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Stoichiometric defects are those defects in which the ratio of cations to anions remains the same are represented by the molecular formula.

• Two properties of graphite

1. It is a good conductor of electricity.

2. Graphite is a good solid lubricant.

(ii) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.

(a)

(ii) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.

• Like liquids, amorphous solids are isotropic. 
• Like liquids, they posses fluidity.

For fee lattice, edge length,

a = 2 √2 x 0.144 nm = 0.407 nm

(c) face centred cubic (4 atoms per unit cell)

Bravais Lattices

Face centred and end centred unit cells:

Option : (a) Paramagnetism

Hexagonal close-packed lattice has the highest packing efficiency

Tooth decay starts when the pH of the mouth is lower than 5.5 as Tooth enamel, made up of calcium phosphate does not dissolve in water, but is corroded . The best way to prevent this is to clean the mouth after eating food using toothpastes, which are generally basic, for cleaning the teethcan neutralise the excess acid and prevent tooth decay.

Turmeric is a natural indicator that turns red when treated with the base. Since soaps are basic in nature they convert the yellow colour of turmeric to red.

PH in plants and animals:

Our body works within the pH range of 7.0 to 7.8. Living organisms can survive only in a narrow range of pH change. When pH of rain water is less than 5.6, it is called acid rain. When acid rain flows into the rivers, it lowers the pH of the river water. The survival of aquatic life in such rivers becomes difficult.

PH in our digestive system:

It is very interesting to note that our stomach produces hydrochloric acid. It helps in the digestion of food without harming the stomach. During indigestion the stomach produces too much acid and this causes pain and irritation. To get rid of this pain, people use bases called antacids. One such remedy must have been suggested by you at the eginning of this Chapter. These antacids neutralise the excess acid.

Magnesium hydroxide (Milk of magnesia), a mild base, is often used for this purpose.

PH in our tooth!

Tooth decay starts when the pH of the mouth is lower than 5.5. Tooth enamel, made up of calcium phosphate is the hardest substance in the body. It does not dissolve in water, but is corroded when the pH in the mouth is below 5.5. Bacteria present in the mouth produce acids by degradation of sugar and food particles remaining in the mouth after eating. The best way to prevent this is to clean the mouth after eating food. Using toothpastes, which are generally basic, for cleaning the teeth can neutralise the excess acid and prevent tooth decay.

Self defense by animals and plants through chemical warfare

Bee-sting leaves an acid which causes pain and irritation. Use of a mild base like baking soda on the stung area gives relief. Stinging hair of nettle leaves inject methanoic acid causing burning pain.

Mass of metal = 0.6 

Mass of metal oxide = 1 g 

Mass of oxygen = 1 – 0.6 = 0.4 g 

0.4 g of oxygen combines with 0.6 g of metal. 

∴ 8 g of oxygen will combine with = 0.6/0.4 x 8 

Equivalent mass of the metal = 12 g eq^-1 .

(a) 2 

C₂H₆ – Ethane – molar mass = 24 + 6 = 30 g 

30 g of C₂ H₆ contains = 1 mole. 

∴ 60 g of C₂H₆ will contains = 2 x 1 = 2 moles.

Mass of ethane = 9 g 

Molar mass of ethane C₂H₆ = 30 g mol^-1 

No. of moles = Mass/ Mola mass 

= 9/30 

= 0.3 mol.

No. of moles = (Mass of the substance) / (Molar mass of the substance) = W/M

Molar Mass of ethane (C₂H₆ ) = 24 + 6 = 30 

Number of moles in 60 g of ethane = 60/30 = 2 moles.

Sulphuric acid = H₂SO₄ 

Molar mass of Sulphuric acid = 2 + 32 + 64 = 96 

Basicity of Sulphuric acid = 2 

Equivalent mass of acid = (Molar mass of an acid) / Basicity

= 96/2

= 48 g eq^-1

3H₂ + N₂ → 2NH₃ 

A per stoichiometric equation, 

No. of moles of hydrogen required for 2 moles of ammonia 3 moles 

No. of moles of hydrogen required for 20 moles of ammonia = \(\frac{3}{2}\) x 20 = 30 moles.

(c) 6 moles 

3H₂ + N₂ → 2NH₂ 

To get 2 moles of ammonia, 3 mole of H₂ is required. 

To get 4 moles of ammonia = 3/2 x 4 

= 6 moles of H₂ is required.

0.869 g of CuSO₄ .aH₂O gave a residue of 0.556 g of Anhydrous CuSO₄ .

