ISC Chemistry Model Question Papers

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SOME BASIC CONCEPTS IN CHEMISTRY

Dalton’s Atomic theory:

In 1803, John Dalton, an English school teacher, proposed the Atomic theory of matter based on his experimental observations. The main postulates of Dalton’s atomic theory are as follows.

1. All matter is made up of extremely small particles called atoms.

2. Atoms of the same elements are alike but differ from atoms of other elements.

3. Atoms are indivisible.

4. Each atom has a definite weight.

5. Atoms of more than one element combine to form molecules.

Although Dalton’s theory failed to explain certain things, it is the cornerstone of modern chemistry. It had a profound influence on scientific thought. It underwent expansion, clarification and modification in an attempt to explain two things:

Divisibility of an atom, and
Existence of positive and negative charges.

Later on, scientists like Thomson, Rutherford and Bohr put forward their views and the discovery of electrons, protons and neutrons, provided evidence that disproved that the atom is indivisible.

LAWS OF CHEMICAL COMBINATION

Law of Conservation of Mass (Given by Antoine Lavoisier in 1789).

It states that matter (mass) can neither be created nor destroyed.

Eg If a chemical reaction takes place in a sealed vessel that permits no matter to enter or to escape, the mass of the vessel and its contents after the reaction will be identical to the earlier mass.

Law of Definite Proportions or Law of Constant Composition:

This law was proposed by Louis Proust in 1799, which states that:

'A chemical compound always consists of the same elements combined together in the same ratio, irrespective of the method of preparation or the source from where it is taken'.

Example

Carbon dioxide may be obtained by the following methods:

a. by burning carbon

b. By reaction between a metal carbonate and a dilute acid.

c. By heating calcium carbonate or sodium bicarbonate.

Analysis of carbon dioxide, prepared by any of the above methods, shows that it contains only carbon and oxygen, combined together in the same proportion by weight, i.e., 12: 32 or 3: 8.

Law of Multiple Proportions

Proposed by Dalton in 1803, this law states that:

When two elements A and B combine to form two or more compounds, then different weights of B which combine with a fixed weight of A bears a simple numerical ratio to one another.

Example 1:

Carbon combines with oxygen to form two different oxides, namely, carbon monoxide (CO) and carbon dioxide (CO₂). The proportions by weight of the two elements are

Carbon monoxide - C: O :: 12 : 16

Carbon dioxide - C: O :: 12 : 32

There, the weights of oxygen that combine with a fixed weight of carbon (12g) are in the ratio 16g : 32g i.e. 1:2, a simple numerical ratio.

Example 2

Nitrogen combines with oxygen to form different oxides. The compositions by weight of these oxides are shown in table.

Compositions by weight of oxides of nitrogen

No. Name of Oxide Wt. of nitrogen(g) Wt. of Oxygen(g)

1 Nitrous oxide (N₂O) 28 16

2 Nitric oxide (2NO) 28 32

3 Nitrogen trioxide (N₂O₃) 28 48

4 Nitrogen tetraoxide (N₂O₄) 28 64

5 Nitrogen pentoxide (N₂O₅) 28 80

It can be seen from the table that different weights of oxygen that combines with a fixed weight of nitrogen (28 g) are in the ratio,16g : 32 g : 48g : 64g : 80g i.e., in the simple numerical ratio of 1 : 2 : 3 : 4 : 5.

Law of Reciprocal Proportions:-

Proposed by Berzelius

When two elements combine separately with a definite mass of a third element, then the ratio of their masses in which they do so is either the same or some whole number multiple of the ratio in which they combine with each other.

Let us consider three elements hydrogen, sulphur and oxygen. Hydrogen combines with oxygen to form H₂O whereas sulphur combines with it to form SO₂. Hydrogen and sulphur can also combine together to form H₂S.

The formation of these compounds is shown in figure.

In H₂O, the ratio of masses of H and O is 2: 16.

In SO₂ , the ratio of masses of S and O is 32: 32.

