## Topic outline

• ### Chapter 3 - Atoms And Molecules – 3

• Atoms And Molecules

Or

What is the relationship between a mole and Avogadro’s Number?

Ans. The number of particles (atoms, molecules or ions) present in 1 mole of any substance is fixed, with a value of 6.022 × 1023. This is an experimentally obtained value. This number is called the Avogadro Constant or Avogadro Number (represented by N0), named in honour of the Italian scientist, Amedeo Avogadro.

1 mole (of anything) = 6.022 × 1023 in number

Q58. What is an ion? Explain with the help of example.

Ans. Compounds composed of metals and nonmetals contain charged species. The charged species are known as ions. An ion is a charged particle and can be negatively or positively charged. A negatively charged ion is called an ‘anion’ and the positively charged ion, a ‘cation’. Take, for example, sodium chloride (NaCl). Its constituent particles are positively charged sodium ions (Na+) and negatively charged chloride ions (Cl–).

Q59. 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.

Ans. Mass of sample compound = 0.24 g

Mass of boron = 0.096 g

Mass of oxygen = 0.144 g

Percentage of boron = mass of boron/mass of sample compound x 100

= 0.096g/0.24g x 100

= 40%

Percentage of oxygen = mass of oxygen/mass of sample compound x 100

= 0.144g/0.24g x 100

= 60%

Q60. A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

Ans. 1 g of gold sample contains = 90/100 = 0.9g gold

Number of moles of gold = Mass of gold/Atomic mass of gold

= 0.9/197

= 0.0046

1 mole of gold contains = 6.022 x 1023

0.0046 mole of gold = 0.0046 x 6.022 x 1023

= 2.77 x 1021

Q61. Express each of the following in kilograms

(a) 5.84×10-3 mg

(b) 58.34 g

(c) 0.584g

(d) 5.873×10-21g

Ans. (a) 106mg = 1 kg

5.84×10-3 mg = 5.84×10-3 mg/106

= 5.84×10-9 kg

(b) 103g = 1 kg

58.34 g = 58.34 g/103

= 5.834x10-2 kg

(c) 103g = 1 kg

0.584 g = 0.584 g/103

= 5.84 x 10-4 kg

(d) 103g = 1 kg

5.873×10-21g = 5.873×10-21g /103

= 5.873×10-24kg

Q62. The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.

Ans. Number of electrons in Na atom = 11

Number of electrons in Na+ = 10

For 1 mole of Na atom and Na+ the difference in electrons = 1 mole

For 100 moles of Na atoms and Na ions the difference = 100 moles of electrons

Mass of 100 moles of electrons = 0.0548002 g

Mass of 1 mole of electron = 0.0548002g/100

Mass of 1 electron = 0.0548002g/(100x6.022 1023)

= 9.1 x 10-28g

Q63. 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) Copper (II) bromide – CuBr2

(b) Aluminium (III) nitrate – A1(NO3)3

(c) Calciuim (II) phosphate – Ca3(PO4)2

(d) Iron (III) sulphide – Fe2S3

(e) Mercury (II) chloride – HgCl2

(f) Magnesium (II) acetate – Mg(CH3COO)2

Q64. Give the formulae of the compounds formed from the following sets of elements

(a) Calcium and fluorine

(b) Hydrogen and sulphur

(c) Nitrogen and hydrogen

(d) Carbon and chlorine

(e) Sodium and oxygen

(f) Carbon and oxygen

Ans. (a) Calcium and fluorine – CaF2

(b) Hydrogen and Sulphur – H2S

(c) Nitrogen and hydrogen – NH3

(d) Carbon and chlorine – CCl4

(e) Sodium and oxygen – Na2O

(f) Carbon and oxygen – CO2

Q65. (a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called ————.

(b) A group of atoms carrying a fixed charge on them is called ————.

(c) The formula unit mass of Ca3 (PO4)2 is ————.

(d) Formula of sodium carbonate is ———— and that of ammonium sulphate is ————.

Ans. (a) Law of conservation of mass

(b) Polyatomic ion

(c) (3xCa)+(2xP)+(8xO)

=(3x40)+(2x31)+(8x16)

=(120+62+128)

=310u

(d) Na2CO3, (NH4)2SO4

Q66. Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them. (You may use appendix-III).

(a) Ammonia

(b) Carbon monoxide

(c) Hydrogen chloride

(d) Aluminium fluoride

(e) Magnesium sulphide

Ans.

 Compounds Chemical formula Ratio by mass (a) Ammonia NH3 N:H = 13:3 (b) Carbon monoxide CO C:O = 12:16 = 3:4 (c) Hydrogen chloride HCl H:Cl = 1:35.5 (d) Aluminium fluoride AlF3 Al:F = 27:57 = 9:19 (e) Magnesium sulphide MgS Mg:S = 24:32 = 3:4

Q67. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Ans. Molecular mass of sulphur (S8) = 8 x atomic mass of sulphur

= 8 x 32 = 256 g

Thus, 16 g of sulphur will contain = 1/256x16

= 0.0625 moles of Sulphur molecules

But, 1 mole of Sulphur contains = 6.022 x 1023 molecules

So, 0.0625 moles of Sulphur contains = 6.022 x 1023 x 0.0625

= 0.376375 x 1023

= 03.76375 x 1022

Q68. Which of the following correctly represents 360 g of water?

(i) 2 moles of H20

(ii) 20 moles of water

(iii) 6.022 × 1023 molecules of water

(iv) 1.2044×1025 molecules of water

(a) (i) (b) (i) and (iv)

(c) (ii) and (iii) (d) (ii) and (iv)

Ans. (d) (ii) and (iv)

Explanation:

1 mole of water = molar mass of water = 18g

(i) 2 mole of water = 18 x 2 = 36g

(ii) 20 moles of water = 18g x 20 = 360g

(iii) 6.022 × 1023 molecules of water = 1 mole = 18g

(iv) 6.022 × 1023 molecules of water = 1 mole = 18g

1.2044×1025 molecules of water = (1.2044×1025 x 18g)/ 6.022 × 1023

= 360g

Q69. What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Ans. (a) 1 mole of oxygen atoms = Molecular mass of O in grams

= 16 g

Mass of 1 mole of oxygen atoms = 16 g

Mass of 0.2 mole of oxygen atoms = 16 x 0.2 g = 3.2 g

(b) 1 mole of water (H2O) = Molecular mass of H2O in grams

= (2 x 1) + 16

= 18 g

Mass of 1 mole of water = 18 g

Mass of 0.5 mole of water = 18 x 0.5 g = 9 g

Q70. What is the mass of—

(a) 1 mole of nitrogen atoms?

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

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

Ans. 1 mole of atoms = Gram atomic mass

(a) Atomic mass of nitrogen = 14 u

Therefore, 1 mole of nitrogen atoms = 14g

(b) Atomic mass of aluminium = 27 u

Therefore, 4 moles of aluminium atoms = 4 x 27 = 108 g

(c) Atomic mass of sodium = 23 u

Atomic mass of sulphur = 32 u

Atomic mass of oxygen = 16 u

Therefore, 10 moles of sodium sulphite (Na2SO3

= 10[(2x23)+32+(3x16)]

= 1260 g