Some Basic Concept of Chemistry

Some Basic Concepts of Chemistry NEET Chemistry PDF Notes | High-Yield PYQs

Topic 1: Basic Constituents of Matter

Subtopic A: Atoms and Molecules

Definition: Chemistry is the science of atoms and molecules. It deals with the composition, structure, and properties of matter.
Fundamental Particles: Collectively, atoms, molecules, ions, electrons, protons, and neutrons are known as fundamental particles.

Topic 2: Atomic and Molecular Masses

Subtopic A: Atomic Weight (Relative Atomic Weight)

Core Formula: Atomic Weight = Actual mass of one atom of element / (1/12th mass of one atom of Carbon-12)
Crucial Detail: Relative Atomic Weight is a pure number, without any unit.
Atomic Mass Unit (amu): The quantity 1/12th mass of an atom of C¹² is known as 1 amu.
1 amu = 1.67 × 10^-24 g = 1.67 × 10^-27 kg.
Actual Mass: Actual mass of one atom = Atomic weight × amu.

Subtopic B: Molecular Weight and Average Atomic Mass

Molecular Weight: Relative molecular weight is also a pure number without units.
Actual Mass of Molecule: Actual mass = Molecular weight × amu.
Average Atomic Mass: Calculated using isotopic composition: M_avg = Σ (Isotopic Mass × % Abundance) / 100.

PYQ Problem:
An element X, has the following isotopic composition: ²⁰⁰X: 90%; ¹⁹⁹X: 8.0%; ²⁰²X: 2.0%. The weighed average atomic mass of the naturally occurring element X is closest to: AIPMT 2007
(1) 201 amu (2) 202 amu (3) 199 amu (4) 200 amu

Solution:
Step 1 Formula Application: M_avg = ((200 × 90) + (199 × 8) + (202 × 2)) / 100
Step 2 Calculation: M_avg = (18000 + 1592 + 404) / 100 = 19996 / 100 = 199.96
Conclusion: The closest value is 200 amu, making (4) the correct option.

PYQ Problem:
Boron has two stable isotopes, ¹⁰B (19%) and ¹¹B (81%). Calculate average atomic weight of boron in the periodic table: AIPMT 1990
(1) 10.80 (2) 10.2 (3) 11.2 (4) 10.0

Solution:
Step 1 Formula Application: M_avg = ((10 × 19) + (11 × 81)) / 100
Step 2 Calculation: M_avg = (190 + 891) / 100 = 10.81
Conclusion: The correct option is (1).

Topic 3: Mole Concept

Subtopic A: Definition and Key Formulas

Definition: A mole is the amount of substance containing as many entities as there are atoms in exactly 0.012 kg (12 g) of the Carbon-12 isotope.
Avogadro's Constant (N_A): 6.022137 × 10²³ entities/mol.
Formula - Moles (n):
n = Weight (g) / Atomic or Molecular Weight.
n = Volume of gas at STP/NTP (L) / 22.4.
n = Number of particles / N_A.
Atomicity: Total number of atoms in one molecule. Mole of atoms = Atomicity × mole of molecules.

PYQ Problem:
Equal masses of H₂, O₂ and methane have been taken in a container of volume at temperature 27°C in identical conditions. The ratio of the volumes of gases H₂: O₂: methane would be: NEET 2014
(1) 8:16:1 (2) 16:8:1 (3) 16:1:2 (4) 8:1:2

Solution:
Step 1 Avogadro's Principle: Volume ratio is equal to the mole ratio at constant T and P. Let mass be w.
Step 2 Molar Masses: H₂ = 2, O₂ = 32, CH₄ = 16.
Step 3 Calculation: n_H2 : n_O2 : n_CH4 = w/2 : w/32 : w/16. Multiply by 32.
Step 4 Result: 16 : 1 : 2.
Conclusion: The correct option is (3).

PYQ Problem:
The highest number of helium atoms is in NEET 2020
(1) 4 g of helium (2) 2.271098 L of helium at STP (3) 4 mol of helium (4) 4 u of helium

Solution:
Step 1 Evaluate Option 1: 4 g He = 1 mole = N_A atoms.
Step 2 Evaluate Option 2: 2.27 L at STP ≈ 0.1 mole = 0.1 N_A atoms.
Step 3 Evaluate Option 3: 4 mol He = 4 N_A atoms.
Step 4 Evaluate Option 4: 4 u He = 1 atom of He.
Conclusion: 4 moles contain the highest number of atoms. Option (3) is correct.

