**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 × 10

^{23}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 × 10

^{23}(Avagadro constant). Mol is also known as chemist dozen.

The value 6.022 × 10

^{23}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 × 10

^{23}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 × 10

^{23}atoms.

**Q4: 1.0 mole of Chlorine molecule (Cl**

_{2}) contains**(i) how many number of molecules.**

**(ii) how many number of atoms.**

**(iii) how much it weighs.**

Answer: (i) 1.0 mol of Cl

_{2}contains 6.022 × 10

^{23}molecules.

(ii) One molecule of Cl

_{2}contains 2 toms of Cl. ∴ Cl

_{2}contains 2 × 6.022 × 10

^{23}atoms

i.e. 12.44 × 10

^{23}atoms

(iii) Molar mass of Cl is 35.5 gm/mol. 1.0 mol of Cl

_{2}weighs = 2 × 35.5 = 71.1 g

*Mole in terms of number,*

**1 Mole of particle =**

**6.022 × 10**

^{23}particles**Q5: Which of the following is correct option?**

**1.0 mole of NH**

_{3}(ammonia) contains ...(a) 6.022 × 10

^{23}molecules

(b) 4 mol of atoms

(c) 1 mol of Nitrogen atoms

(d) 3 × 6.022 × 10

^{23}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 × 10

^{23}C-12 atoms

1 amu = 1g per mol = 1/N

_{A}= 1/ (6.022 × 10

^{23}) = 1.66 × 10

^{-24}g.

**Q7: What is the mass of one**

^{12}C**atom? Express it in grams as well as in amu.**

Answer: one mole of C-12 atoms = 12 g = mass of 6.022 × 10

^{23}C-12 atoms

∴ mass of 1 C-12 atom = 12g ÷ 6.022 × 10

^{23}= 1.994 × 10

^{-23}g

or = 12g ÷ 6.022 × 10

^{23}= 12g × 1.66 × 10

^{-24}g = 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 (CH

_{4}) = (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 H

_{2}O is 18u (2u + 16u), therefore, molar mass of H

_{2}O is 18 g mol

^{-1}.

*Mole in terms of mass,*

**No. of Moles = Mass ÷ Molar mass**

**Q11: In 14g of N**

_{2}(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 × 10

^{23}= 3.011 × 10

^{23}molecules

(iii) One molecule of N

_{2}contains 2 N atoms.

No. of atoms = 2 × 3.011 × 10

^{23}= 6.022 × 10

^{23}atoms = 1 N

_{A}atoms.

**Q12: One gram molecule H**

_{2}O contains how many moles of molecules and number of molecules?Answer: 1 gram molecule H

_{2}O = 1 mol of molecule H

_{2}O = 6.022 × 10

^{23}molecules of H

_{2}O

**Q13: How many moles of molecule are present in 15g of C**

_{2}H_{4}(ethylene). Given atomic mass of C = 12.0 amu and of H = 1.0 amu.Answer: Molecular mass of C

_{2}H

_{4}= 2 × 12.0 amu + 4 × 1.0 amu = 24 + 4 = 28amu

∴ Molar mass of C

_{2}H

_{4}= 28 g mol

^{-1}

No. of moles of molecule C

_{2}H

_{4}= 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 (HgBr

_{2}) in ionic form in an aqueous solution is

HgBr

_{2}→ Hg

^{+2}+ 2Br

^{-1}

∴ 1 molecule of HgBr

_{2 }contains 2 Br

^{-1 }ions.

⇒ 1 mol of HgBr

_{2 }contains 2 mol Br

^{-1 }ions.

∴ 3 molecule of HgBr

_{2 }contains (2×3) = 6 Br

^{-1 }ions. = 6 × 6.022 × 10

^{23}= 1.807 × 10

^{24 }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. CH

_{4}), 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 CH

_{4}= 22.4 litres at STP gram molecular volume = 1 gram molecule of CH

_{4}

= 6.022 × 10

^{23}molecules of CH

_{4}= 16g of CH

_{4}

*Mole in terms of molar volume, at STP*

**No. of Moles =**

**Volume of gas (dm**

^{3}) ÷ 22.4**(dm**

^{3}**mol**

^{-1}**)**

**Q16: What is the volume in litres at S.T.P. of 3 moles of hydrogen sulfide H**

_{2}**S gas?**

Answer: At S.T.P., 1 mole = 22.4 litres molar volume

3 moles = 3 × 22.4 = 67.21 litres of Hydrogen Sulphide (H

_{2}S).

**Q17: A balloon is filled with He gas at N.T.P. The volume of the balloon is 2.24 dm**

^{3}, find the amount of He gas in terms of moles is present.Answer: At N.T.P., 22.4 dm

^{3}contains 1 mol of Helium gas.

2.24 dm

^{3}will have = 2.24 × 1/ 22.4 = 0.1 mol.

**Q18: Calculate the molar mass of glucouse (C**

_{6}H_{12}O_{6}_{}).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 × 10

^{23}molecules of glucouse (C

_{6}H

_{12}O

_{6}) 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) H

_{2}O

(ii) CO

_{2}

(iii) CH

_{4}

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 (H

_{2}O) = (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 CO

_{2}= (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 CH

_{4}= (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 CuSO

_{4}contains = 1 mole of Cu + 1 mol of S + 4 moles of O.

Molar mass of CuSO

_{4}= (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**

^{12}C atom in g?Answer: 1 mole of carbon atoms = 6.022 × 10

^{23}atoms of carbon

= 12 g of carbon

Mass of one

^{12}C atom = 12 ÷ 6.022 × 10

^{23}= 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

(ii) 52 u of He

**(iii) 52 g of He.**

Answer:

(i) 1 mole of Ar = 6.022 × 10

^{23}atoms of Ar

52 mol of Ar = 52 × 6.022 × 10

^{23 }atoms of Ar

=

**3.131 × 10**atoms of Ar

^{25 }(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 × 10

^{23}atoms of He

52 g of He atoms of He = 6.022 × 10

^{23}× 52 / 4 =

**7.8286 × 10**atoms of He

^{24}(In progress...)

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

ReplyDeleteThis comment has been removed by the author.

ReplyDeletevery useful...............thanks a lot

ReplyDeleteThanks it helped me a lot....

ReplyDeletebahut help hui yarr

ReplyDeletethis is amazing...

ReplyDeleteGood.help

ReplyDeleteGood.help

ReplyDeleteThis comment has been removed by the author.

ReplyDeletehelpful.....but cud have given few tough Qs

ReplyDeletehelpful.....but cud have given few tough Qs

ReplyDeletePlease answer this ques.

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

19098807711.82141045154743784880 years

Delete1.91×10^10

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

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

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

ReplyDeletejust bullshit

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteIt's a simple give some difficult one

ReplyDeleteIts a best methad for prepration of exam.........

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

ReplyDeleteThis comment has been removed by the author.

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

ReplyDeleteIts tooooooooooo simple please give some hard questions

ReplyDeletegood questions thanks

ReplyDeletethanks for all these ques with solutions

ReplyDeletePl check And. to Q3 and Q4(iii)

ReplyDelete