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A prospectus is ‘any document described or issued as a prospectus including any notice, circular, advertisement or other document inviting deposits from the public or inviting offers from the public for the subscription or purchase of any shares or – debentures of, a body corporate’. In other words, it is an invitation to the public to apply for shares or debentures of the company or to make deposits in the company.

It is issued by a public company which is seeking to raise the required funds from the public by means of issue of shares and debentures. It is not necessary for every company to file a prospectus. A statement in ‘ lieu of prospectus is filed with the Registrar of Companies if the company has adopted Table A of the Companies Act instead of Articles of Association. Private companies are not required to file a prospectus.

Formation of a company is a complex activity, involving these stages which are as follows:

1. Promotion Identification of opportunities, analysis of its prospects and initiating steps to form a company is known as promotion of a company. 

2. Incorporation Registration Of company as body corporate under Companies Act, 1956 is known as incorporation. 

3. Subscription of Capital A public company’s raising funds from the public by means of issue of shares and debentures is known as capital subscription. 

4. Commencement of Business The registrar issues certificate of commencement of business which is a conclusive evidence of completion of formation requirement of a company.

Correct Answer is: (b) A public company seeking investment from public 

(i) Square of a real number is always greater than the number when the magnitude of the number is greater than one.

(ii) In a rectangle, the diagonals are equal.

(iii) There are three and only three angles in a triangle.

We have to prove that the product (n + 1) is divisible by 2. 

Now we have two cases. Either is even or odd. Let us examine each case. Suppose n is even. Then we can write n = 2 m, for same natural number m. And, then 

n(n + 1) = 2m (2m + 1) which is clearly divisible by 2.

 Next, suppose n is odd. Then n + 1 is even and we can write n + 1 = 2r, for some natural number 2. 

We have n(n + 1) = (2r - 1 2r = 2r (2r - 1) which is clearly divisible by 2. 

So, we can say that the natural number n(n + 1) is divisible by 2.

(i) This statement is true.

(ii) This statement is false ; for example, 9 is not a prime number

(iii) This statement is true.

(iv) This statement is true

(v) This statement is false, for example (1/2)³ = 1/8 and 1/8 is not greater than 1/2.

(vi) This statement is true.

Correct Answer is: (c) Promotion, Incorporation, Capital Subscription, Commencement of Business

(i) Hindi is the mother tongue of India. 

(ii) India is progressing rapidly.

(iii) (a + b)³ = a³ + 3a²b + 3ab² + b³ 

(iv) 2 is the real part of 2 + 4i.

(v) 4 is multiple of 20. 

(vi) 2nd October is the birthday of Mahatma Gandhi.

(vii) Republic day of India held o 26th January.

(viii) The roots of the equation x⁴ + 4x³ + 6x² + 4x + 1= 0 are equal.

(ix) New-York is in England.

Correct Answer is: Nature of a Company, class-11, 

Both numbers have 3 significant figures.

Each of the digits of a number that are used to express it to the required degree of accuracy or precision, starting from the first non-zero digit, are significant figures.

Water containing half H₂O and half D₂O will have mass = average of masses of,H₂O (18) and D₂O (20)
Mass of water = \(\frac{(18 + 20)}{ 2}\) = \(\frac{38}{2}\) = 19

Molar mass of HNO₃ = (1 × atomic mass of H + 1 × atomic mass of N + 3 atomic mass of O)
= (1 × 1) + (14 x 1) + (16 × 3)
= 1 + 14 + 48 = 63 g/mol

Atomicity of a gas is the number of atoms of an element present in one molecule of that element. For example, helium is monoatomic and hydrogen is diatomic.

Molecular mass is a number equal to the sum of the atomic masses of the atoms in a molecule. The molecular mass gives the mass of a molecule relative to that of the 12 C atom, which is taken to have a mass of 12. 

Example: molecular mass of H₂O = 18.

The answer is (d) 180 u

Gram molar mass of C₁₂H₂₂O₁₁ (Sucrose) is 342 g So, 342 g of sucrose contains 1 mole of it. Therefore, 10 g will contain 10/342 = 0.029 mole

Gram atomic mass of Oxygen is 16 g.
So, 16 g of Oxygen contains 1 mole of it
Therefore, 80 g of Oxygen will contain = 80/ 16 = 5 moles.

Molecular mass of sodium sulphate (Na₂SO₄) is 142 g mol.
The elements present in sodium sulphate are sodium, sulphur and oxygen.
Mass of Sodium : 23  × 2 = 46 g
Mass of Sulphur : 32  × 1 = 32 g
Mass of Oxygen : 16  × 4 = 64 g
Mass percentage of sodium = \(\frac{32}{142}\)  × 100
= 32.38 or 32.4%
Mass percentage of sulphur = \(\frac{32}{142}\) × 100
= 22.57 or 22.6
Mass percentage of oxygen = \(\frac{64}{142}\) × 100 .
= 45.05%

Molar mass of C₂H₆ = (2 x atomic mass of C + 6 × atomic mass of H) = ( 2 × 12) + (6 × 1) = 24 + 6
= 30 g/mol

The answer is (c) 3

(i) Mass of 1 mole = molar mass
Mass of 1 mole of MgSO₄ = 120 g
Mass of 2 moles of MgSO₄ = 120 g x 2 = 240 g
(ii) Mass of 1 mole = molar mass
Mass of 1 mole of KCI = 74.5 g
Mass of 2 mole of KCI = 74.5 x 2 = 149 g

(i) Molar mass of ethyl alcohol (C₂H₅OH) is 46 g/mol
So, 46 g of ethyl alcohol will have 6.023  × 10²³ molecules,
(ii) Molar mass of nitrogen pentaoxide (N₂O₅) is 108 g/mol
So, 54 g of N₂O₅ will contain 1/ 2 mole
i.e., 6.023  × 10²³  × (1/2) molecules.
= 3.0115  × 10²³ molecules
Hence, 54 g of nitrogen pentaoxide will have minimum number of molecules.