PYQ Problem:
Which one of the followings has maximum number of atoms? NEET 2020
(1) 1 g of Mg(s) [Atomic mass of Mg = 24] (2) 1 g of O₂(g) [Atomic mass of O = 16] (3) 1 g of Li(s) [Atomic mass of Li = 7] (4) 1 g of Ag(s) [Atomic mass of Ag = 108]

Solution:
Step 1 Identify Formula: Number of atoms = (Mass / Molar Mass) × Atomicity × N_A.
Step 2 Calculate for each:
Mg: 1/24 × N_A
O₂: 1/32 × 2 × N_A = 1/16 × N_A
Li: 1/7 × N_A
Ag: 1/108 × N_A
Step 3 Comparison: The smallest denominator gives the largest value. Li (7) is the smallest.
Conclusion: 1 g of Li has the maximum atoms. Option (3) is correct.

PYQ Problem:
In which case is number of molecules of water maximum? NEET 2018
(1) 18 mL of water (2) 0.18 g of water (3) 10^-3 mol of water (4) 0.00224 L of water vapours at 1 atm and 273 K

Solution:
Step 1 Evaluate Option 1: Density of water = 1 g/mL. Mass = 18 g = 1 mole = N_A molecules.
Step 2 Evaluate Option 2: 0.18 g = 0.01 mole = 0.01 N_A molecules.
Step 3 Evaluate Option 3: 10^-3 mole = 0.001 N_A molecules.
Step 4 Evaluate Option 4: 0.00224 L at STP = 0.00224 / 22.4 = 10^-4 moles.
Conclusion: 18 mL of water equals 1 mole, which is the maximum. Option (1) is correct.

Subtopic B: Gram Atomic and Gram Molecular Mass

Definition: Numerical value of atomic/molecular mass expressed in grams.
Equivalent terms: Gram atomic mass = mass of 1 gram atom = mass of 1 mole atom = mass of N_A atoms.
Crucial detail: "1 gram molecule" means 1 mole of molecules, NOT a mass of 1 gram.

Topic 4: Vapour Density (V.D.)

Subtopic A: Concept and Relation to Molecular Weight

Definition: Ratio of density of gas and hydrogen at the same temperature and pressure.
Formula: V.D. = Density of gas / Density of H₂.
Key Relationship: Molecular Weight (gas) = 2 × V.D.
Constants: Density of H₂ gas at NTP = 0.000089 g/mL. V.D. is a unitless term.

PYQ Problem:
0.24 g of a volatile gas, upon vaporisation, gives 45 mL vapour at NTP. What will be the vapour density of the substance? (Density of H₂ = 0.089 g/L): AIPMT 1996
(1) 95.93 (2) 59.93 (3) 95.39 (4) 5.993

Solution:
Step 1 Definition of V.D: V.D. = Mass of certain volume of gas / Mass of same volume of H₂.
Step 2 Calculate mass of 45 mL H₂: Mass = Volume × Density = 0.045 L × 0.089 g/L = 0.004005 g.
Step 3 Calculate V.D.: V.D. = 0.24 / 0.004005 = 59.93.
Conclusion: Option (2) is correct.

Topic 5: Percentage Composition, Empirical & Molecular Formula

Subtopic A: Percentage Formula

Formula: % mass = (Number of atoms × atomic mass / molecular mass) × 100.

Subtopic B: Empirical and Molecular Formula

Empirical Formula (E.F.): Expresses the simplest whole-number ratio of atoms in 1 molecule.
Molecular Formula (M.F.): Represents the actual number of atoms present in 1 molecule.
Relationship: M.F. = n × E.F.
n = Molecular formula mass / Empirical formula mass.
Steps for Empirical Formula Determination:
1. Find % by weight of each element.
2. Divide % by atomic weight to get atomic ratio.
3. Divide each atomic ratio by the minimum value to get the simplest ratio.
4. If fractional, multiply by a suitable integer to make it a whole number.
5. Write the Empirical formula using this ratio.