∴ Weight of a H₂O molecule = 0.869 – 0.556 

= 0.313 g 

Molecular weight of H₂O = (1 x 2) + 16 

= 2 + 16 

= 18

No. of moles of water = Mass / (Molecular mass) 

CuSO₄.5H₂O – Molecular mass = 63.5 + 32 + 64 + 90 = 249.5 g 

249.5 g of CuSO₄.5H₂O on heating gives 159.5 g of CuSO₄ . 

0.869 g of CuSO₄.aH₂O on heating gives = 159.5/249.5 x 0.869 = 0.556 g of anhydrous CuSO₄ . ∴ a = 5 

The molecular formula of hydrated copper sulphate = CuSO₄.5H₂O

(b) + 6 

K₂Cr₂O₂ 

2 + 2x – 140 

2x – 12 = 0 

2x = + 12 

x = + 6

2 x + 3 y → 41 + m

1. 2x reacts with 3y to give products. 

5x reacts with l5y means, y is the excess because 8 moles of x should react withn 4 x 3y = 12y moles of y to give products. In this reaction 15y moles are used. 

Therefore, 3 moles of y is excess and it is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 x 41 i.e. 161 and 4m as product. 

8x + 12y → 161 + 4m

3. At the end of the reaction, the excess reactant left in 3 moles of y.

Empirical Formula: 

\(\bullet\)  Empirical formula is the simplest formula. 

 \(\bullet\) It shows the ratio of number of atoms of different elements in one molecule of the compound. 

 \(\bullet\) It is calculated from the percentage of composition of the various elements in one molecule. 

 \(\bullet\) For example, Empirical formula of Benzene = CH.

Molecular Formula:

 \(\bullet\) Molecular Formula is the actual formula. 

 \(\bullet\) It shows the actual number of different types of atoms present in one molecule of the compound. 

 \(\bullet\) It is calculated from the Empirical formula Molecular Formula = (Empirical formula)_n

 \(\bullet\) Molecular formula of Benzene = C₆H₆

Empirical Formula: 

\(\bullet\)  It is the simplest formula. 

 \(\bullet\)  It shows the ratio of number of atoms of different elements in one molecule of the compound

Molecular Formula: 

 \(\bullet\)  It is the actual formula. 

 \(\bullet\)  It shows the actual number of different types of atoms present in one molecule of the compound.

(a) AuCl₃

AuCl₃ undergoes reduction. So, it is an oxidising agent.

Empirical formula shows the ratio of number of atoms of different elements in one molecule of the compound.

Steps for finding the Empirical formula: 

The percentage of the elements in the compound is determined by suitable methods and from the data collected; the empirical formula is determined by the following steps.

1. Divide the percentage of each element by its atomic mass. This will give the relative number of atoms of various elements present in the compound. 

2. Divide the atom value obtained in the above step by the smallest of them so as to get a simple ratio of atoms of various elements. 

3. Multiply the figures so obtained, by a suitable integer if necessary in order to obtain whole number ratio. 

4. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound. 

5. Percentage of Oxygen = 100 – (Sum of the percentage masses of all the given elements).

Molecular mass and empirical formula are used to deduce molecular formula of the compound. 

Steps to calculate molecular formula:

\(\bullet\)  If Empirical formula is found out from the percentage composition of elements

 \(\bullet\)  Empirical formula mass can be found from the empirical formula 

 \(\bullet\)  Molecular mass is found out from the given data 

 \(\bullet\)  Molecular formula = (Empirical formula)_n

 \(\bullet\)  where, n = (Molecular mass) / (Empirical formula mass)

(b) 58.5 

NaCl = Salt Molar mass 

= 23 + 35.5 

= 58.5 

Equivalent mass of Salt = Molar mass of Salt.

(b) CH₂ 

Alkene C₂H_2n Molecular formula 

E.F. = M.F./2 

∴ Empirical formula = CH₂

(c) gram molecular mass

Answer: (b) – 1

(d) Cl₂O₇

Cl₂O₇

2x - 14 = 0

2x = +14

x = +7

Na₂CO₃ mass = 424 x 10⁶ g 

Molecular mass of Na₂CO₃ 

= (23 x 2) + 12 + (16 x 3) 

= 46+ 12 +48 

= 106 g

No. of moles of Na₂ CO₃ = (Mass of Na₂CO₃) / (Molecular mass of Na₂CO₃)

= (424 x 10⁶ g) / 106 g

= 4 x 10⁶ moles 

Methyl alcohol mass = 320 x 10⁶ g 

Molecular mass of CH₃OH = 12 + (1 x 4)+ 16 

= 12 + 4 + 16 

= 32 g 

No. of moles of Methyl alcohol = (Mass of Methyl alcohol) / (Molecular mass of Methyl alcohol) 

= (320 x 10⁶ g) / 32 g

= 10 x 10⁶ moles. 

