Solution-Class-11-Science-Chemistry-Chapter-1-Some Basic Concepts of Chemistry-Maharashtra Board

(a) 88.9

(d) Na₂S, NaF

(a) - 40⁰

(b) Ampere

(d) gaseous volumes

(b) 2 g H₂

(a) 4.5

(a)  6.022 x 10²⁰

(c) 1g Li (s)

Avogadro’s law : It states that equal volumes of different gases under identical conditions of pressure and temperature contain equal number of molecules.

Explanation: Consider the reaction between gaseous hydrogen and oxygen forming water vapour.

2H₂(g)      +   O₂(g)    à  2H₂O(g)

100 mL               50 mL             100 mL

2 volumes       1 volume         2 volumes

Since volume V of a gas is directly proportional to number of gas molecule (Vµn),

2 n molecules : n molecule : 2 n molecules OR

2 molecules : 1 molecule : 2 molecules

Thus 2 molecules of hydrogen combine with 1 molecule of oxygen to give 2 molecules of water vapour.

12 g of carbon represents one mole of carbon containing 6.022  x 10²³ atoms of carbon. 12 u of carbon represents one carbon atom.

Atomic mass of hydrogen is 1.0 u. Hence its mass is 1.0 x 1.66054 x 10^-24 g = 1.66054 x 10^-24 g.

(a) NH₃ = 14u + 3 x 1u = 17u.

(b) CH₃COOH = 2 x 12u + 4 x 1u + 2 x 16u = 60u.

(c) C₂H₅OH = 2 x 12u + 6 x 1u + 16u = 46u

One mole of a substance contains 6.022 x 10²³ particles.

The SI unit of amount of a substance is ’mol'

Molar volume of a gas : The volume of one mole of a gas is called molar volume. At STP condition (0°C and 1 atm), the molar volume of any gas is 22.4 dm³.

Law of conservation of mass : This law states that during chemical combination of matter, the mass is neither created nor destroyed.

Explanation : Antoine Lavoisier a French scientist while studying several combustion reactions, determined accurately the masses of the reactants before reactions and the masses of the products after the completion of the reactions. He observed that the total masses of the reactants before the reaction is equal to the total masses of the products.

For example

1) When hydrogen gas burns and combines with oxygen to yield water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed.

Hydrogen + Oxygen —> Water

   2 g               16 g              18 g

2) When solid carbon burns and combines with oxygen to form carbon dioxide, The total mass of carbon (12 g) and oxygen taken initially (32 g) is equal to the mass of Carbon dioxide (44 g) formed.

C   + O₂   —> CO₂

(s)        (g)            (g)

(12g)   (32g)  —> (44g)

Applications :

• From the masses of the reactants taken, the masses of products that would be obtained in a chemical reaction can be predicted.
• Amounts of the reactants can be decided from the required amounts of the products.

The law of multiple proportions : This law states that when two elements chemically combine to form two or more compounds with different compositions by weights, then the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Explanation : Hydrogen combines with oxygen to form two compounds, namely water (H₂O) and hydrogen peroxide (H₂O₂).

Hydrogen + Oxygen ——> Water

     2 g               16 g                  18 g

Hydrogen + Oxygen -—> Hydrogen Peroxide

     2 g                32 g                 34 g

It is found that the two masses of oxygen namely 16 g and 32 g which combine with a fixed mass of hydrogen (2 g) are in the ratio of small hole numbers, i.e. 16 : 32 or 1 : 2

An aqueous solution of sugar

A mixture of oil and water.

Copper -Cu is an element.

Water, H₂O is a compound.

1 mole of NH₃ and 1 mole of HNO₃ contain 6.022 x 10²³ molecules each. Hence the ratio of molecules is 1 : 1.

At STP, 22.4 litre of H₂ contains 1 mol of H₂

∴ At STP 0.448 litre of H2 will contain,

0.448/22.4 = 0.02 mol of H₂

Answer is - Moles of H₂ = 0.02 mol.

Mass of hydrogen atoms = 18 x 1.008 = 18.144 u

Answer is -Mass of hydrogen atoms = 18.144 u.

