General Topics in Chemistry Chapter - Chemistry JEE Advanced

General Topics in Chemistry is an integral segment in the JEE Advanced syllabus, encompassing a wide array of fundamental concepts and principles in the realm of chemistry. This chapter serves as the cornerstone, providing students with a comprehensive understanding of the basic laws, theories, and principles that govern the behavior of matter and its transformations. It covers topics such as stoichiometry, atomic structure, chemical bonding, chemical kinetics, and thermodynamics. Aspiring engineers and scientists preparing for JEE Advanced will find this chapter to be a vital stepping stone, laying the groundwork for advanced topics and serving as a bridge to more complex chemical concepts and applications.

Important Topics of General Topics in Chemistry

• Stoichiometric Concepts

• Mole Concept

• Limiting Reagent Concept

• Oxidation and Reduction

• Gravimetric Analysis

• Volumetric Analysis

Stoichiometric Concepts

• Stoichiometry is the science of calculating the amounts of different reactants and products in a chemical process. 

• Stoichiometric coefficients are the numbers used to balance a chemical equation. 

• Stoichiometric calculation refers to the use of these numerals to solve issues based on chemical equations.

• The mole relationship between different reactants and products is necessary in stoichiometric calculations since the mass-mass, mass-volume, and volume-volume relationships between different reactants and products may be determined from them.

• Gravimetric Analysis and Volumetric Analysis are two approaches for doing stoichiometric calculations. 

• The amount of chemical species is determined in the first approach by measuring mass, whereas in the second way, it is calculated by measuring volume. 

• Stoichiometric calculations can be performed using two key concepts: (i) Mole concept and (ii) the Equivalent concept.

Mole Concept

• A mole (symbol mol) is the amount of stuff that includes the same number of atoms, molecules, ions, electrons, or other elementary entities as there are carbon atoms in 12 grams of 12C. 

• Avogadro's number is the number of atoms in 12 grams of 12C. (N0). 

• N0 = 6.023 x 1023 is the value reported for N0.

• Depending on the values supplied, the number of moles of a chemical can be computed using the following methods:
  1. Number of moles of molecules: (Weight in gram)/(Molecular Mass)
  2. Number of moles of atoms: (Weight in gram)/(Atomic Mass)
  3. NUmber of moles of gases: (Volume at STP)/(Standard Molar Volume)
  4. Number of moles of particles: (Number of particles)/(Avogadro Number).
  5. Number of moles of solute = Molarity x Volume of solution in litres. 

Principle of Atom Conservation

• Atoms are conserved in chemical reactions, hence moles of atoms must be conserved as well. This is known as the atomic conservation principle. 

• When balancing the chemical equation, we can break it down into a series of mathematical equations using POAC. 

• Let’s take the reaction,
  KClO3(s) → KCl(s) + O2(g) .
  - Applying POAC for ‘K’ atoms,
  ∴ Moles of K atoms in KClO3(s) = Moles of K atoms in KCl(s).
  - As 1 mole of KClO3 contains 1 mole of K atoms. In the same manner, 1 mole of KCl contains 1 mole of K atoms. 

• It's worth noting that POAC can be used for atoms that are preserved during a chemical reaction.

Limiting Reagent Concept

• In reactions involving more than one reactant, the limiting reagent must be identified first, out of all the reactants that are entirely utilised (limiting reagent). 

• Knowing the initial amount of equivalents or milli equivalents of each reactant can also help determine the limiting reactant. 

• The limiting reagent is the one with the fewest equivalents or milli equivalents. 

• Equivalent approaches for determining the limiting reactant do not necessitate chemical equation balancing.

Gravimetric Analysis

• In general, when performing necessary computations, the stages are as follows:

1. For the chemical change, write down a balanced molecular equation.

2. Write the number of moles of each reactant and product below the formula.

3. Using the formula underneath the corresponding formula, write out the relative masses of the reactants and products. These are the theoretical reactant and product amounts.

4. The unknown factor or factors are determined using the unitary technique, mole concept, or proportionality method.

• In general, we do not acquire the theoretical amount of product while performing a reaction in the laboratory. 

