Atoms And Molecules

Q81. Convert into mole.

(a) 12 g of oxygen gas

(b) 20 g of water

(c) 22 g of carbon dioxide. 

Ans. (a) Atomic mass of oxygen (O) = 16u

1 mole of O₂ = Molecular mass of O₂ in grams

                     = 2 x 16 g = 32 g

So, the mass of 1 mole of oxygen gas is 32 g

Now, 32 g of oxygen gas = 1 mole

Therefore, 12 g of oxygen gas = 1/32 x 12

                                    = 0.375 mole

Thus, 12 grams of oxygen gas are equal to 0.375 mole.

(b) Atomic mass of hydrogen (H) = 1u

Atomic mass of oxygen (O) = 16 u

1 mole of H₂O = Molecular mass of H₂O in grams

                     = 2 x Mass of ‘H’ + Mass ‘O’

                     = (2 x 1) + 16

                     = 18 g

So, the mass of 1 mole of water is 18 g

Now, 18 g of water = 1 mole

Therefore, 20 g of water = 1/18 x 20

                                    = 1.11 mole

Thus, 20 grams of water are equal to 1.11 mole.

(c) Atomic mass of carbon (C) = 12u

Atomic mass of oxygen (O) = 16 u

1 mole of CO₂ = Molecular mass of CO₂ in grams

                     = Mass of ‘C’ + 2 x Mass ‘O’

                     = 12 + (2 x 16) 

                     = 44 g

So, the mass of 1 mole of carbon dioxide is 44 g

Now, 44 g of carbon dioxide = 1 mole

Therefore, 22 g of carbon dioxide = 1/44 x 22

                                                  = 1/2

                                                  = 0.5 mole

Thus, 22 grams of carbon dioxide are equal to 0.5 mole.

Q82. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?

Ans. To calculate the number of atoms in 100 grams of sodium:

23g of Na contains = 6.022 x 10²³

1g of Na contains = 6.022 x 10²³/23

100g of Na contains = 6.022 x 10²³ x 100/23

                             = 2.618 x 10²⁴ atoms

To calculate the number of atoms in 100 grams of iron:

56g of Fe contains = 6.022 x 10²³

1g of Fe contains = 6.022 x 10²³/56

100g of Fe contains = 6.022 x 10²³ x 100/56

                             = 1.075 x 10²⁴ atoms

Q83. Which of the following statements is not true about an atom?

(a) Atoms are not able to exist independently

(b) Atoms are the basic units from which molecules and ions are formed

(c) Atoms are always neutral in nature

(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch

Ans. (d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch. This is statement is not true.

Explanation:

Atoms aggregate in large numbers to form the matter. But we cannot see the matter with naked eyes.

Q84. Which of the following would weigh the highest?

(a) 0.2 mole of sucrose (C₁₂ H₂₂ O₁₁)

(b) 2 moles of CO₂

(c) 2 moles of CaCO₃

(d) 10 moles of H₂O

Ans. (c) 2 moles of CaCO₃ would weigh the highest.

Explanation:

(a)Mass of 1 mole of sucrose (12 x12) + (1 x22) + (16×11)= 342g

Mass of 0.2 mole of sucrose = 342 x 0.2 = 68.4g

(b)Mass of 1 mole of CO₂ = 12 + (16 x 2) = 44g  

Mass of 2 moles of CO₂ = 44 x 2 = 88g

(c)Mass of 1 mole of CaCO₃ = 40 + 12 + (16 x 3) = 100g

Mass of 2 moles of CaCO₃ = 100 x 2 = 200g

(d)Mass of 1 mole of H₂0 = 2 + 16 = 18g

Mass of 10 moles of H₂0 = 18 x 10 = 180g

Q85. Which of the following has maximum number of atoms?

(a) 18g of H₂O

(b) 18g of O₂

(c) 18g of CO₂

(d) 18g of CH₄

Ans. (d) 18g of CH₄

Explanation:

Number of atoms 

= (mass substance x no. of atoms x 6.022 x 10²³)/molar mass

(a) Number of atoms in 18g of H₂O

= (18 x 6.022 x 10²³ x 3)/18 

= 18.066 x 10²³

= 1.8066 x 10²⁴

(b) Number of atoms in 18g of O₂

= (18 x 6.022 x 10²³ x 2)/32 

= 6.774 x 10²³

(c) Number of atoms in 18g of CO₂

= (18 x 6.022 x 10²³ x 3)/44 

= 7.390 x 10²³

(d) Number of atoms in 18g of CH₄

= (18 x 6.022 x 10²³ x 5)/16 

= 3.387 x 10²⁴

Q86. Which of the following contains maximum number of molecules?

