JEE Syllabus
General topics:
The concept of atoms and molecules;
Dalton's atomic theory;
Mole concept;
Chemical formulae;
Balanced chemical equations;
Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions;
Concentration in terms of mole fraction, molarity, molality and normality.
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Main topics in TMH Book Chapter
ATOM AND MOLECULAR MASSES
LAWS OF CHEMICAL COMBINATION
QUANTITATIVE INFORMATION FROM A CHEMICAL EQUATION
EXPRESSING CONCENTRATION OF A SUBSTANCE
CONVERSION OF CONCENTRATION UNITS
CONCEPT OF EQUIVALENT
BALANCING CHEMICAL EQUATION
RULES TO COMPUTE OXIDATION NUMBER
BALANCING REDOX REACTIONS VIA OXIDATION NUMBERS
BALANCING REDOX REACTIONS VIA ION ELECTRON (OR HALF EQUATION) METHOD
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Sub-topics and concepts
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ATOM AND MOLECULAR MASSES
* Relative Atomic Mass of an Element
Relative Molecular Mass of a Compound
Atomic Mass Unit
Atomic Mass
Molecular Mass
Mole of a Substance
Amount of a Substance: is expressed in moles.
Molar Mass: The average mass per unit amount of a substance.
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LAWS OF CHEMICAL COMBINATION
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Law of Conservation fo Masses
Law of Constant Composition
Law of Multiple Proportion
QUANTITATIVE INFORMATION FROM A CHEMICAL EQUATION
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Stochiometric coefficients or numbers: The numbers which appear before the chemical symbols in a chemical equation.
Chemical equation gives information about moles of various reactants and products. Hence molar masses involved in the reaction and molar masses of products.
EXPRESSING CONCENTRATION OF A SUBSTANCE
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Mass percentage of substance in a system
Mole fraction of a substance in a system
Molarity = Amount of a substance (in mol)/Volume of solution expressed in dm^3
It is applicable to solutions only.
The unit of molarity is mol dm^-3. It is commonly abbreviated by the symbol M and is spelled as molar.
Molality = Amount of a a substance (in mol)/Mass of solvent expressed in kg
It is also applicable to solutions only
CONVERSION OF CONCENTRATION UNITS
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Mole fraction into molarity
Mole fraction inot molality
Molality not molarity
Molarity into mole fraction
Molality into mole fraction
Molarity into molality
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CONCEPT OF EQUIVALENT
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In TMH Book it is given: "One equivalent of a substance in a reaction is defined as the amount of substance which reacts or liberates 1 mol of electrons (or H^+ or OH^- ions). In a reaction, a substance always reacts with another substance in equivalent amounts."
More detailed treatment is given in Madan and Bisht, ISC Chemistry.
Equivalent weight of a substance (element or compound) is defined as 'the number of parts by weight of it, that will combine with or displace directly or indirectly, 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen, 35.5 parts by weight of chlorine, 108 parts by weight of silver or the equivalent parts by weight of any other elements."
The meaning will be more clear as we study experiments that determine equivalent weights. Since substances react in the ratio of their equivalent weights, their knowledge and experimental determination is extremely important. A number of methods, with different methods being applicable to different materials are available. Two methods are mentioned below.
a. Hydrogen displacement method: The equivalent weight of metals like calcium, zinc, tin, magnesium etc., which react with dilute acids to produce hydrogen can be determined by this method. In this method, a known quantity of the metal is treated with excess of the dilue sulphuric acid or hydrochloric acid till whole of the metal disappears by way of reaction with the acid. the volume of hydrogen gas so produced is determined and reduced to STP. The weight of the hydrogen at STP is determined from the volume of hydrogen at STP by making use of molar volume concept. From the weight of hydrogen produced and the weight of metal taken for the experiment, equivalent weight of the metal is determined.
Example: o.205 g of a metal on treatment with a dilute acid gave 106.6 ml of hydrogen(dry) at 740.6 mm pressure and 17 degrees centrigrade temperature. Calculate the equivalent weight of the metal.
volume of hydrogen at 740.6 mm pressure and 17 degree centigrade (290 K) = 106.6 ml
Volume of hydrogen at STP 760 mm and 273 K = (740.6/290)*(273/760)*106.6
= 97.72 ml
22.4 litres of H-2 = 22400 ml of H-2 at STP weighs 2g
So 97.72 ml of hydrogen weights (97.72/22400)*2 g
0.205 g of metal gives (97.72/22400)*2 g = .008725 g
To get one g of hydrogen, metal required = (1/.008725)*.205 = 23.5 g
Hence equivalent weight of metal = 23.5
23.5 parts of metal by weight react with 1 part by weight of hydrogen.
b. Oxide Formation Method
The equivalent weight of metals like copper, magnesium, mercury, zinc etc., which form their oxides relatively easily can be found by this method.
In this method a known weight of the metal is taken an converted into its oxide say by heating the metal in the atmosphere of oxygen.
Calculation: Let the weight of the metal = x g.
Let the weight of oxide formed = y g
Therefore y - x g of oxygen combines with x g of the metal.
1 g of oxygen combines with [1/(y-x)]*x g of metal
8 g of oxygen combines with [1/(y-x)]*8x g of metal. this quantity is the equivalent weight of metal.
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BALANCING CHEMICAL EQUATION
RULES TO COMPUTE OXIDATION NUMBER
BALANCING REDOX REACTIONS VIA OXIDATION NUMBERS
BALANCING REDOX REACTIONS VIA ION ELECTRON (OR HALF EQUATION) METHOD
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web sites
Calculations - Mole concept and some more
http://www.docbrown.info/page04/4_73calcs.htm
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JEE Question 2007 paper I Linked Comprehension
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 ×10^23 ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry electrochemistry and radiochemistry. The following example illustrates a typical case, involving
chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200;1 Faraday = 96500 coulombs).
1. The total number of moles of chlorine gas evolved is
(A) 0.5
(B) 1.0
(C) 2.0
(D) 3.0
Answer: B
2. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is
(A) 200
(B) 225
(C) 400
(D) 446
Answer: D
In presence of Hg cathode sodium ion will discharge in place of hydrogen gas due to over voltage in the form of amalgams.
Weight of amalgam = 2 × (23 + 200) = 446 g
3. The total charge (coulombs) required for compete electrolysis is
(A) 24125
(B) 48250
(C) 96500
(D) 193000
Answer: D
Total charge = 2 × 96500 = 193000 C
This question is better answered after studying the electro chemistry chapter.
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JEE Question 2007 paper II
Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is
(A) 3
(B) 4
(C) 5
(D) 6
Answer D
From Compounds of metals chapter
Mohr's salt is a double salt. FeSO-4.(NH-4)-2SO-4.6H-2O
It is obtained by mixing freshly prepared ferrous sulphate in solution with equal molar amounts of ammonium sulphate and then allow the solution to crystallize.
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