(i) Molar mass of water (H₂O) is 18 g/mol 18 g of water contains 1 mole which contains 6.023 × 10²³ molecules of it.
Therefore, 36 g water will contain 2 moles i.e, 6.023 × 10²³ × 2 molecules
= 12.046 × 10²³ molecules
(ii) Molar mass of carbon monoxide (CO) is 28 g/mol So, 28 g of CO will contain 1 mole i.e, 6.023 × 10²³  molecules
Therefore 36 g of water will contain maximum number of molecules.

(i) 3 significant figures
(ii) 4 significant figures

The answer is 3.60 × 10³

In SI system, the unit and symbol for quantity of matter are mole and mol respectively.

1 mole is the amount of substance that contains the same number of entities as there are atoms in exactly 12 g of the 12 C isotope.
∴ Number of atoms in 12 g of ¹²C-isotope
= 6.023 × 10²³ atoms/mol
The term ‘mole’ is given by the “Amedeo Avaogadro”. So it is also known as Avogadro’s number (NA).
∴ 1 mole =N_A= 6.023 × 10²³ atoms, molecules, ions etc.

2 g of hydrogen molecules = 1 mol
1 g of hydrogen molecules = 1/2 mol
20 g of hydrogen molecules = 1/2 × 20 = 10 mol

8.0 g has two significant figures which shows precision up to 1 decimal place where as 8.000 g, has 4 significant figures which shows precision up to 3 decimal place.

(d) 3.011 × 10²³ atoms of carbon

Molarity: It is the number of moles, of the solute dissolved per litre of the solution. It is represented by the symbol ‘M’. It is given as:
M = Mass of solute/Volume of solution in litre.
Unit of Molarity is mol/L.
Molality: It is the number of moles of the solute dissolved in 1 kg of the solvent. It is represented by the symbol ‘m’. It is given as :
m = Moles of solute / Weight of solvent in kg
Unit of molality is mol/kg.
Normality: Normality is number of gram equivalents of the solute dissolved per litre of solution. It is denoted by N.
Normality = No. of gram equivalent of solute / Volume of solution
The unit of Normality is gram equivalent per litre,

Gram molecular weight of oxygen = 32 g
32 g of O₂ contains = 6.023 × 10²³ molecules
Hence, 64 g of O₂ will contain 6.023 × 10²³ × \(\frac{64}{32}\)
= 6.023 × 10²³ × 2
= 12.046 × 10²³ molecules

Equivalent weight of acid = molecular weight of acid/basicity

For example, Equivalent weight of H₂SO₄ = \(\frac{98}{2}\) = 49 g/eq 

Equivalent weight of base = molecular weight of base/acidity

Example, Equivalent weight of NaOH = \(\frac{40}{1}\) = 40 g/eq 

Equivalent weight of salt = molecular weight of salt/charge present on ionic form

Example, Equivalent weight of Na₂CO₃= \(\frac{106}{2}\) = 53 g/eq

Empirical formula = C₃H₄N
Empirical formula mass = (3 × 12) + (4 × 1) + 14 = 54
Thus, Molecular formula of the compound = (Empirical formula) x n
= (C₃H₄N)₂ = C₆H₈N₂

Equivalent weight of H₂SO₄ = Molecular mass / number of replaceable H⁺
= \(\frac{98}{2}\) = 49

l = 3; m = -3, -2, -1, 0, +1, +2, +3.

n = 3; l = 0, 1, 2.

Using the data, in the case of methane,

Mass of carbon/Mass of hydrogen = 334.6/111.2 = 3.009

Since there are 4 atoms of hydrogen combined with one carbon atom, therefore,

Mass of one carbon atom/Mass of one hydrogen atom = 3.009 x 4 = 12.036 ≈ 12.04

From the calculations made in the part (b), we find that the empirical formula for acetylene can be CH₂, C₂H₄, C₃H₆, C₄H₈ etc.

The formulae of the compounds formed from the sets of carbon and oxygen are: 

i. Na₂CO₃ – Sodium carbonate 

ii. H₂CO₃ – Carbonic acid 

iii. NaHCO₃ – Sodium bicarbonate

FCH₂COOH > ClCH₂COOH > C₆H₅CH₂COOH > CH₃COOH > CH₃CH₂OH

Greater the number of unpaired electrons greater will be the paramagnetic character 

(a) sp³ ,sp³

(b) sp³ sp²

Molecular orbital (MO) theory was developed by F. Hund and R.S. Mulliken in 1932

sp³< sp² <sp 

sp² -Trigonal planar e.g, BH₃ 

sp³ - Tetrahadral eg CH₄