CBSE Class 11 - Chemistry - CH1 - Mole Concept

Q1: Define mole

Answer: A mole (or mol) is defined as the amount of substance which contains equal number of  particles (atoms / molecules / ions) as there are atoms in exactly 12.000g of carbon-12.

One mole of carbon-12 atom has a mass of exactly 12.000 grams and contains 6.02 × 10²³ atoms.

A mol is just a number like a dozen. A dozen equals to 12 eggs , a gross of Pencil equals to 144 Pencil. Similarly, mol is equal to 6.022 × 10²³(Avagadro constant). Mol is also known as chemist dozen.




The value 6.022 × 10²³ is known as Avogadro Constant (N_A), after the Italian scientist who first recognized the importance of the mass/number relationship

Q2: Why was there need to use this number called mole?

Answer: Atoms and molecules are extremely small in size and their numbers in even a small amount
of any substance is really very large. To handle such large numbers, a unit of similar magnitude is required.

In SI system, mole (symbol, mol) was introduced as seventh base quantity for the amount of a substance.

Q3: 1 mol of chlorine atom contains
(a) 6.022 × 10²³ atoms
(b) one  atom
(c) 35.5 g of Cl
(d) All of the above.

Answer: Both (a) and (b) are correct. 1 mol of Cl atom = 36.5 g of Cl (molar mass)
   = 6.02 × 10²³ atoms.

Q4: 1.0 mole of Chlorine molecule (Cl₂) contains
(i) how many number of molecules.
(ii) how many number of atoms.
(iii) how much it weighs.

Answer: (i) 1.0 mol of Cl₂ contains 6.022 × 10²³molecules.
(ii)  One molecule of Cl₂ contains 2 toms of Cl. ∴ Cl₂ contains 2 × 6.022 × 10²³ atoms
       i.e. 12.44 × 10²³ atoms

(iii) Molar mass of Cl is 35.5 gm/mol. 1.0 mol of Cl₂ weighs = 2 × 35.5 = 71.1 g

Mole in terms of number,

1 Mole of particle = 6.022 × 10²³ particles

Q5: Which of the following is correct option?
1.0 mole of NH₃ (ammonia) contains ...
(a) 6.022 × 10²³ molecules
(b) 4 mol of atoms
(c) 1 mol of Nitrogen atoms
(d) 3 × 6.022 × 10²³ of H atoms

Answer: All of the above options are correct.

Q6: What is atomic mass unit (amu)?

Answer: Atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. One atomic mass unit also called one Dalton.

Mass of one mole of C-12 atoms = 12 g = mass of 6.022 × 10²³ C-12 atoms

1 amu = 1g per mol = 1/N_A = 1/ (6.022 × 10²³) = 1.66 × 10^-24g.

Q7: What is the mass of one ¹²C atom? Express it in grams as well as in amu.

Answer: one mole of C-12 atoms = 12 g = mass of 6.022 × 10²³ C-12 atoms
∴ mass of 1 C-12 atom =  12g ÷ 6.022 × 10²³ = 1.994 × 10^-23g
or   = 12g ÷ 6.022 × 10²³  = 12g × 1.66 × 10^-24g = 12 amu.


Q8: What is molar mass?

Answer: The mass of one mole of an element or one mole of compound is referred  as molar mass.
It is expressed as g mol^-1.

Example:
    molar mass of Mg  = 24 g mol^-1.
    molar mass of methane (CH₄) = (12 + 4) g mol^-1 = 16 g mol^-1.

Q9: What is Gram atomic mass or molar mass of an element?

Answer:  Gram atomic mass or molar mass of an element is mass of 1 mol of atoms or atomic mass expressed in grams. For example, atomic mass of Mg = 24u, therefore, molar mass of Mg is 24 grams per mol. Molar mass of an element is also called one gram atom.


Q10: What is Gram molecular mass or molar mass of molecular substance?

Answer: Gram molecular mass or molar mass of a molecular substance is the mass of 1 mol of molecules or molecular mass expressed is grams. For example, molecular mass of H₂O is 18u (2u + 16u), therefore, molar mass of H₂O is 18 g mol^-1. 

Mole in terms of mass,

No. of Moles = Mass ÷ Molar mass 

Q11: In 14g of N₂(nitrogen gas), calculate 
(i) number of moles (take molar mass of 28 g mol^-1)
(ii) number of molecules
(iii) number of atoms

Answer:
(i) No. of moles = 14g ÷ 28 g mol^-1 = 0.5 moles
(ii) No. of molecules = (number of moles) × N_A = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules
(iii) One molecule of N₂ contains 2 N atoms.
 No. of atoms  = 2 × 3.011 × 10²³ = 6.022 × 10²³ atoms = 1 N_A atoms.

Q12: One gram molecule H₂O contains how many moles of molecules and number of molecules?

Answer:  1 gram molecule H₂O = 1 mol of molecule H₂O = 6.022 × 10²³ molecules of  H₂O

Q13: How many moles of molecule are present in 15g of C₂H₄ (ethylene). Given atomic mass of C = 12.0 amu and of H = 1.0 amu.

Answer: Molecular mass of C₂H₄ = 2 × 12.0 amu + 4 × 1.0 amu = 24 + 4 = 28amu
∴ Molar mass of C₂H₄ = 28 g mol^-1
No. of moles of molecule C₂H₄ = mass ÷molar mass = 15 ÷ 28 = 0.536 mol.

Q14: Calculate number of bromide ions in 3 moles of  mercury(II) bromide.

