10}^{22}}  atoms of  ‘X’ = 0.930 g

Mass\text{ }of6.022\times {{10}^{23}}atoms\,\,of\text{X}=\frac{0.930}{{{10}^{22}}}\times 6.022\times {{10}^{23}}=56g

Molar mass of X= 56 g           

11. What is the mass of 4 moles of aluminium atoms? (Atomic mass of Al = 27u)

Ans: The atomic mass of aluminium is given to be 27u. This means that I mole of aluminium atoms has a mass of 27 grams.

Now, 1 mole of aluminium atoms = 27 g

So, 4 moles of aluminium atoms =27\times 4\,g=108\,g

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

12. Write the molecular formulae for the following compounds

            (a) Copper (II) bromide                       (b) Aluminium (III) nitrate

            (c) Calcium (II) phosphate                  (d) Iron (III) sulphide

            (e) Mercury (II) chloride                     (f) Magnesium (II) acetate

Ans.     (a) CuB{{r}_{2}}

(b) Al{{(N{{O}_{3}})}_{3}}

(c) C{{a}_{3}}{{(P{{O}_{4}})}_{2}}

(d) F{{e}_{2}}{{S}_{3}}

(e) HgC{{l}_{2}}

(f) Mg{{(C{{H}_{3}}COO)}_{2}}

13. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur

Ans: Mass of 1 mole of Sulphur ({{S}_{8}})=8\times 32\,g=256\,g also as 1 mole =6.022\times {{10}^{23}} number of particles of that substance

So, 256 g of sulphur contains  =6.022\times {{10}^{23}} molecules of sulphur

Therefore 16 g of sulphur will contain    = \frac{6.022\times {{10}^{23}}}{256}\times 16

=3.76\times {{10}^{22}}  molecules (approx)

14. Calculate number of atoms in the following :-

(atomic mass of He = 4u)

(i) 16 moles of He       (ii) 16 u of He              (iii) 16 g of He

Ans.:  (i) Number of atoms in 1mole of He =6.022\times {{10}^{23}}atoms

Number of atoms in 16 moles of He=6.022\times {{10}^{23}}\times 16=96.352\times {{10}^{23}}atoms

(ii) Number of atoms in 4u of He  = 1atom

Number of atoms in 16u of He=  \frac{1}{4}\times 16=4\,\,atoms of He

(iii) 16g of He ~~~~~~~~~~~~~~~~~No.of\,Moles=\frac{16}{4}=4\,moles

Number of atoms in 1 mole of He =6.022\times {{10}^{23}}atoms

Number of atoms in 4 moles of He = =6.022\times {{10}^{23}}\times 4=24.088\times {{10}^{23}}atoms

15. When 3.0 g of carbon is burnt in 8.00 g 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 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Ans.:

\underset{\begin{smallmatrix} \\ \text{3}\text{.0g} \end{smallmatrix}}{\mathop{\text{Carbon}\,}}\,\,+\,\,\underset{\begin{smallmatrix} \\ \text{8}\text{.0g} \end{smallmatrix}}{\mathop{\text{Oxygen}}}\,\xrightarrow{{}}\underset{\begin{smallmatrix} \\ \text{11}\text{.0g} \end{smallmatrix}}{\mathop{\text{Carbon}\,\,\text{dioxide}}}

Since 3.0 g carbon reacts with only 8.0 g of oxygen, 42g of oxygen will remain unreacted and only 8.0 g will react and form 11.0 g of carbon dioxide.

\underset{\begin{smallmatrix} \\ \text{3}\text{.0g} \end{smallmatrix}}{\mathop{\text{Carbon}\,}}\,\,+\,\,\underset{\begin{smallmatrix} \\ \text{50}\text{.0g} \end{smallmatrix}}{\mathop{\text{Oxygen}}}\,\xrightarrow{{}}\underset{\begin{smallmatrix} \\ \text{11}\text{.0g} \end{smallmatrix}}{\mathop{\text{Carbon}\,\,\text{dioxide}}}\,\,+\underset{\begin{smallmatrix} \\ \text{42}\text{.0g} \end{smallmatrix}}{\mathop{\text{Oxygen}}}\,

The answer is governed by:

Law of constant proportion i.e. in carbon dioxide carbon and oxygen combine in a fixed ratio by mass (i.e. 3:8).

Leave a Reply

Your email address will not be published. Required fields are marked *