Therefore, the ratio of masses of H and S which combine with a fixed mass of oxygen (say 32 parts) will be 4: 32 i.e., 1 : 8

When H and S combine together, they form H₂S in which the ratio of masses of

H and S is 2:32 i.e., 1: 16

The two ratios (i) and (ii) are related to each other as 2 : 1

i.e., they are whole number multiple of each other.

Thus, the ratio of masses of H and S which combines with a fixed mass of oxygen is a whole number multiple of the ratio in which H and S combine together.

Gay Lussac’s Law of Gaseous Volumes

Given by Gay Lussac in 1808, according to this law when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

e.g. H₂(g) + Cl₂(g) ---→2HCl(g)

1V 1V 2V

All reactants and products have simple ratio 1:1:2.

Avogadro Law

In 1811, Given by Avogadro, according to this law equal volumes of gases at the same temperature and pressure should contain equal number of molecules.

Numerical problems based on the above laws:-

(1) In an experiment 5.0g of CaCO₃ on heating gave 2.8 g of CaO and 2.2 g of CO₂. Show that these results are in accordance to the law of conservation of mass.

Solution:-

CaCO₃ à CaO + CO₂

Mass of CaCO₃ = 5.0g

Mass CaO = 2.8g

Mass of CO₂ = 2.2g

Since the mass of reactants is equal to the total mass of products, it verifies the law of conservation of mass.

(2) 1.375 g of CuO were reduced by H₂ and 1.098 g of Cu were obtained. In another experiment, 1.178 g of Cu were dissolved in nitric acid and the resulting copper nitrate was converted into CuO by ignition. The weight of CuO formed was 1.476 g. Show that these results prove the law of constant proportion.

Solution

Experiment 1:

Weight of CuO = 1.375 g

Weight of Cu = 1.098 g

Weight of oxygen = (1.375 – 1.098) g

= 0.277 g

Ratio of copper oxygen = 1.098 : 0.277

= 3.96 : 1

Experiment 2:

Weight of CuO = 1.476

Weight of Cu = 1.178

Weight of oxygen = 1.476 – 1.178

= 0.298

Ratio of copper: oxygen = 1.178 : 0.298

= 3.96 : 1

In both experiments the ratio of Copper: oxygen is some (3.96: 1). Hence it illustrates the law of definite proportions.

(3) In an experiment 34.5 g oxide of a metal was heated so that O₂ was liberated and 32.1 g of metal was obtained. In another experiment 119.5 g of another oxide of the same metal was heated and 103.9 g metal was obtained and O₂ was liberated. Calculate the mass of O₂ liberated in each experiment. Show that the data explain the law of multiple proportions.

Solution:

Experiment 1

Weight of the metal oxide = 34.5 g

Weight of the metal = 32.1 g

Weight of oxygen liberated = 2.4 g

32.1 g metal combines with 2.4 g oxygen.

1 g of the metal combine with = 2.4/32.1

= 0.075 g

Experiment 2

Weight of the oxide taken = 119.5 g

Weight of the metal formed = 103.9 g

Weight of oxygen liberated = 15.6 g

103.9 g of metal combines with = 15.6 g oxygen.

1 g of metal combines with = 15.6/103.9

= 0.15014 oxygen

Therefore different weights of oxygen, that combine with the fixed weight of the metal (1 g) are in the ratio

0.1501 : 0.075

2 : 1

Hence the data is in accordance with Law of Multiple proportions.

(4) One gram of hydrogen combines with 15.88 g of sulphur. One gram of hydrogen combines with 7.92 g of oxygen, one gram of sulphur combines with 0.998 g of oxygen. Show that these data illustrate the law of reciprocal Proportions.

Solution:

In hydrogen – Sulphur and Hydrogen – Oxygen combinations

Weight of hydrogen = 1.0 grams

Weight of sulphur = 15.88 grams

Weight of oxygen = 7.92 grams

In Hydrogen – oxygen combinations the ratio masses of H and O is

1:8 ..… (1)

In Hydrogen – sulphur combinations the ratio masses of H and S is

1:16 ..... (2)

So the ratio (1) and (2) are related to each other as

1/8 : 1/16 ……(3)

They are whole number multiple of each other.

In oxygen – sulphur combinations.

