Chemistry

Class Exercise

Class Exercise - 1 Express the following numbers to three significant figures. (i) × (ii) g (iii) g(iv) Solution (i)6.02 × (ii)5.36 g (iii) g (iv)13.2

Class Exercise - 2 What is the sum of g and 1.02 g? Solution = 3.39 g

Class Exercise - 3 Express the result of the following calculation to the appropriate number of significant figures 816 × Solution 816 × = 20.0 Product rounded off to 3 significant figures because the least number of significant figure in this multiplication is three. Rounded off to 235.7

Class Exercise - 4 Solve the following calculations and express the results to appropriate number of significant figures. (i) 1.6 × × 10 2 – 2.16 × 10 2 (ii) Solution (i) 1.6 × × 10 3 Rounded off to 1.8 × 10 3

Class Exercise - 4 Rounded off to 1.6 × 10 3 or 16 × 10 2 (ii) = × 10 3 (rounded off to 7.5 × 10 3 )

Class Exercise - 5 Convert 10 feet 5 inches into SI unit. 10 feet 5 inches = 125 inches 1 inch = 2.54 × m Solution Rounded off to 317 × 10 –2 m 125 inches = 2.54 × × 125 m = × m

Class Exercise - 6 A football was observed to travel at a speed of 100 miles per hour. Express the speed in SI units. Solution 1 mile = 1.60 × 10 3 m 100 miles per hour = 4.4 × × 10 5 m/s = 4.4 × 10 m/s = 44 m/s

Class Exercise - 7 What do the following abbreviations stand for? (i) O (ii) 2O (iii) O 2 (iv) 3O 2 Solution (i)Oxygen atom (ii)2 moles of oxygen atom (iii)Oxygen molecule (iv)3 moles of oxygen molecule

Class Exercise - 8 Among the substances given below choose the elements, mixtures and compounds (i) Air(ii) Sand (iii) Diamond(iv) Brass Solution (i)Air- Mixture (ii)Sand (SiO 2 )- Compound (iii)Diamond(Carbon) - Element (iv)Brass (Alloy of metal) - Mixture

Class Exercise - 9 Classify the following into elements and compounds. (i)H 2 O (ii)He (iii)Cl 2 (iv)CO (v)Co Solution Element: He, Cl 2, Co Compound: H 2 O and CO

Class Exercise – 10 Explain the significance of the symbol H. Solution (i)Symbol H represents hydrogen element (ii)Symbol H represents one atom of hydrogen atom (iii)Symbol H also represents one mole of atoms, that is, × atoms of hydrogen. (iv)Symbol H represents one gm of hydrogen.

BASIC CONCEPTS OF CHEMISTRY Session - 2

Session Opener

Session Objectives

1.The law of conservation of mass 2.The law of definite proportions 3.The law of multiple proportions 4.The law of reciprocal proportions 5.Gay Lussac’s law of gaseous volumes 6.Dalton’s atomic theory 7.Modern atomic theory Session Objectives

Law of conservation of mass Total mass of the products remains equal to the total mass of the reactants. H 2 + Cl 2 2 HCl 2g 71g73g

Question

8.4 g of sodium bicarbonate on reaction with 20.0 g of acetic acid (CH 3 COOH) liberated 4.4 g of carbon dioxide gas into atmosphere. What is the mass of residue left? Illustrative Problem = m m = 24 g It proves the the law of conservation of mass. Solution:

A chemical compound always contains same elements combined together in same proportion of mass. Law of definite proportions Ice water H 2 O 1 : 8River water H 2 O 1 : 8 Sea water H 2 O 1 : 8

Question

Two gaseous samples were analyzed. One contained 1.2g of carbon and 3.2 g of oxygen. The other contained 27.3 % carbon and 72.7% oxygen. The experimental data are in accordance with (a)Law of conservation of mass (b)Law of definite proportions (c)Law of multiple proportions (d)Law of reciprocal proportions Illustrative Problem

% of C in the 1 st sample Which is same as in the second sample. Hence law of definite proportion is obeyed. Solution

the mass of one of the elements which combines with fixed mass of the others, bear a simple whole number ratio to one another. Law of multiple proportions 8:16:24:32:401:2:3:4:5 = 14:814:1614:2414:3214:40 Two elements combinetwo or more compounds Ratio of oxygen combining with 14 parts of nitrogen

The ratio of the weights of two elements A and B which combine separately with a fixed weight of the third element ‘C’ is either the same or some simple multiple of the ratio of the weights in which A and B combine directly with each other. Statement of law of reciprocal Proportions k may be 1

Law of reciprocal proportions SO 2 O H S H 2 S H 2 O H 2 S 2 : 32, 1 : 16 SO 2 32 : 32, 1 : 1

Gases reacts in volume which bear a simple ratio to one another and to the volume of the products under similar conditions of temperature and pressure. Gay Lussac’s law of gaseous volumes

Gay Lussac’s Law of gaseous volumes N 2 O2O2 2NO2H 2 O2O2 2H 2 O 1 volume nitogen gas 1 volume oxygen gas 2 volume nitrogen oxide + 2 volume hydrogen gas 1 volume oxygen gas 2 volume steam + 1 : 1 : 2 2 : 1 : 1

Ask yourself Why Gay–Lussac’s law is not applicable to solids and liquids ? because they have negligible volumes as compared to gases.

Do you know The laws of chemical combinations are based on quantitative results of chemical reactions.