Answer:  Mercury(II) bromide (HgBr₂) in ionic form in an aqueous solution is
       HgBr₂ →  Hg⁺² + 2Br^-1
 ∴ 1 molecule of HgBr₂ contains 2 Br^-1 ions.
 ⇒ 1 mol of HgBr₂ contains 2 mol Br^-1 ions.
  ∴ 3 molecule of HgBr₂ contains (2×3) = 6 Br^-1 ions. = 6 × 6.022 × 10²³   = 1.807 × 10²⁴ ions

Q15: What is Molar Volume?

Answer: One mole of any gas contains 22.4 liters at N.T.P or S.T.P. (0°C and 1atm/760 mm of Hg), the volume is called molar volume of standard molar volume. For molecular gases (e.g. CH₄), it is expressed  as gram molecular volume and for atomic gases (e.g. He gas), it is expressed as gram atomic volume.

e.g. One mol of CH₄ = 22.4 litres at STP gram molecular volume = 1 gram molecule of CH₄
       =  6.022 × 10²³  molecules of CH₄ = 16g of  CH₄

Mole in terms of molar volume, at STP

No. of Moles = Volume of gas (dm³) ÷ 22.4 (dm³ mol^-1)

Q16: What is the volume in litres at S.T.P. of 3 moles of hydrogen sulfide H₂S gas?
Answer: At S.T.P., 1 mole    = 22.4 litres molar volume
3 moles = 3 × 22.4 = 67.21 litres of Hydrogen Sulphide (H₂S).


Q17: A balloon is filled with He gas at N.T.P. The volume of the balloon is 2.24 dm³, find the amount of He gas in terms of moles is present.

Answer:  At N.T.P., 22.4 dm³ contains 1 mol of  Helium gas.
2.24 dm³ will have = 2.24 × 1/ 22.4 = 0.1 mol.


Q18: Calculate the molar mass of glucouse (C₆H₁₂O₆).

Answer:
   Atomic mass of C = 12u
   Atomic mass of H = 1 u
   Atomic mass of O = 16 u

 No. of C atoms glucouse = 6
 No. of H atoms glucouse = 12
 No. of O atoms glucouse = 6


Molar mass of glucouse = 6 × 12 + 12  × 1 + 6  × 16 = 180g
⇒ 1 mole  or 6.022 × 10²³ molecules of glucouse (C₆H₁₂O₆) weighs = 180g


Q19: A 12.0 g sample of gas occupies 19.2 L at STP. What is the molecular weight of this gas?

Answer: According to Avagadro's law, at n.t.p. 1 molar volume of gas = 22.4 L

At n.t.p. 19.2 L gas weighs = 12.0 g
At n.t.p. 22.4 L gas weighs = 12.0 x 22.4 /19.2 = 14.0g / mol


Q20(NCERT): Calculate the molecular mass of the following:
(i) H₂O
(ii) CO₂
(iii) CH₄

Answer:
Molecular mass is the sum of atomic masses of the elements present in a molecule. It is obtained by multiplying the atomic mass of each element by the number of its atoms and adding them together.

(i) Molecular mass of Water (H₂O) = (2× Atomic mass of H) + (1 × Atomic mass of Oxygen)
     = 2 × 1.008u + 1 × 16.00u = 2.016 u + 16.00 u = = 18.016 u  = 18.02 u

(ii) Molecular mass of CO₂ = (1 × Atomic mass of C ) + (2 × Atomic mass of O)
     =  1 × 12.011u + 2 × 16.00u = = 12.011 u + 32.00 u  = 44.01 u

(iii) Molecular mass of CH₄ = (1 × Atomic mass of C ) + (4 × Atomic mass of H)
     = 1  × 12.011u + 4 × 1.008u = = 12.011 u + 4.032 u = 16.043 u

Q21(NCERT): How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Answer: 1 mole of CuSO₄ contains  = 1 mole of Cu + 1 mol of S + 4 moles of O.
Molar mass of CuSO₄ = (63.5g) + (32.00g) + 4(16.00g)
       = 63.5 + 32.00 + 64.00 = 159.5 g

Since, 1 mole of CuSO4 contains 1 mole of Copper.

⇒ 159.5 g of CuSO4 contains 63.5 g of copper.
100g of CuSO4  will have Cu = (63.5 × 100) / 159.5 = 39.81g

Q22(NCERT): What will be the mass of one ¹²C atom in g?
Answer:  1 mole of carbon atoms = 6.022 × 10²³  atoms of carbon
                                = 12 g of carbon
 Mass of one ¹²C atom = 12 ÷  6.022 × 10²³  = 1.994 × 10^-23 g

Q23 (NCERT): Calculate the number of atoms in each of the following 
(i) 52 moles of Ar
(ii) 52 u of He 
(iii) 52 g of He.

Answer:
(i) 1 mole of Ar = 6.022 × 10²³ atoms of Ar
     52 mol of Ar = 52 × 6.022 × 10²³ atoms of Ar
              = 3.131 × 10²⁵  atoms of Ar

(ii) 1 atom of He = 4 u of He
 ⇒ 4 u of He = 1 atom of He
     1 u of He atom of He = 1/4 atoms of He
   52u of He atom of He = 52 × 1/4 =  13 atoms of He

(iii) 4 g of He = 6.022 × 10²³ atoms of He
      52 g of He atoms of He = 6.022 × 10²³ × 52 / 4 = 7.8286 × 10²⁴ atoms of He


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