Weight of sulphur = 1 gram Weight of oxygen = 0.998 gram

The ratio masses of H and S is 1:1 ..... (4)

(1), (2)and (3) are simple multiples of each other therefore, the law of reciprocal proportions holds good.

(5) 100cm³ of propane was burnt in excess oxygen to form carbon dioxide and water. Calculate: (i) the volume of oxygen used up, and (ii) the volume of carbon dioxide formed.

[Hint: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O].

Solution:

C₃H₈(g) + 5O₂(g) 3CO₂(g) +4H₂O

1 volume 5 volumes 3 volume 4 volumes

By Gaylussac’s Law:-

100 cm³ 500 cm³ 300 cm³ 400 cm³

Volume of O₂ used = 500cm³

Volume of CO₂ formed= 300 cm³

Atomic Mass Unit

One atomic mass unit is defined as a mass exactly equal to one twelfth the mass of one carbon -12 atom. And 1 amu = 1.66056×10^–24 g.

Atomic Mass

Atomic mass of an element is defined as the average relative mass of an atom of an element as compared to the mass of an atom of carbon -12 taken as 12.

Gram Atomic Mass

The quantity of an element which has mass in grams is numerically equal to its atomic mass. In simple terms, atomic mass of an element expressed in grams is the gram atomic mass or gram atom.

Eg Atomic mass of Oxygen – 16 amu

Gram atomic mass of or one gram atom of oxygen – 16g

Relative Molecular Mass

The relative molecular mass of a substance is the mass of one molecule of that substance as compared with the mass of an atom of C-12 taken as 12.

It may also be defined as the sum of relative atomic masses of all atoms in a molecule.

Eg Relative Molecular Mass of H₂SO₄ = 2×1 + 32 + 4×16 = 98

Gram Molecular Mass or Gram Mole is the molecular mass expressed in grams.

Molar Volume

The volume occupied by one gram mole of a gaseous substance is called Molar Volume.

One Mole of a gaseous substance occupies 22.4 litres at STP.

Eg One mole of hydrogen i.e 2g of hydrogen gas occupies 22.4L at STP.

One mole of Carbon dioxide i.e 44g of CO₂ gas occupies 22.4 L at STP

Determination of Molecular mass :-

By Victor Meyer’s Method:-

A known weight of a volatile compound is vaporised and the volume of vopours thus formed is measured. If the weight of the compound is w g and volume of its vapours is Vml at STP, the molecular mass can be calculated as follows:-

V ml of substance at STP weighs – w g

22400 ml of substance at STP weighs =

Chemical Equivalents

Equivalent weight of a substance is defined as the weight which will combine or displace 1.008 parts by mass of hydrogen, 35.5 parts by mass of chlorine, 8 parts by mass of oxygen, and 108 parts by mass of silver.


S.No			Value of x	Equivalent weight
1.	Acid	H₂SO₄	Basicity = 2 (Replaceable H⁺ ions)	98/2 = 49
2.	Base	Ca(OH)₂	Acidity = 2 (Replaceable OH^- ions)	74/2 = 37
3.	Salt	Na₂CO₃	Charge on all the anions =2 (2Na⁺ or CO₃^2-)	106/2 = 53
4.	Oxidising agent	KMnO₄	No. of electrons gained = 5 MnO₄^- +8H⁺+5e^- →Mn²⁺ + 4H₂O	158/5 = 31.6
5.	Reducing agent	Na₂S₂O₃.5H₂O	No. of electrons lost per molecule = 1 2S₂O₃^2- → S₄O₆^2- + 2e^-	248/1 = 248

Mole concept

A Mole is a number which represents 6.023×10²³ particles.

This number is also called Avagadro’s number.

Ø Weight of one mole atoms of an element in grams is equal to its gram atomic weight.

Ø Weight of one mole of molecules of a compound is equal to its gram molecular weight.

Molecular weight of an substance in grams	Number of moles	Number of molecules	Volume at S.T.P
2 gms of H₂	1 mole of H₂	6.023 x 10²³	22.4 L
32 gms of O₂	1 mole of O₂	6.023 c 10²³	22.4 L
44 gms of CO₂	1 mole of CO₂	6.023 x 10²³	22.4 L
18 gms of H₂O	1 mole of H₂O	6.023 x 10²³

Examples:-

(1) Calculate the mass of one molecule of one molecule of H₂SO₄.

