# ATOMS AND MOLECULES : NCERT Exercise Questions

Q.1     A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g if boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Sol.     Mass of boron = 0.096g(Given)
Mass of oxygen = 0.144g (Given)
Mass of sample = 0.24g (Given)
Thus, percentage of boron by weight in the compound = ${{0.096} \over {0.24}} \times 100\%$
= 40%
Thus, percentage of oxygen by weight in the compound = ${{0.144} \over {0.24}} \times 100\%$
= 60 %

Q.2     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 combinations will govern your answer ?
Sol.      Carbon + Oxygen $\to$ Carbon dioxide 3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g of oxygen, then 3g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11g of carbon dioxide will be formed.  The above answer is governed by the law of constant proportions.

Q.3     What are polyatomic ions? Give examples?
Sol.     A polyatomic ion is a group of atoms carrying a charge (positive or negative).For example, ammonium ion $\left({NH_4^ + } \right)$, hydroxide ion $\left( {O{H^ - }} \right)$, carbonate ion $\left( {CO_3^{2 - }} \right)$,sulphateion $\left( {SO_4^{2 - }} \right)$.

Q.4      Write the chemical formula of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Sol.      (a) Magnesium chloride $\to MgC{l_2}$
(b) Calcium oxide $\to CaO$
(c) Copper nitrate $\to Cu{\left( {N{O_3}} \right)_2}$
(d) Aluminium chloride $\to AlC{l_3}$
(e) Calcium carbonate $\to CaC{O_3}$

Q.5     Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Sol. .

 Compound Chemical formula Elements present Quick lime Cao Calcium,oxygen Hydrogen bromide HBr Hydrogen,bromine Baking powder NaHCO3 Sodium,hydrogen, carbon, oxygen Potassium sulphate k2SO4 Potassium,sulphur,oxygen

Q.6     Calculate the molar mass of the following substances:
(a) Ethyne, ${C_2}{H_2}$
(b) Sulphur molecule, ${S_8}$
(c) Phosphorus molecule, ${P_4}$ (atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, $HN{O_3}$
Sol.     (a) Molar mass of ethyne, ${C_2}{H_2}$ = 2 × 12 + 2 × 1 = 28g
(b) Molar mass of sulphur molecule, ${S_8}$ = 8 × 32 = 256g
(c) Molar mass of phosphorus molecule,${P_4}$ = 4 × 31 = 124g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g
(e) Molar mass of nitric acid, $HN{O_3}$ = 1 + 14 + 3 × 16 = 63g

Q.7     What is the mass of --
(a) 1 mole of nitrogen atoms?
(b) 4 mole of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite $\left( {N{a_2}S{O_3}} \right)$ ?
Sol.     (a) The mass of 1 mole of nitrogen atoms is 14g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g
(c) The mass of 10 moles of sodium sulphite $\left( {N{a_2}S{O_3}} \right)$ is
10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g = 1260g

Q.8     Convert into mole.
(a) 12g of oxygen gas
(b) 12g of water
(c) 22g of carbon dioxide
Sol.     (a) 32 g of oxygen gas = 1 mole
Then, 12g of oxygen gas = ${{12} \over {32}}$ mole = 0.375 mole
(b) 18g of water = 1 mole
Then, 20 g of water = ${{20} \over {18}}$ mole = 1.11 moles (approx)
(c) 44g of carbon dioxide = 1 mole
Then, 22g of carbon dioxide = ${{22} \over {44}}$ mole = 0.5 mole

Q.9     What is the mass of :
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Sol.     (a) Mass of one mole of oxygen atoms = 16g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g
(b) Mass of one mole of water molecule = 18g
Then, mass of 0.5 mole of water molecules = 0.5 × 18g = 9g

Q.10     Calculate the number of molecules of sulphur $\left( {{S_8}} \right)$ present in 16g of solid sulphur.
Sol.        1 mole of solid sulphur $\left( {{S_8}} \right)$ = 8 × 32g = 256g
i.e., 256g of solid sulphur contains = 6.022 × ${10^{23}}$ molecules
Then, 16g of solid sulpur contains ${{6.022 \times {{10}^{23}}} \over {256}} \times 16$ molecules
= 3.76 × ${10^{22}}$ molecules (approx)

Q.11     Calculate the number of aluminium ions present in 0.051g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)
Sol.        1 mole of aluminium oxide $\left( {A{l_2}{O_3}} \right)$ = 2 × 27 + 3 × 16 = 102g
i.e., 102g of ${A{l_2}{O_3}}$ = 6.022 × ${10^{23}}$ molecules of $A{l_2}{O_3}$
Then, 0.051 g of  ${A{l_2}{O_3}}$ contains = ${{6.022 \times {{10}^{23}}} \over {102}} \times 0.051$ molecules
= 3.011 ×  ${10^{20}}$  molecules of  ${A{l_2}{O_3}}$
The number of aluminium ions $\left( {A{l^{3 + }}} \right)$ present in one  molecules of aluminium oxide is 2.
Therefore, The number of aluminium ions  $\left( {A{l^{3 + }}} \right)$ present in
3.11 × ${10^{20}}$ molecules (0.051g) of aluminium oxide $\left( {A{l_2}{O_3}} \right)$ = 2 × 3.011 × ${10^{20}}$
= 6.022 × ${10^{20}}$

• Anonymous

• Anonymous

This website is awesome
Keep it up DRONSTUĎÝ

• This website is very nice and it is very helpful

• khushi

• Saba khan

Super

• Saba khan

• Saba khan

It is very helpful for me

• super

• thanks

nice

• thanks

thanks

• Anushka

wow it's wonderful

• Jahnavi

Awesome

• Jahnavi

Better

• Anonymous

Super

• Tanishq

I understood all the question

• Tanishq

Nice website

• Thank u should muxx

Very useful in lock down for completion of notes.☺

• good lectures

• sakhawat ali malik.

i think it is easy .

• sakhawat ali

• Feriha khan

brilliant

• Kiruthika

Easy to get marks

• Very good website keep it u

Very good keep it up

• I was confused over last ques but DroneStudy helped me to understand clearly, step by step.... So Thank You DroneStudy

 Get Courses by IITian KOTA teachers at BEST Price ApplyNOW