Questions

Carbon is found to form two oxides, which contains 42.8% and 27.27% of carbon respectively. Find out which of the laws of chemical combination is proved correct by this data? Illustrative Problem In the first oxide, Carbon :Oxygen = : = : 1 In the second oxide, Carbon :Oxygen = : = : 1 = 2:1 = : Solution:

Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Find out which of the laws of chemical combination is proved correct by this data? Illustrative Problem Hence k=1 which proves law of reciprocal proportion. Solution:

Ask your self The balanced chemical reaction is an expression of Law of multiple proportion Law of conservatio n of mass Law of constant proportion None of the above

Illustrative example Zinc sulphate crystals contain 22.6% of zinc and 43.6% of water. How much Zinc should be used to produce 13.7 gm of zinc sulphate crystals and how much water they will contain? Solution: 100 gm of ZnSO 4 will have 43.9 gm water and 22.6 gm zinc (Hint: Law of constant composition)

Solution

1.Matter is made up of indivisible particles called atoms. 2.Atoms of the same element are similar with respect to shape, size and mass. 3.Atoms combine in simple whole number ratio to form molecule. 4.An atom can neither be created nor destroyed. Dalton’s Atomic Theory

5.Atoms of two elements may combine in different ratios to form more than one compound. SO 2 1:2 SO 3 1:3 S+O 2

Limitations of Dalton’s Atomic Theory It could not explain: Why atoms of different elements have different masses, sizes, etc. Nature of binding force between atoms and molecules.

Limitations of Dalton’s Atomic Theory 1.Can not explain causes of chemical combination 2.Can not explain law of combining volume 3.It does not give idea about structure of atom 4.It does not distinguish between the ultimate particle of an element and a compound 5.It does not give idea about isotopes and isobars.

Modern Atomic Theory 1.Atom is divisible 2.Same atom may have different atomic masses like 1 H, 2 H and 3 H. 3.Different atoms may have same atomic mass like 40 Ca and 40 Ar. 4.Atom is the smallest particle that takes part in a chemical reaction. 5. The mass of an atom can be changed into energy.

Class Test

Class Exercise - 1 Percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This proves the law of (a) constant proportions(b) reciprocal proportions (c) multiple proportions(d) none of these Solution Definition

Class Exercise - 2 A balanced chemical equation is in accordance with (a) Boyle’s law (b) Avogadro’s law (c) Gay Lussac’s law (d) law of conservation of mass Solution Definition

Class Exercise - 3 Different samples of water were found to contain hydrogen and oxygen in the approximate ratio of 1 : 8. This shows the law of (a) multiple proportion (b) constant proportion (c) reciprocal proportion (d) none of these Solution Definition

Class Exercise - 4 Law of multiple proportion is illustrated by the following pair of compounds. (a) HCl and HNO 3 (b) KOH and KCl (c) N 2 O and NO (d) H 2 S and SO 2 Solution Definition

Class Exercise - 5 The oxides of nitrogen contain 63.65%, 46.69% and 30.46% nitrogen respectively. These data proves the law of (a) definite proportions(b) multiple proportions (c) reciprocal proportions(d) conservation of mass Solution Definition

Class Exercise g carbon combine with 64 g sulphur to form CS g of carbon also combine with 32 g oxygen to form CO g sulphur combine with 10 g oxygen to form SO 2. These data proves the (a) law of multiple proportions (b) law of definite proportions (c) law of reciprocal proportions (d) Cray Lussac’s Law of gaseous volume

Solution Ratio of the weights of S and O combining with fixed weight of C is 64 : 32 = 2 : 1. Ratio of weights of S and O combining directly = 10 : 10 = 1 : 1. The two ratios are simple multiple of each other. This proves the law of reciprocal proportions.

Class Exercise - 7 Nitrogen forms five stable oxides with oxygen of the formula, N 2 O, NO, N 2 O 3, N 2 O 4, N 2 O 5. The formation of these oxides explain the (a) law of definite proportions (b) law of partial pressure (c) law of multiple proportion (d) law of reciprocal proportions The mass of oxygen which combine with the fixed mass of nitrogen (= 14 g) is N 2 O, NO, N 2 O 3, N 2 O 4, N 2 O 5 are 8, 16, 24, 32, 40 g respectively. They are in the ratio of 1 : 2: 3 : 4 : 5. This proves the law of multiple proportions. Solution

Class Exercise - 8 Two metallic oxides contains 27.6% and 30.0% oxygen respectively. If the formula of the first oxide is M 3 O 4, then that of the second will be (a) MO(b) MO 2 (c) M 2 O 5 (d) M 2 O 3 Solution M = 56

Hence oxide is M 2 O 3. Class Exercise - 8 Two metallic oxides contains 27.6% and 30.0% oxygen respectively. If the formula of the first oxide is M 3 O 4, then that of the second will be (a) MO(b) MO 2 (c) M 2 O 5 (d) M 2 O 3 Solution:

Class Exercise - 9 One litre of nitrogen combines with three litre of hydrogen to form two litre of ammonia under the same conditions of temperature and pressure. This explain the (a) law of constant composition (b) law of multiple proportion (c) law of reciprocal proportions (d) Gay Lussac’s law of gaseous volumes Solution The ratio of volumes of N 2, H 2 and NH 3 is 1 : 3 : 2, which is a simple ratio. This proves Gay Lussac’s law of gaseous volumes.

Class Exercise g of copper metal when treated with nitric acid followed by ignition of the nitrate gave 2.70 g of copper oxide. In another experiment 1.15 g of copper oxide upon reduction with hydrogen gave 0.92 g of copper. Show that the above data illustrate the law of definite proportions. Solution % of Cu in copper oxide in 1 st case % of oxygen = 20% of Cu in copper oxide in 2 nd case % of oxygen = 20%

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