__12____

6.023×10²³

(ans) Molecular Mass of H₂SO₄ = 98

∴ Mass of one molecule of H₂SO₄ = = 1.627×10^-22 g

(2) Calculate the number of moles and number of molecules in 60g of glucose.

(ans) Molecular mass of glucose = 180

180 g of glucose = 1 mole of glucose

∴ 60 g of glucose = 60/ 180 = 0.33 moles

Number of molecules in one mole of glucose = 6.023 x 10²³

∴ number of molecules in in 0.33 moles of glucose = 0.33×6.023×10²³ = 1.98 x 10²³

(3) Calculate the volume occupied by 88g of CO₂ at STP.

(ans) Molar Mass of CO₂ = 44

44g of CO₂ = 1 mole of CO₂

∴ 88g of CO₂ = 2 moles of CO₂

One mole of CO2 occupies = 22.4 L at STP

∴ 2 moles of CO2 will occupy = 2× 22.4 = 44.8 L

Volumetric calculations:-

DIFFERENT METHODS FOR EXPRESSING CONCENTRATIONS OF SOLUTIONS

The concentration of a solution expresses the amount of solute dissolved in a given quantity of a solvent.

Qualitatively, concentration can be expressed as :-

(a) Dilute : A solution at contains a relatively small amount of dissolved solute.

(b) Concentrated : A solution that contains a realtively large amount of dissolved solute.

Quantitatively, concentration may be expressed in following ways:-

1. Mass percentage (w/W)

This method expresses the percentage of solute in a given mass of solution.

Eg 10% NaOH means that for every 100g of this solution there will be 10 g of NaOH and 90g of water.

Mass percent = _Mass of the solute(g) × 100

Mass of solution(g)

2. Mass/Volume percentage (w/V)

This method expresses the concentration of solute (in g) per 100mL of solution.

Eg 10% (w/V) glucose solution is prepared by dissolving 10g of glucose in water and diluting it to 100mL.

Mass/Volume percent = _Mass of the solute(g) × 100

Volume of solution(mL)

3. Volume percentage (V/V)

This method is used for solutions that are formulated from rwo liquids. It is the volume of liquid in 100mL of solution.

Eg Alcohol 70% by volume is made by mixing 70mL of alcohol with water to make the total volume 100mL.

Volume percent = _Volume of the liquid(mL) × 100

Volume of solution(mL)

4. Mole fraction :-

It is the ratio of the no of moles of solute to the total no of moles of in solution.

Eg if n_A is the no of moles of solute dissolved in n_B moles of solvent , then

Mole fraction of the solute (x_A) = n_{A___}

n_A + n_B

Mole fraction of the solvent (x_B) = n_{B _____}

n_A + n_B

Ø The sum of mole fractions for all components of a mixture is equal to one.

Ø No of moles of solute = mass of solute in grams ………….[a]

Molar mass of solute

5. Molarity

It is the no. of moles of solute present in one litre of solution.

Eg: 1Molar solution of sodium hydroxide means one mole of NaOH dissolved in one litre of solution. One mole of NaOH is equal to its molar mass in grams.(40g)

Molarity (M) = No. of moles of solute (Unit = mol/L)

Volume of solution (L)

Molarity(M) = Concentration of solute (g/L)

Molar mass of solute

Ø Molarity changes with change in temperature.

6. Molality

It is the no. of moles of solute present in one kilogram of solvent.

Eg : 1Molal solution of sodium hydroxide means one mole of NaOH dissolved in one kg of water.

Molality (m) = No. of moles of solute

Mass of solvent (kg)

Ø Molality does not change with change in temperature.

7. Normality

It is the no. of gram-equivalents of a solute present in one litre of solution.