(a) Number of atoms of iodine (I) = 254/127 = 2 atoms

(b) Mass of 1 mol of I atoms = 127 x 6.022 x 10²³ u

= 127 x 6.022 x 10²³ X 1.66054 x 10⁻²⁴ g

= 127 g mol⁻¹

 ∴ Number of moles of I = 254/127 = 2 mol

∴ Number of I atoms = 2 x 6.022 x 10²³ = 1.2044 x 10²⁴ atoms

 Answer is- (a) Number of I atoms = 2

                   (b) Number of I atoms = 1.2044 x 10²⁴

(a) Number of moles of carbon = \(\frac{5×10^{-3}}{12}\) = 4.166 x 10⁻⁴ mol

(b) Number of moles of carbon = \(\frac{12×10^{-3}}{12}\) = 1 x 10⁻³ mol

∴ Number of carbon atoms = 1 x  10⁻³ x 6.022 x 10²³ = 6.022 x 10²⁰

Answer is-

(a) Number of moles of carbon = 4.166 x 10⁻⁴ mol

(b) Number of carbon atoms = 6.022 x 10²⁰

Molar mass of glucose (C₆H₁₂O₆)

= 6 x 12 + 12 x 1+ 6 x 16 = 180 g mol⁻¹

 250 g of glucose costs Rs. 40

 ∴ 180 g of glucose will cost,

 = \(\frac{40×180}{250}\) = 28.8

Answer is - Cost of glucose per mole = Rs. 28.8.

Consider a boron sample containing 19.60% of ¹⁰B and 80.40% of ¹¹B isotopes.

Average atomic mass of boron = \(\frac{19.60×10.013+80.40×11.0009}{100}\) = \(\frac{196.2548+885.1236}{100}\)

= 10.813784 ≅ 10.81 u

Answer is- Average atomic mass of boron = 10.81 u

(a) \(\frac{°C}{5}=\frac{°F-32}{9}\)

∴ ⁰F = \(\frac{9×°C}{5}+32\)=(\frac{9×40}{5}+32\) = 72 + 32 =104 ⁰F

(b) ⁰F = \(\frac{9×°C}{5}+32\)=(\frac{9×30}{5}+32\)  = 54 + 32 = 86 ⁰F

Answer is – (a) 104 ⁰F (b) 86 ⁰F

Molar mass of acetic acid (CH₃COOH) = 60 g mol⁻¹

Number of moles of CH₃COOH = n = W/M = 22/60 = 0.3660 mol

Number of molecules of CH₃COOH

= n x N_A = = 0.3660 x 6.022 x 10²³

= 2.2076 x 10²³

Answer is- Number of moles of CH₃COOH = 0.3660 mol

Number of molecules of CH₃COOH = 2.2076 x 10²³

Mass of carbon + Mass of oxygen = Mass of CO₂

 Mass of oxygen

= Mass of CO₂ — Mass of carbon

= 88 − 24 = 64 g

Answer is- Mass of oxygen used = 64 g.

(a) Number of molecules of N₂ = 0.4 x 6.022 x 10²³

∴ Number of N atoms = 0.4 x 6.022 x 10²³ x 2 = 4.8176 x 10²³

(b) Number of moles of S = W/M = \(\frac{1.6}{32}\) mol

∴ Number of atoms of S = \(\frac{1.6}{32}\)  x 6.022 x 10²³ = 3.011 x 10²²

Answer is-

(a) Number of N atoms = 4.8176 x 10²³

(b) Number of atoms of S = 3.011 x 10²²

Mass of oxygen in metal oxide = 3.2 − 2 = 1.2 g

Mass of oxygen in second case = 2.27 − 1.42 = 0.85 g

 0.85 g oxygen combines with 1.42 g metal

∴ 1.2 g oxygen will combine with, \(\frac{1.42×1.2}{0.85}\) = 2.0 g metal.

This observation represents a law of definite proportion.

(a)  1 mol of acetaldehyde, CH₃CHO contains 2 mol of carbon

 ∴ 2 mol of acetaldehyde contains 2 x 2 = 4 mol carbon.

(b)  1 mol CH₃CHO contains 4 mol H

∴ 2 mol CH₃CHO contain 2 x 4 = 8 mol H

(c)  1 mol CH₃Cl-IO contains 1 mol oxygen

∴ 2 mol CH₃CHO contain 2 mol oxygen

(d) Number of molecules of CH₃CHO = 2 x 6.022 x 10²³ = 1.2044 x 10²⁴ molecules

Answer is- (a) 4 mol C, (b) 8 mol H (c) 1 mol oxygen (d) 1.2044 x 10²⁴ molecules.