PYQ Problem:
An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C = 12, H = 1] NEET 2021
(1) CH (2) CH₃ (3) CH₂ (4) CH₄

Solution:
Step 1 Mass Percentages: Carbon = 78%, Hydrogen = 22%.
Step 2 Atomic Ratios: C = 78 / 12 = 6.5. H = 22 / 1 = 22.
Step 3 Simplest Ratio: Divide by 6.5. C = 6.5 / 6.5 = 1. H = 22 / 6.5 ≈ 3.38.
Step 4 Whole Number Conversion: Multiply by 3. C = 3, H = 10 (Wait, standard alkane). Let's re-evaluate simplest integer ratio for standard compounds. 78/12 = 6.5, 22/1=22. Ratio = 1 : 3.38. Closest empirical fit given options: if CH₃, C=12, H=3 (80% C). if CH, C=12, H=1 (92%). The closest integer approximation for 78% C is CH₃.
Conclusion: Option (2) is correct.

PYQ Problem:
The percentage of C, H and N in an organic compound are 40%, 13.3% and 46.7% respectively then empirical formula AIPMT 2007
(1) CH₂N (2) CH₄N (3) CH₅N (4) CH₃N

Solution:
Step 1 Moles: C = 40/12 = 3.33. H = 13.3/1 = 13.3. N = 46.7/14 = 3.33.
Step 2 Simplest Ratio: C = 3.33/3.33 = 1. H = 13.3/3.33 ≈ 4. N = 3.33/3.33 = 1.
Step 3 Formula: CH₄N.
Conclusion: Option (2) is correct.

Topic 6: Stoichiometry and Stoichiometric Calculations

Subtopic A: Basic Stoichiometry

Concept: A balanced chemical equation gives quantitative relationships in moles, masses, molecules, and volumes.
Crucial Detail: Mass is not expressed in the ratio of stoichiometric coefficients (only moles/volumes are).

Subtopic B: Limiting Reagent (L.R.) Concept

Definition: The reactant which is completely consumed in a reaction is the Limiting Reagent.
Identification Method: Calculated when more than one initial quantity is given.
Formula for checking L.R.: Least value of (Given value (moles, vol, or molecules) / Stoichiometric Coefficient) indicates the L.R.

Subtopic C: Combustion Reactions

Balancing order: First balance C atoms, then H atoms, finally balance Oxygen atoms.

PYQ Problem:
When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g), each at STP, the moles of HCl(g) formed is equal to: NEET 2014
(1) 2 mol of HCl(g) (2) 0.5 mol of HCl(g) (3) 1.5 mol of HCl(g) (4) 1 mol of HCl(g)

Solution:
Step 1 Balanced Equation: H₂(g) + Cl₂(g) → 2HCl(g).
Step 2 Moles Given: Moles of H₂ = 22.4/22.4 = 1 mol. Moles of Cl₂ = 11.2/22.4 = 0.5 mol.
Step 3 Identify L.R.: For H₂: 1/1 = 1. For Cl₂: 0.5/1 = 0.5. Cl₂ is the limiting reagent.
Step 4 Calculation: 1 mol Cl₂ gives 2 mol HCl. So, 0.5 mol Cl₂ gives 0.5 × 2 = 1 mol HCl.
Conclusion: Option (4) is correct.

Topic 7: Laws of Chemical Combinations

Subtopic A: The Five Laws

(a) Law of Mass Conservation: Given by Lavoisier. Mass can neither be created nor destroyed in a chemical/physical reaction. Total mass reactants = Total mass products + Unreacted mass.
(b) Law of Definite Proportion: Given by Proust. A compound obtained from different sources has the same mass ratio of each component.
(c) Law of Multiple Proportion: Given by Dalton. When two elements form >1 compound, different masses of one combining with fixed mass of another bear a simple ratio.
(d) Law of Gaseous Volume: Given by Gay Lussac. In gaseous reactions, reactants combine in simple volume ratios at same T, P. Note: Relates volume to moles, not mass.
(e) Avogadro's Law: Equal volumes of all gases contain equal number of molecules at same temperature and pressure.

Topic 8: Equivalent Weight

Subtopic A: Calculation Methods

Definition: Parts by mass of a substance that combine with/displace 1.008 parts H, 8 parts O, 35.5 parts Cl, or 108 parts Ag.
Formulas:
Eq. wt. (element) = Atomic weight / Valency factor.
Eq. wt. (acid/base) = Molecular weight / Basicity or Acidity.
Eq. wt. (salt) = Molecular weight / Total charge on cation/anion.

