3. Atoms and Molecules IX Chapter 3

Page 32

Q.No. 1:

In a reaction 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g

of carbon dioxide, 0.9g water and 8.2 g of sodium ethanoate. Show that these observations are in

agreement with the law of conservation of mass.

Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Answer:

In a reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate,

carbondioxide, and water.

 Sodium + Ethanoic → Sodium + Carbon + Water

Carbonate acid ethanoate dioxide

Mass of sodium carbonate = 5.3g (Given)

Mass of ethanoic acid = 6g (Given)

Mass of sodium ethanoate = 8.2g (Given)

Mass of carbon dioxide = 2.2 (Given)

Mass of water = 0.9g (Given)

Now, total mass before the reaction = (5.3 + 6)g

= 11. 3g

and total mass after the reaction = (8.2 + 2.2 + 0.9)g

= 11.3g

Therefore, Total mass before the reaction = Total mass after the reaction

Hence, the given observations are in agreement with the law of conservation of ma

Page 33

Q.No. 2:

Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen

gas would be required to react completely with 3g of hydrogen gas?

Answer:

It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8. Then, the mass of

oxygen gas required to react completely with 1g of hydrogen gas is 8g. Therefore, the mass of

oxygen gas required to react completely with 3g of hydrogen gas is 8 × 3g = 24 g.

Q.No. 3:

Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?

Answer:

The postulate of Dalton's atomic theory which is a result of the law of conservation of mass is

“Atoms are indivisible particles, which can neither be created nor destroyed in a chemical

reaction”.

Q.No. 4:

Which postulate of Dalton's atomic theory can explain the law of definite proportions?

Answer:

The postulate of Dalton's atomic theory which can explain the law of definite proportion is “The

relative number and kind of atoms in a given compound remains constant”.

Page 35

Q.No. 1:

Define atomic mass unit.

Answer:

Mass unit equal to exactly one- twelfth the mass of one atom of carbon - 12 is called one atomic

mass unit. It is written as 'u'.

Q.No. 2:

Why is it not possible to see an atom with naked eyes?

Answer:

The size of an atom is so small that it is not possible to see it with naked eyes. Also, atom of an

element does not exist independently.

Page 39

Q.No. 1:

Write down the formula of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium suphide

(iv) magnesium hydroxide

Answer:

(i) Sodium oxide → Na2O

(ii) Aluminium chloride → AlCl3

(iii) Sodium suphide → Na2S

(iv) Magnesium hydroxide → Mg(OH)2

Q.No. 2:

Write down the names of compounds represented by the following formula:

(i) Al2(SO4)3

(ii) CaCl2

(iii) K2SO4

(iv) KNO3

(v) CaCO3

Answer:

(i) Al(SO4)3→ Aluminium sulphate

(ii) CaCl2→ Calcium chloride

(iii) K2SO4→ Potassium sulphate

(iv) CaCO3→ Calcium carbonate

Q.No. 3:

What is meant by the term chemical formula?

Answer:

The chemical formula of a compound means the symbolic representation of the composition of a

compound. From the chemical formula of a compound, we can know the number and kinds of

atoms of different elements that constitute the compound. For example, from the

chemical formula CO2 of carbon dioxide, we come to know that one carbon atom and two oxygen

atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

Question 4:

How many atoms are present in a

(i) H2S molecule and

(ii) PO4

3-

ion?

Answer :

(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.

(ii) In a PO4

3-

ion, five atoms are present; one of phosphorus and four of oxygen.

Page 40

Question 1:

Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Answer:

Molecular mass of H2 = 2 × Atomic mass of H

= 2 × 1 = 2u

Molecular mass of O2 = 2 × Atomic mass of O

= 2 × 16 = 32u

Molecular mass of Cl2 = 2 × Atomic mass of Cl

= 2 × 35.5 = 71 u

Molecular mass of CO2 = Atomic mass of C + 2 × Atomic mass of O

= 12 + 2 × 16 = 44 u

Molecular mass of CH4 = Atomic mass of C + 4 × Atomic mass of H

= 12 + 4 × 1 = 16 u

Molecular mass of C2H6 = 2 × Atomic mass of C + 6 × Atomic mass of H

= 2 × 12 + 6 × 1 = 30u

Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H

= 2 × 12 + 4 × 1 = 28u

Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H

= 14 + 3 × 1 =17 u

Molecular mass of CH3OH Atomic mass of C+4 ×Atomic mass of H+Atomic mass of O

= 12 + 4 × 1 + 16 = 32 u

Q.No. 2:

Calculate the formula unit masses of ZnO, Na2O, K2CO3, given masses of Zn = 65u, Na = 23u, K

= 39u, C = 12u, and O = 16u.

Answer :

Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O

= 65 + 16 = 81 u

Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O

= 2 × 23 + 16 = 62u

Formula unit mass of K2CO3

= 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O

= 2 × 39 + 12 + 3 × 16 = 138u

Page 42

Question 1:

If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?