(a) 1g CO₂

(b) 1g N₂

(c) 1g H₂

(d) 1g CH₄

Ans. (c) 1g H₂

Explanation:

(a) Molar mass of CO₂ = 44g

Number of molecules in 44g of CO₂ = 6.022 x 10²³

Number of molecules in 1g of CO₂ = (6.022 x 10²³)/44

                                                   = 1.37 x 10²²

(b) Molar mass of N₂ = 28g

Number of molecules in 28g of N₂ = 6.022 x 10²³

Number of molecules in 1g of N₂ = (6.022 x 10²³)/28

                                                   = 2.15 x 10²²

(c) Molar mass of H₂ = 2g

Number of molecules in 2g of H₂ = 6.022 x 10²³

Number of molecules in 1g of H₂ = (6.022 x 10²³)/2

                                                   = 3.011 x 10²³

(d) Molar mass of CH₄ = 16g

Number of molecules in 16g of CH₄ = 6.022 x 10²³

Number of molecules in 1g of CH₄ = (6.022 x 10²³)/16

                                                 = 3.76 x 10²²

Q87. 3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are

(a) 6.68 × 10²³

(b) 6.09 × 10²²

(c) 6.022 × 10²³

(d) 6.022 × 10²¹

Ans. (a) 6.68 × 10²³

Explanation:

Molar mass of sucrose C₁₂H₂₂C₁₁ = 12x12+1x22+16x11 =342g

342g = 1 mole of sucrose

3.42g = 0.01 mole of sucrose

1 mole of sucrose contains, O atoms = 11 x 6.022 x 10²³

0.01 mole of sucrose contains, O atoms = 0.01 x 11 x 6.022 x 10²³

                                                           = 6.6242 x 10²²

18g of H₂O = 1 mole of water

1 mole of water contains, O atoms = 6.022 x 10²³

Adding number of O atoms present in 3.42g of sucrose and 18g of water, we get, (6.6242 x 10²² + 6.022 x 10²³)

= 10²² (6.6242 + 60.22)

= 66.844 x 10²²

= 6.68 x 10²³

Q88. Verify by calculating that

(a) 5 moles of CO₂ and 5 moles of H₂O do not have the same mass.

(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3:5.

Ans. (a) Molar mass of CO₂ = 12 + 2 x 16 = 44g

1 mole of CO₂ has a mass = 44g

5 moles of CO₂ have a mass = 44 x 5 = 220g

Molar mass of H₂O = 2 x 1 + 16 = 18g

1 mole of H₂O has a mass = 18g

5 mole of H₂O have a mass = 5 x 18g = 90g

Thus, verified that 5 moles of CO₂ and 5 moles of H₂O do not have the same mass.

(b) Molar mass of calcium = 40g

40 g of calcium = 1 mole

240 g of calcium = 1/40 x 240 = 6 mol

Molar mass of magnesium = 24 g

24 g of magnesium = 1 mole

240 g of magnesium = 1/24 x 240 = 10 mol

Ratio = 6:10 = 3:5

Q89. Find the ratio by mass of the combining elements in the following compounds. 

(a) CaCO₃ (d) C₂H₅OH (b) MgCl₂ (e) NH₃ (c) H₂SO₄ (f) Ca(OH)₂ 

Ans. (a) CaCO₃  

=Ca(1) : C(1) : O(3)

= 40:12:16x3

= 40:12:48

= 10:3:12

(b) MgCl₂  

=Mg(1) : Cl(2) 

= 24:35.5x2

= 24:71

(c) H₂SO₄

=H(2) : S(1) : O(4)

= 2x1:32:16x4

= 2:32:64

= 1:16:32

(d) C₂H₅OH

=C(2) : H(6) : O(1) 

= 12x2:1x6:16x1

= 24:6:16

= 12:3:8

(e) NH₃

=N(1) : H(3) 

= 14:1x3

= 14:3

(f) Ca(OH)₂

=Ca(1) : O(2) : H(2) 

= 40:16x2:1x2

= 40:32:2

= 20:16:1