Subtopic B: Methods for Determination

Method	Formula	Application
Hydrogen displacement	E = (W_metal × 1.008) / W_H2	Active metals
Oxide formation	E = (W_element × 8) / W_oxygen	Direct/indirect oxidation
Chloride formation	E = (W_element × 35.5) / W_chlorine	Direct/indirect chlorination
Metal displacement	m₁ / E₁ = m₂ / E₂	Active metal displaces less active
Silver salt	E = (W_RCOOAg × 108) / W_Ag	Organic (carbonic) acids

Subtopic C: Law of Chemical Equivalence

Principle: In a reaction, equal number of gram equivalents of reactants react to give equal number of gram equivalents of products.
Number of gram equivalents = Weight / Equivalent Weight = Moles × Valency Factor.

Topic 9: Determination of Atomic and Molecular Weight

Subtopic A: Atomic Weight Determination

Dulong and Petit's Law: Applicable to solids (except Be, B, Si, C).
Atomic mass × specific heat (Cal g^-1 °C^-1) ≈ 6.4.
Law of Isomorphism: Isomorphous substances form crystals of same shape/size (e.g., ZnSO₄ · 7H₂O and MgSO₄ · 7H₂O).
Volatile Chloride Method: x = (2 × V.D.) / (E + 35.5).

Subtopic B: Molecular Weight Determination

Victor Mayer's method: Used to determine molecular weight of volatile compounds.

Topic 10: Concentration Terms

Subtopic A: Essential Definitions

Solution: Homogeneous mixture of ≥ 2 chemically non-reacting substances.
Dilute: Small amount of solute.
Concentrated: Large amount of solute.

Subtopic B: Formulas and Relationships

Normality (N): N = Gram equivalents of solute / Volume of solution (L).
Relation: N = M × n-factor.
Molarity (M): M = Moles of solute / Volume of solution (L).
Molality (m): m = Moles of solute / Weight of solvent (kg). Independent of temperature.
Mole Fraction (X): X_B = n / (n + N). Sum of mole fractions = 1.
Relation to Molality: m = (X_B × 1000) / (X_A × M_A).
Percent by Mass (w/W): (Mass of solute / Mass of solution) × 100.
Parts Per Million (ppm): (Mass of solute / Mass of solution) × 10⁶.

PYQ Problem:
The density of 1 M solution of a compound 'X' is 1.25 g/mL. The correct option for the molality of solution is (Molar mass of compound X = 85 g): NEET 2023
(1) 0.705 m (2) 1.200 m (3) 1.165 m (4) 0.855 m

Solution:
Step 1 Extract Given Data: M = 1 mol/L, Density (d) = 1.25 g/mL. Molar Mass (M_w) = 85 g/mol.
Step 2 Mass of Solution: 1 L of solution = 1000 mL × 1.25 g/mL = 1250 g.
Step 3 Mass of Solute: 1 mole × 85 g/mol = 85 g.
Step 4 Mass of Solvent: 1250 - 85 = 1165 g = 1.165 kg.
Step 5 Molality Calculation: m = Moles of solute / Mass of solvent (kg) = 1 / 1.165 ≈ 0.858 m.
Conclusion: Option (4) is closest and correct.

Topic 11: Significant Figures

Subtopic A: Rules for Calculation

Rule for Multiplication/Division: The final result should be reported having the same number of significant figures as the original number with the least number of significant figures.
Zeroes:
All non-zero digits are significant.
Non-zero digits to the right of the decimal point are significant.
Zeroes to the left of the first non-zero digit are not significant.

PYQ Problem:
The density of the solution is 2.15 g/mL, then mass of 2.5 mL solution in correct significant figures is: NEET 2022
(1) 53.75 g (2) 5375 × 10^-3 g (3) 5.4 g (4) 5.38 g

Solution:
Step 1 Formula: Mass = Density × Volume.
Step 2 Calculation: 2.15 g/mL × 2.5 mL = 5.375 g.
Step 3 Significant Figures Rule: The number with the least sig figs is 2.5 (2 sig figs). Therefore, the answer must be rounded to 2 significant figures.
Step 4 Rounding: 5.375 rounded to 2 sig figs is 5.4 g.
Conclusion: Option (3) is correct.

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