Answer:

One mole of carbon atoms weighs 12g (Given)

i.e., mass of 1 mole of carbon atoms = 12g

Then, mass of 6.022× 1023 number of carbon atoms = 12g

Therefore, mass of 1 atom of carbon =

12

6.022× 10– 23 g g

= 1.9926 × 10– 23 g

Q.No. 2:

Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass

of Na = 23u, Fe =56 u)?

Answer 2:

Atomic mass of Na = 23u (Given)

Then, gram atomic mass of Na = 23g

Now, 23g of Na contains = 6.022×1023 number of atoms

Thus, 100g of Na contains =

6.022×1023 ×10⁻²³

number of atoms

= 2.6182 × 1024 number of atoms

Again, atomic mass of Fe = 56u (Given)

Then, gram atomic mass of Fe = 56g

Now, 56 g of Fe contains = 6.022×1023 number of atoms

Thus, 100 g of Fe 6.022×1023 ×100

56

number of atoms

= 1.0753 × 1024 number of atoms

Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Q.No. 5:

Give the names of the elements present in the following compounds:

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

Answer:

Compound Chemical formula Elements present

Quick lime CaO Calcium, oxygen

Hydrogen bromide HBr Hydrogen, bromine

Baking powder NaHCO3 Sodium, hydrogen, carbon,

oxygen

Potassium sulphate K2SO4 Potassium, sulphur, oxygen

Q.No. 6:

Calculate the molar mass of the following substances:

(a) Ethyne, C2H2

(b) Sulphur molecule, S8

(c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO3

Answer:

(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28g

(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256g

(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124g

(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g

(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63g

Q.No. 7:

What is the mass of

(a) 1 mole of nitrogen atoms?

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

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

Answer:

(a) The mass of 1 mole of nitrogen atoms is 14g.

(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g

(c) The mass of 10 moles of sodium sulphite (Na2SO3) is 10 × [2 × 23 + 32 + 3 × 16]g

= 10 × 126g = 1260g

Q.No. 8:

Convert into mole.

(a) 12g of oxygen gas

(b) 12g of water

(c) 22g of carbon dioxide

Answer:

(a) 32 g of oxygen gas = 1 mole

Then, 12g of oxygen gas = 12/32 mole = 0.375 mole

(b) 18g of water = 1 mole

Then, 20 g of water = 20/18 mole = 1.11 moles (approx.)

(c) 44g of carbon dioxide = 1 mole

Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole

Q.No. 9:

What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Answer:

(a) Mass of one mole of oxygen atoms = 16g

Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g

(b) Mass of one mole of water molecule = 18g

Then, mass of 0.5 mole of water molecules = 0.5 × 18g = 9g

Question 10:

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

Answer 10:

 1 mole of solid sulphur (S8) = 8 × 32g = 256g

i.e., 256g of solid sulphur contains = 6.022 × 1023 molecules

Then, 16g of solid sulpur contains 6.022×1023

256

× 16 molecules

= 3.76 × 1022 molecules (approx)

Q.No. 11:

Calculate the number of aluminium ions present in 0.051g of aluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al

= 27u)

Answer:

1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g

i.e., 102g of Al2O3 = 6.022 × 10⁻²³ molecules of Al2O3

Then, 0.051 g of Al2O3 contains =

6.022 × 10⁻²³ 

102

× 0.051 molecules

= 3.011 × 1020

 molecules of Al2O3

The number of aluminium ions (Al3+

) present in one molecules of aluminium oxide is 2.

Therefore, The number of aluminium ions (Al3+

) present in

3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020

= 6.022 × 1020

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4. Moving Charges and Magnetism

Magnetic Field

Region around a current carrying conductor in which electro-magnetism effect is produced, is said to be Magnetic Field.

It is generally denoted by B. Magnetic field is a vector quantity. The dimension of magnetic field is [MA⁻¹T⁻²]. The SI unit of magnetic field is NA⁻¹m⁻¹ or Tesla.

B = F_m/qvsinθ

Here,
Fm = Magnetic Force
q = Magnitude of the charge
v = Velocity of the charge
θ = Angle between velocity and magnetic force.

Magnetic field vertically upward and downward in a plane are generally denoted by conventional sign (.) and (x) respectively. 

Total Magnetic field from different sources is the vector sum of the different magnetic field.
Therefore,

B = B1 + B2 + B3 + ...................... 
Super position principle is used during the sum of magnetic field.

Lorentz Force

The field in which both electric and magnetic field are in existence, the total force experienced by charge q due to motion is said to be Lorentz Force.

Lorentz Force = Force on charge due to electric field + Force on charge due to Magnetic field
F = Fe + Fm
Here,
F = Lorentz Force
Fe = Electric Force
Fm = Magnetic Force

Therefore,
Lorentz Force = F = qE + qvBsinθ

Special Cases of Magnetic Force

Magnetic Force = Fm = qvBsinθ
Case I
When θ = 0⁰ or 180⁰
Fm = 0
Charge will be moved either parallel or antiparallel of magnetic field.