Molar mass of MgO = 24 + 16 = 40 g mol⁻¹

∴ (i) Number of moles of MgO in 80g = n = W/M = 80/40 = 2 mol

(ii) Number of moles of MgO in 10g = 10/40 = 0.25 mol

Answer is –

(i) Number of moles of MgO = 2 mol

(ii) Number of moles of MgO = 0.25 mol

1 mol CO₂ at STP occupies 22.4 dm³

∴ (i) 5 mol CO₂ at STP occupies, 5 x 22.4 = 112 dm³

   (ii) 0.5 mol CO₂ occupies Volume of CO₂ at STP = 0.5 x 22.4 = 11.2 dm³

Answer is-

(i) Volume of CO₂ = 112 dm³

(ii) Volume of CO₂ = 11.2 dm³

2KClO₃ ——> 2KCl + 3O₂

2 mol                      3 mol

Volume of 3 mol O₂ at STP = 3 x 22.4 = 67.2 dm³

∴ 67.2 dm³ of O₂ is liberated at STP by 2 mol KClO₃

∴ 6.72 dm³ of O₂ at STP will be liberated by, \(\frac{6.72}{67.2}×2\) =  0.2 mol KClO₃

 Mass of KClO₃ required = 0.2 x 122.5 = 24.5 g

Answer is- Mass of KClO₃ required =24.5 g.

Molar mass of urea, (NH₂)₂CO = 60 g mol⁻¹

Number of moles of urea = W/M = 5.6/60 = 0.09333 and

∴ Number of molecules of urea = n x N_A = 0.09333 x 6.022 x 10²³

= 0.5620 x 10²³

∴ Number of H atoms = 4 x 0.5620 x 10²³ = 2.248 x 10²³

Number of N atoms = 2 x 0.5620 x 10²³ = 1.124 X 110²³

Number of C atoms = 1 x 5.620 x 10²² = 5.620 x 10²²

Number of O atoms = 1 x 5.620 x 10²² = 5.620 x 10²²

Answer is-

Number of H atoms = 2.248 x 10²³

Number of N atoms = 1.124 x 10²³

Number of C atoms = 5.620 x 10²²

Number of O atoms = 5.620 x 10²²

S    +    O₂     →   SO₂

32 g                   1 mole

32 g sulphur produces 1 mole SO₂

∴ 16 g sulphur will produce 0.5 mol SO₂

Molar mass of SO₂ = 64 g mol⁻¹

 ∴ Mass of SO₂ produced = 0.5 x 64 = 32 g

Answer is- Mass of SO₂ produced = 32 g.

Average atomic mass : An element may have different isotopes i.e., atoms having same number of electrons and protons but different number of neutrons.

The isotopes have different atomic masses and they exist in different proportion or abundance.

The atomic mass of an element is the weighted average of atomic masses of its isotopes. This is called average atomic mass,

Isotopes with relative abundances and atomic masses as shown against each of them.

From the above data, the average atomic mass of carbon

= (12 u) (98.892/100) + (13.00335 u) (1.108/100) + (14.00317) (2 × 10^-10/100)

= 12.011 u

Similarly, average atomic masses for other elements can be calculated.

Molar Mass :The mass of one mole of a substance (element / compound) in grams is called its molar mass. The molar mass of any element in grams is numerically equal to atomic mass of that element in u.

Similarly molar mass of any substance, existing as polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u.

Mole : One mole of a substance is defined as the amount of a substance that contains the number of particles like atoms, molecules, ions or electrons equal to the number of carbon atoms present in 0.012 kg of Carbon-12 i.e., 6.0221367 x 10²³ particles.

One mole of a substance contains 6.022 x 10²³ molecules while one mole (or one gram atom) of an element contains 6.0221367 x 10²³ atoms.

Formula mass : It is the sum of atomic masses of the atoms present in the formula of the substance.

Explanation : Consider the formula mass of NaNO₃.

Formula mass of NaNO₃

= average atomic mass of Na + average atomic mass of N + 3 x average atomic mass of O

=23 u + 14 u + 3 x 16u = 85u

Molar volume of a gas : The volume of one mole of a gas is called molar volume. At STP condition (0°C and 1 atm), the molar volume of any gas is 22.4 dm³.

Volume of one mole of a gas at STP :

One mole of any gas occupies a volume of 22.4 dm³ at standard temperature and pressure condition.

Number of moles of a gas (n) = \(\frac{\text{Volume of the gas at STP}}{\text{Molar volume of gas}}\)

= \(\frac{\text{Volume of the gas at STP}}{22.4}\)

∴ Number of molecules = number of moles x N_A

= number of moles x 6.022 x 10²³ mol^-1

A matter may be an element, like Cu, Ag, etc. or may be a compound like NaCl, H₂O, etc.