• The actual yield is the amount of product that is really obtained. 

• The percentage yield can be estimated using the actual and theoretical yields as follows:
  %yield = (actual yield)/(theoretical yield)

Oxidation and Reduction

• The acquisition of oxygen, the loss of hydrogen, the loss of electrons (de-electronation), and the growth of O.N. are all examples of oxidation. 

• Loss of oxygen, the gain of hydrogen, the gain of electrons, and the fall in oxidation number are all examples of reduction.

• Now that you understand what oxidation-reduction is, you're ready to learn how to balance a redox (oxidation-reduction) reaction.

• This is significant since redox reactions are at the heart of many stoichiometry difficulties.

• Oxidation-reduction is a chemical reaction that occurs in a variety of ways. 

• A reaction in which electrons are removed from an atom is called oxidation, while a reaction in which electrons are added to an atom is called reduction.

• The concept of oxidation state is required to characterise these changes. 

• The oxidation state of each atom is believed to be the charge on each ion in ionic species. In NaCl, for example, Na occurs as Na+ and Cl as Cl-. As a result, the oxidation state of Na+ in NaCl is +1, while the oxidation state of Cl- is -1.

• In covalent molecules, however, the oxidation state or oxidation number is defined as the charge that an atom would have in a molecule if all of the bonds connected with that atom were entirely ionic.

Volumetric Analysis

• Volumetric analysis is the process of determining the strength of an unknown solution by comparing it to a known solution (Titration). 

• Acid-base titration (neutralisation reaction), redox-titration, iodometric titrations, precipitation reactions, and other types of titrations exist. 

• The important concepts of volumetric analysis are discussed below:
  Molarity: The number of moles of solute present in one litre of a solution is known as its molarity.
  $M = \frac{Weight \ of \ solute \times 1000}{Molecular \ Weight \ of \ Solute \times Volume \ of \ solution}$
  Normality: The number of equivalents of solute present in one litre of solution is known as the normalcy.
  $N = \frac{Weight \ of \ solute \times 1000}{Equivalent \ Weight \ of \ Solute \times Volume \ of \ solution}$

Molality: The amount of moles of solute present in one kilogramme of solvent is known as molality.
$N = \frac{Weight \ of \ solute \times 1000}{Molecular \ Weight \ of \ Solute \times Weight \ of \ solution}$

Relationship between Molarity and Molality
- $m = \frac{Molarity \ of \ solution \times 1000}{1000 \times \left ( Density \ Weight \ of \ Solute \ - Molarity \ of\ solution \right ) \times mass \ of \ solute} $

JEE Advanced General Topics in Chemistry Solved Examples

Question 1: A 2.21 gm sample of CaCO3 and MgCO3 is burned to a constant weight of 1.152 gram. What is the mixture's composition? Calculate the volume of CO2 released at 0 degrees Celsius and 76 pounds per square inch of pressure.

Solution: 

• Two compounds CaCO3 and MgCO3 are burnt. The two reactions are given as:
  CaCO3 → CaO + CO2;
  MgCO3 → MgO + CO2.

• Let CaCO3 be x gm and MgCO3 be y gm.
  ∴ The total = x gm + y gm = 1.152 gm.

• Now, 100 gm of CaCO3 gives 56 gm of CaO. (Mol. Wt. of CaCO3 = 100 gm and CaO = 56 gm)
  ∴ x gm of CaCO3 gives = (56/100)*x gm of CaO. 

• Similarly, 84 gm of MgCO3 gives 40 gm of MgO. (Mol. Wt. of MgCO3 = 84 gm and MgO = 40 gm)
  ∴ y gm of MgCO3 gives = (40/84)*y gm of MgO. 

• Hence, the weight of residue = (56x/100) + (40y/84) = 1.152 gm.
  On solving the two x and y containing equations, we get,
  X = 1.10 gm, y = 1.02 gm.

• Moles of CO2 formed from both the reactions = Moles of CaCO3 + Moles of MgCO3.

So, Moles of CO2 = 0.0241 gm.