Case II

When  θ = 90⁰ 
Fm = qvB
Charge will be moved along the perpendicular direction of magnetic field.

Magnetic Force on a current carryin conductor

                 Consider about a conductor which is placed in a magnetic field B along z axis. The direction of the magnetic field is along x axis. So that magnetic force will be along y axis according to Fleming's left hand rule.
                   We know that a large numbers of electrons are present in free state in a conductor which move opposite of current with drift velocity.

Mathematical Calculation

Let the length of the conductor = l
cross - sectional area of the conductor = A
drift velocity = v
current = I
charge = -e
No. of electrons per unit volume of the conductor = n

Now, according to Lorentz Magnetic Force
Fm = -evBsinθ
Now consider a small length dl.

Volume of conductor for this length = Adl
No. of electrons = nAdl
Charge = -enAdl

Magnetic Force for this length 
dFm = -enAdlvBsinθ

drift velocity for dl = v = -dl/dt             (The direction of dl is opposite to drift velocity)

Therefore,
dFm = -enAdlBsinθ.-dl/dt ---------------- (i)

enAdl/dt = I

Now, from (i)

dFm = I(dlxB)

For whole conductor

Fm = IlBsinθ

Direction of Fm, I and B are determined by Fleming's left hand rule.

Special Cases

Case I

When θ = 0⁰ or 180⁰
Charged particles move either parallel or anti parallel of magnetic field.

Case II

When θ = 90⁰ 

Charged particles move along the perpendicular of magnetic field.

Under this circumstances the charged particles move in circular path.

                  Consider about a charged particles of mass m and charge q moves in a circular path of radius r.

Centripetal Force experienced by the particle = mv²/r

Magnetic Force = qvB

qvBsinθ = mv²/r

r = mv/qB

Let the charged particle takes T time to move one rotation.
Therefore,
T = 2πr/v
Now putting the value of r
T = 2πm/qB

Frequency ν = 1/T = qB/2πm

Case III

When θ is other than 0⁰, 90⁰ or 180⁰

Let the angle be θ 

Magnetic field B is along x axis, current is along z axis and magnetic force Fm is along y axis.

Under this circumstances the charged particle moves in hellical path.

Vertical Components of drift velocity of charged particle = vsinθ
Horizontal Components of drift velocity of charged particle = vcosθ

Particle along vertical components moves along circular path.

Centripetal Force = m(vsinθ)²/r
Magnetic Force = qBvsinθ

Centripetal Force = Magnetic Force
r = mvsinθ/qB

T = 2πr/ vsinθ

Now putting the value of r

T = 2πm/qB

Frequency ν = 1/T = qB/2πm

Motion in Combined Electric and Magnetic Field

Consider about a charge q moves in a field in which both electric and magnetic field are perpendicular to each other.

Electric force acts along electric field and magnetic force opposite to electric force. Direction of magnetic field is determined by Fleming's right hand rule.

Mathematical Calculation

Fe = Fb

qE = qvB.sin90⁰
qE = qvB
v = qE/qB
v = E/B

Magnetic Field due to a current element, Biot Savart Law

Biot Savart Law states that 
"The magnitude of the magnetic field is proportional to the current, the element length and inversely proportional to the square the distance of the point from current element at which magnetic field is determined. The direction of the magnetic field is perpendicular to the plane containing element length and distance."

          Consider about a finite conductor XY carrying current I. dl is the current element. There is a point P at a distance r from the current element. At this point magnetic field dB is to be determined. The angle between dl and displacement vector r is θ.

The relevant figure is drawn below

Now according to Biot Savart Law

dB∝ Idlxr/r³
dB = 𝛍ₒIdlxr/4𝛑r³

Here 
𝛍ₒ/4𝛑 is a constant of proportionality ( The medium is vacuum)

Therefore, magnitude of the magnetic field
।dB। = 𝛍ₒIdlsinθ/4𝛑r²

The value of 𝛍ₒ/4𝛑 = 10⁻⁷ Tm/A
𝛍ₒ is the permeability of free space or vacuum.

There are some similarities, as well as differences, between Biot Savart's Law and Coulomb's Law. Which are as follows

1. Both depend inversely on the square of distance from the source of interest.
2.The principle of superposition applies to both magnetic and electric fields.
3. Both magnetic field and electric field is linear to their sources Idl (current element) and q (source of electric charge) respectively.
4. The electric field is produced by scalar source q (electric charge) whereas magnetic field is produced by vector source Idl (current element)
5. The electric field is along the displacement vector whereas magnetic field is perpendicular to the plane containing the displacement vector r and current element Idl.
6. The magnetic field at any point in the direction of dl is zero.

Relation between permitivity of free space (𝛆ₒ) and permeability of free space (𝛍ₒ) and speed of light (c).

𝛆ₒ𝛍ₒ = 4𝛑𝛆ₒ.𝛍ₒ/4𝛑

= 10⁻⁷/9x10⁹
= 1/9x10¹⁶
= 1/(3x10⁸)²
= 1/c²

Chapter 4 Class XII