Some Basic Concepts of Chemistry for NEET UG

The mole is chemistry's counting unit. Once you can convert between mass, moles and particles, and balance equations, every quantitative chemistry problem becomes manageable.

ChemistryClass 1174 MCQs3 lessons12 flashcards4 memory tricks

Practise Some Basic Concepts of Chemistry free

What you'll learn

  1. The mole and molar mass
    Convert between mass, moles, number of particles and gas volume.
  2. Empirical and molecular formulas
    Find the empirical formula from percentage composition, and the molecular formula from molar mass.
  3. Stoichiometry and the limiting reagent
    Use a balanced equation to find amounts of products and identify the limiting reagent.

Key formulas

Moles from massn = mass/molar mass
ParticlesN = n × N_A
Gas volume at STP (273 K, 1 atm)V = n × 22.4 L
MolarityM = moles of solute / volume of solution (L)
Molalitym = moles of solute / mass of solvent (kg)
Mole fractionx_A = n_A/(n_A + n_B)

Memory tricks

Limiting reagentAt a pav bhaji stall, the number of plates depends on whichever runs out first, the pav or the bhaji.

The limiting reagent is used up first and decides how much product forms. Divide the moles of each reactant by its coefficient; the smallest answer is the limiting one.

Empirical versus molecular formulaEmpirical formula is the family recipe ratio; molecular formula is the full wedding batch.

Glucose: empirical CH₂O, molecular C₆H₁₂O₆. Molecular formula = n × empirical formula, where n = molar mass ÷ empirical-formula mass.

Molarity versus molalityMolarity uses litres, and volume swells in a Delhi summer. Molality uses kilograms, which don't care about the weather.

Molarity = mol per litre of solution, which changes with temperature. Molality = mol per kg of solvent, which does not.

The moleA mole is the chemist's dozen, just a very, very big one.

1 mol = 6.022 × 10²³ particles, and one mole of a gas at STP occupies 22.4 L (22.7 L at the 1 bar standard).

Try 3 NEET-style questions

Q1. The mass of water produced by the complete combustion of 16 g of methane (CH₄ + 2O₂ → CO₂ + 2H₂O) is

  1. 36 g
  2. 18 g
  3. 44 g
  4. 16 g
Show answer

Answer: (1) 36 g

16 g of CH₄ is 1 mol. It produces 2 mol of H₂O, which is 2 × 18 = 36 g.

Q2. The mass of one atom of carbon-12 is about (N_A = 6.022 × 10²³ mol⁻¹)

  1. 1.99 × 10⁻²³ g
  2. 12 g
  3. 7.2 × 10²⁴ g
  4. 1.66 × 10⁻²⁴ g
Show answer

Answer: (1) 1.99 × 10⁻²³ g

Mass of one atom = 12 g / 6.022 × 10²³ ≈ 1.99 × 10⁻²³ g.

Q3. The molar mass of water (H₂O) is

  1. 16 g/mol
  2. 18 g/mol
  3. 20 g/mol
  4. 2 g/mol
Show answer

Answer: (2) 18 g/mol

M(H₂O) = 2 × 1 + 16 = 18 g/mol.

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