∴ Methyl alcohol is more produced in terms of moles.

(b) 143.5 g

Mass of AgCl = 108 + 35.5 

= 143.5 g

(d) 18 

Ca = Atomic mass = 40 

Ca → Ca²⁺ + 2e^- 

41 g of Ca = 1 mole (for Ca²⁺ Atomic mass remains same) 

1 mole of Ca²⁺ = 40 g 

∴ 0.45 mole of Ca²⁺ = 40/(1 x 0.45) = 18g

(c) 23 g

1 gram atom of Na 

Na = Atomic mass 23 g (or) 23 amu 

1 gram atom of Na = 1 mole = 23 g.

(a) Fluorine 

(a) 20.8 g 

Na₂CO₂ .10H₂O = 1 mole 

1 mole of Na₂CO₃ . 10H₂O contains 13 oxygen atoms. 

Mass of 13 oxygen atoms = 13 x 16 = 208 

1 mole of Na₂CO₃ .10H₂O contains 208 g of oxygen. 

∴ 0.10 mole of Na₂CO₃.10H₂O contains 208/1 x 0.12 = 08 g.

(a) \(\frac{14}{6.023\times10^{23}}g\)

(a) 2 x 6.023 x 10²³ 

Methane (CH₄) – Molar mass = 12 + 4 = 16g. 

16 g contains 6.023 x 10²³ molecules. 

∴ 32 g of methane will contain =\(\frac{6.023\times10^{23}}{16}\) x 32²

= 2 x 6.023 x 10²³

(c) 3.0115 x 10²³ 

44 g of CO₂ contains 6.023 x 10²³ molecules. 

∴ 22 g of CO₂ will contain = \(\frac{6.023 \times10^{23}}{44}\)x 22 

= 3.0115 x 10²³

(b) 6.023 x 10²³ 

Methane: CH₄ 

Molecular mass 12 + 4 = 16 

16 g of methane contains Avogadro number of molecules 6.023 x 10²³ molecules.

(a) 6.02 x 10²³ 

0.0018 ml – drop of water = 0.0018 g 

H₂O = molecular mass = 18 g. 

Number of molecules in 18 g = 6.023 x 10²³ 

∴ Number of molecules in 0.0018 g 6.023 x 10²³ x 0.0018 

= 6.023 x 10²³ x 10⁵

= 6.023 x 10¹⁹ molecules.

or

Density of water at 25°C = 997.0479 g / L 

Mass of 0.0018 ml (or) 0.00 18 x 10^-3 L 

= D x V 

= 997.05 x 0.0018 x 10^-3 

= 1.795 x 10^-3 g 

Molar mass of water = 18 g 

Mole = Mass/ (Molecular mass) 

= (1.795 x 10^-3) / 18

= 9.971 x 10^-5 

∴ Number of molecules in 0.0018 ml = moles x (Avogadro number) 

= 9.971 x 10^-5 x 6.023 x 10²³ 

= 6 x 10¹⁹ molecules.

L = 1 m

W = 2 revolutions/s

A = 0.065 cm² = 6.5 x 10^-6 m²

F = mg + mlw²

= 14.5 × 9.8 +14.5 × 1 × (2)²

= 200.1 N

∆L = \(\frac{FI}{AY}\) 

[∵ Y = \(\frac{FI}{AΔI}]\)

Young’s modulus of steel is 2 × 10¹¹ Nm^-2

∆L = \(\frac{200.1 \times 1}{6.5 \times 10^{-6} \times 2 \times 10^{11}}m\)

= 1.539 × 10^-4m.

r = radius of cable = 1.5 cm.

= π(0.015)² m²

max stress = 10⁸ Nm^-2

Let F be the maximum force.

stress = \(\frac{F}{A}\)

∴ F = max stress × area.

= 10⁸ × π × (0.015)²

= 7.065 × 10⁴N

= 70,650 N

∴ The maximum load is 70,650 N.

(b) Mass m should be suspended close to B to have equal stresses in both the wires.

(d) Mass m should be suspended close to wire A to have equal strain in both wires.

Correct answer is (d) stress