• ∴ Volume of CO2 at STP = 0.0241 x 22.4 litre = 539.8 ml.

Key points to remember: The relationship between the number of moles and molecular weights and the number of moles and volume at STP. 

Question 2: After treating a 0.56 gm mixture of KBr and NaBr with an aqueous solution of Ag+, the bromide ion was recovered as 0.97 gm of pureAgBr. What was the KBr concentration in the sample?

Solution: 

• The equation for the above reaction is given below:
  KBr + NaBr + Ag+ → AgBr.

• Let the amount of KBr be x gm. Hence, the amount of NaBr will be (0.56 - a) gm.

• Applying the POAC as mentioned above for Br atoms,
  Moles of Br in KBr + Moles of Br in NaBr = Moles of Br in AgBr.
  ∴ (a/119) + (0.59-a/103) = 0.97/188.
  ∴ a = 0.1332 gm.

• Hence, the percentage of KBr in the sample = (0.1332/0.562)*100 = 23.78 gm.

Key Points to remember: Remember the POAC relation and concept. 

Solved Examples from Previous Year Question Papers

Question 1: The concentrated sulphuric acid that is peddled commercially is 95% H2SO4 by weight. If the density of this commercial acid is 1.834 g cm-3, the molarity of this solution is:

(1) 17.8 M

(2) 15.7 M

(3) 10.5 M

(4) 12.0 M

Solution: 

• The given density in the question is = 1.834

• Now, 1 ml solution contains 1.834 g.

∴ 1000 ml solution will contain 1834 g.

• 95% H2SO4 means 100 gm contain 95 gm H2SO4

• Mass of solute = (95/100)×1834

• Molecular weight of H2SO4 = 98

• Molarity = No. of moles/ volume = mass of solute/98

= (95/100)×(1834/98)

= 17.8 M

• Hence option (1) is the answer.

Question 2: When CO2(g) is passed over red hot coke, it partially gets reduced to CO(g). Upon passing 0.5 litres of CO2(g) over red hot coke, the total volume of the gases increased to 700 mL. The composition of the gaseous mixture at STP is:

(1) CO2 = 200 mL: CO = 500 mL

(2) CO2 = 350 mL: CO = 350 mL

(3) CO2 = 0.0 mL: CO = 700 mL

(4) CO2 = 300 mL: CO = 400 mL

Solution: 

• The reaction for the given question is: CO2(g) + C(s) → 2CO(g)

• Now, Total volume = 700 ml = 0.7 L

• Hence, 0.5 + x = 0.7

• Therefore, x = 0.2L = 200 mL

• That's why CO2(g) = 0.5 - 0.2 = 300ml

• And CO(g) = 2x = 400 mL

• Hence option (4) is the answer.

Question 3: An open vessel at 300 K is heated till ⅖ th of the air in it is expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated is :

(1) 750 K

(2) 400 K

(3) 500 K

(4) 1500K

Solution: 

• At constant V and P, n1T1 = n2T2

• n1 = n

• n2 = n-2n/5 = 3n/5

• T1 = 300 K

• 300n = (3n/5) T2

• T2 = 300 × 5/3 = 500 K

• Hence option (3) is the answer.

Practice Questions

Question 1: An element's equivalent weight is 4. The vapour density of its chloride is 59.25. Determine the element's valency.
(a) 4
(b) 3

(c) 2

(d) 1

Answer: (b) 3

Question 2: Two metallic oxides contain 27.6% and 20% oxygen, respectively. If the first oxide's formula is M3PO4, the second's formula will be:

(a) MO

(b) MO2

(c) M2O5

(d) M2O3

Answer: (d) M2O3

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Conclusion

Chemistry is a discipline of science concerned with the atomic and molecular level study of matter and its properties. It is concerned with comprehending the nature of components. We are made of matter and are surrounded by it. We are surrounded by chemistry, and we are surrounded by chemistry. Every second, our bodies go through a slew of chemical reactions. The periodic table is the foundation of elemental chemistry. To have a deeper knowledge of our nature, it is critical to comprehend the fundamental concepts of chemistry.