Normality(N) = Concentration of solute (g/L)

Equivalent mass of solute

8. Relation between Normality and Molarity

Molarity(M) = Concentration of solute (g/L)

Molar mass of solute

Normality(M) = Concentration of solute (g/L)

Equivalent mass of solute

Molarity x Molar Mass = Normality x Equivalent Mass

Molar Mass_____ = n = Normality

Equivalent Mass Molarity

Normality = n × Molarity

Stoichiometry
(Calculation Based On Chemical Equations)
A balanced chemical equation tells the exact mass ratio of the reactants and products in the chemical reaction.
There are three relationships involved for the stoichiometric calculations from the balanced chemical equations which are
1. Mass – Mass

2. Mass – Volume

3. Volume – Volume

1. Mass – Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of the substance involved in the chemical reaction by using a balanced chemical equation.
Example
Calculate the mass of CO₂ that can be obtained by heating 50 gm of limestone.
Solution
Step I – Write a Balanced Equation
CaCO₃ → CaO + CO₂
Step II – Write down the Molecular Masses and Moles of reactant and product
CaCO₃ → CaO + CO₂

100 56 44

Number of moles of 50 gm of CaCO₃ = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO₃ gives 1 mole of CO₂
0.5 mole of CaCO₃ will give 0.5 mole of CO₂
Mass of CO₂ = Moles x Molecular Mass
= 0.5 x×44
= 22 g

2. Mass – Volume Relationship
The major quantities of gases can be expressed in terms of volume as well as masses. According to Avogardro One gm mole of any gas always occupies 22.4 dm³ volume at S.T.P. So this law is applied in mass-volume relationship.
This relationship is useful in determining the unknown mass or volume of reactant or product by using a given mass or volume of some substance in a chemical reaction.
Example
Calculate the volume of CO₂ gas produced at S.T.P by combustion of 20 g of CH₄.
Solution
Step I – Write a Balanced Equation
CH₄ + 2O₂ → CO₂ + 2H₂O
Step II – Write down the Molecular Masses and Moles of reactant and product
CH₄ + 2O₂ → CO₂ + 2H₂O

16 44
Convert the given mass of CH₄ in moles
Number of moles of CH₄ = Given Mass of CH₄ / Molar Mass of CH₄
= 20/16 = 1.25 moles

From Equation
1 mole of CH₄ gives 1 mole of CO₂
1.25 mole of CH₄ will give 1.25 mole of CO₂
No. of moles of CO₂ obtained = 1.25
But 1 mole of CO₂ at S.T.P occupies 22.4 dm³
1.25 mole of CO₂ at S.T.P occupies 22.4 × 1.25 = 28 dm³
Method II
Molecular mass of CH₄ = 16
Molecular mass of CO₂ = 44
According to the equation
16 g of CH₄ gives 44 g of CO₂
1 g of CH₄ will give 44/16 g of CO₂
20 g of CH₄ will give 20 × 44/16 g of CO₂
= 55 g of CO2
44 g of CO₂ at S.T.P occupy a volume 22.4 dm³
1 g of CO₂ at S.T.P occupy a volume 22.4/44 dm³
55 g of CO₂ at S.T.P occupy a volume 55 × 22.4/44
= 28 dm³
3. Volume – Volume Relationship
This relationship determine the unknown volumes of reactants or products from a known volume of other gas.
This relationship is based on Gay-Lussac’s law of combining volume which states that gases react in the ratio of small whole number by volume under similar conditions of temperature & pressure.
Consider this equation
CH₄ + 2 O₂ → CO₂ + 2 H₂O
In this reaction one volume of CH₄ gas reacts with two volumes of oxygen gas to give one volume of CO₂ and two volumes of H₂O
Examples
What volume of O₂ at S.T.P is required to burn 100 litres (dm³) of C₂H₄ (ethylene)?
Solution
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
According to Equation
1 dm³ of C₂H₄ requires 3 dm3 of O2
100 dm³ of C₂H₄ requires 3 × 100 dm³ of O₂
= 300 dm³ of O₂

ISC Chemistry Model Question Papers

Pages

Monday, August 19, 2019

ISC Class XI Model Papers for Chemistry

Monday, February 25, 2019

Physical Chemistry - Solid State

Wednesday, August 1, 2018

Model Question Paper 1 for ISC Chemistry Class XI

Friday, May 11, 2018

Chapter 1- Some Basic Concepts in Chemistry