Class 09 : Atoms Molecules : NCERT Exercise Solution

 

Question 1: A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Answer: Percentage of boron in given compound $=\left(\frac{0.096}{0.24}\right)\times 100=40\%$
Percentage of oxygen in given compound $=\left(\frac{0.144}{0.24}\right)\times 100=60\%$

Question 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 combination will govern your answer?

Answer: We know that atoms of different elements combine in definite proportions to make a particular compound. Hence, 3 g of carbon will always give 11 g of carbon dioxide.

Question 3: What are polyatomic ions? Give examples.

Answer: An ion with more than one atom is called a polyatomic ion. Examples: Sulphate (SO42-), Nitrate (NO32-)

Question 4: Write the chemical formulae of the following.

(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

Answer: (a) MgCl2, (b) CaO, (c) CuNO3, (d) AlCl3, (e) CaCO3

Question 5: Give the names of the elements present in the following compounds.

(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

Answer: (a) Calcium, oxygen, (b) Hydrogen, bromine, (c) sodium, hydrogen, carbon, oxygen, (d) Potassium, sulphur, oxygen

Question 6: Calculate the molar mass of the following substances.

(a) Ethyne, C2H2

Answer: $2\times 12+2\times 1=26$ u

(b) Sulphur molecule, S8

Answer: $8\times 32=256$ u

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

Answer: $4\times 31=124$

(d) Hydrochloric acid, HCl

Answer: $1\times 1+1\times 35.5=36.5$ u

(e) Nitric acid, HNO3

Answer: $1\times 1+1\times 14+3\times 16=63$ u

Question 7: What is the mass of:

(a) 1 mole of nitrogen atoms?

Answer: Atomic mass of nitrogen = 14 u
Hence, mass of 1 mole of nitrogen atoms = 14 g

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27) ?

Answer: Atomic mass of aluminium = 27 u
Hence, mass of 4 moles of aluminium atoms $=4\times 27=108$

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

Answer: Molecular mass of sodium sulphite
$=2\times 23+1\times 32+3\times 16=46+32+48=126u$
Hence, mass of 10 mole of sodium sulphite $=10\times 126=1260g$

Question 8: Convert into mole.

(a) 12 g of oxygen gas

Answer: Mass of 1 mole of oxygen = 32 g
Hence, number of mole in 12 g oxygen $=\frac{12}{32}=0.375M$

(b) 20 g of water

Answer: Molecular mass of water $=2\times 1+1\times 16=18u$
So, mass of 1 mole of water = 18 g
Hence, number of mole in 20 g of water $=\frac{20}{18}=1.11M$

(c) 22 g of carbon dioxide.

Answer: Molecular mass of carbon dioxide $=1\times 12+2\times 16=44u$
So, mass of 1 mole of carbon dioxide = 44 g
Hence, number of mole in 22 g of carbon dioxide $=\frac{22}{44}=0.5M$

Question 9: What is the mass of

(a) 0.2 mole of oxygen atoms?

Answer: Mass of 1 mole oxygen atoms = 16 g
Hence, mass of 0.2 mole of oxygen atoms $=0.2\times 16=3.2g$

(b) 0.5 mole of water molecules?

Answer: Mass of 1 mole of water = 18 g
Hence, mass of 0.5 mole of water $=0.5\times 18=9g$

Quesiton 10: Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Answer: Mass of 1 mole of sulphur $=8\times 32=256g$
Hence, number of mole in 16 g of sulphur $=\frac{16}{256}=0.625M$

Question 11: Calculate the number of aluminium ions present in 0.051 g 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 = 27 u)

Answer: Formula of aluminium oxide = Al2O3
Mass of aluminium atoms in aluminium oxide $=2\times 27=54u$
Molecular mass of aluminium oxide $=2\times 27+3\times 16=54+48=102u$

Here; 2 mole aluminium ions are present in 102 g aluminium oxide
Hence, number of aluminium ions in 0.051 g aluminium oxide
$=\left(\frac{2}{102}\right)\times 0.051=\frac{2}{2000}$ mole
$=\frac{6.022\times {10}^{23}}{1000}=6.022\times {10}^{20}$