## Saturday, 28 July 2012

### 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 (NA), 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/NA = 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

(i) No. of moles = 14g ÷ 28 g mol-1 = 0.5 moles
(ii) No. of molecules = (number of moles) × NA = 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 NA 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).

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

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.

(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

(In progress...)

1. COOL!!!!!!!!!!!!!!!!!!.............I LIKE IT

2. This comment has been removed by the author.

3. very useful...............thanks a lot

4. Thanks it helped me a lot....

5. bahut help hui yarr

6. this is amazing...

7. This comment has been removed by the author.

8. helpful.....but cud have given few tough Qs

9. helpful.....but cud have given few tough Qs

How many years would it take to spend Avagadro's no. of rupees at the rate of 10 lakh rupees per second?

1. 19098807711.82141045154743784880 years

2. 1.91×10^10

11. This was really useful for students.. This was in parts for different types of problems.. But I need more sums to work out ...

12. This was really useful for students.. This was in parts for different types of problems.. But I need more sums to work out ...

13. This was really useful for students.. This was in parts for different types of problems.. But I need more sums to work out ...

14. just bullshit

15. This comment has been removed by the author.

16. It's a simple give some difficult one

17. Its a best methad for prepration of exam.........

18. It contains not even a single hard question. You could have made it a bit more harder.

19. This comment has been removed by the author.

20. It's was ok but tooooo simple I need more difficult and out of questions to practice.

21. Its tooooooooooo simple please give some hard questions

22. good questions thanks

23. thanks for all these ques with solutions

24. Pl check And. to Q3 and Q4(iii)

25. Tq soo much
It ws really vry usfl