Mole Concept in ONE SHOT - Concepts, Tricks & PYQ For NEET 2026

What is the Mole Concept in Chemistry for NEET 2026?

The Mole Concept is used to measure the amount of a substance. It provides a connection between the atoms and molecules, with the grams and litres that we can measure in our lab.

A mole is defined as the amount of a substance that contains 6.022 × 10²³ particles (atoms, molecules, or ions). This number is known as Avogadro’s number. For example, 1 mole of carbon atoms contains 6.022 × 10²³ carbon atoms.

Avogadro’s Number and Definition of Mole (NEET Important)

One mole of any substance contains exactly 6.022 × 10²³ elementary entities (atoms, molecules, ions, electrons, etc.). This number is called Avogadro's Number (Nₐ)

• Avogadro's Number (Nₐ)= $6.022 \times 10^{23}$

• 1 Mole = Nₐ

Molar Mass, Atomic Mass and Basic Terms in Mole Concept

The molar mass of a substance is the mass of one mole of that substance, unit g/mol. Numerically, it equals the molecular or atomic mass in amu.

Important Mole Concept Formulas for NEET 2026

Here are the most important mole concept formulas for NEET 2026 that help in solving numerical questions quickly and accurately, especially in topics like stoichiometry, concentration, and gas calculations.

1. Number of Moles (n) , $n=\frac{w}{M}$ where w= given mass and M = molarmass

2. Number of Entities (N), $N=n \times N_A=\frac{w}{M} \times 6.022 \times 10^{23}$

3. Moles from Volume (Gas at STP), $n=\frac{V}{22.4}$

Where: $V=$ volume of gas at STP (in litres)

4. Moles of Electrons / Charge, $n=\frac{Q}{96500}$, where Q = charge in coulombs, 96500 C = 1 Faraday

5. Percentage Composition, $\%$ of element $=\left(\frac{\text { Mass of element in } 1 \text { mole of compound }}{\text { Molar mass of compound }}\right) \times 100$

6. Mole Fraction, $\chi_A=\frac{n_A}{n_A+n_B+\ldots}$

7. Molarity, $M=\frac{n}{V}$ where M = molarity, n = moles of solute, V = volume of solution in litres

8. Molarlity, $M=\frac{n}{W}$

Stoichiometry and Limiting Reagent for NEET 2026

The empirical and molecular formulas are important concepts in NEET Chemistry that help determine the simplest ratio and actual composition of compounds, often tested through numerical and concept-based questions.

Relationship between the empirical formula and molecular formula

Molecular Formula $=\mathrm{n} \times$ Empirical Formula

Where n is the Molecular Mass / Empirical Formula Mass

Stoichiometry and Limiting Reagent for NEET Chemistry 2026

Stoichiometry and limiting reagent are key concepts in NEET Chemistry that help determine the exact amounts of reactants and products, often tested through numerical and application-based questions.

Stoichiometry:

Stoichiometry is the calculation of reactants and products using balanced chemical equations. The balanced equation gives the mole ratio of all species involved.

Limiting Reagent:

The limiting reagent is the reactant that is completely consumed first and limits the amount of product formed.

How to identify the limiting reactant

• Convert all the rectant in the moles.

• Divide each by its stoichiometric coefficient from the balanced equation.

• The reactant with the smallest quotient is the limiting reagent.

• Use moles of limiting reagent to calculate moles and mass of product.

PERCENTAGE YIELD

$\text { \% Yield = (Actual Yield / Theoretical Yield) × } 100$

PERCENTAGE PURITY

$\text { \% Yield = (Pure substance/ Total mass) × } 100$

Previous Year Questions (PYQs) on Mole Concept for NEET 2026

Solve these Previous Year Questions (PYQs) to understand the exam pattern and important concepts asked in NEET. Detailed solutions will help you improve accuracy and strengthen your problem-solving skills.

Ques 1: How many molecules are present in 4 g of hydrogen gas ($H_2$)? (Molar mass of $H_2$ = 2 g/mol) (NEET 2020)

(A) $6.022 \times 10^{23}$

(B) $12.044 \times 10^{23}$

(C) $3.011 \times 10^{23}$

(D) $1.505 \times 10^{23}$

Correct Answer: (B)

Explanation:

Moles of $H_2$ = 4 ÷ 2 = 2 mol. Number of molecules = $2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23}$.

Ques 2: What is the mass of $3.011 \times 10^{23}$ atoms of nitrogen? (Atomic mass of N = 14 u) (NEET 2019)

(A) 28 g

(B) 7 g

(C) 14 g

(D) 3.5 g

Correct Answer: (B)

Explanation

$3.011 \times 10^{23}$ atoms = 0.5 mol of N. Mass = $0.5 \times 14 = 7$ g.

Ques 3:The empirical formula of a compound is $CH_2O$. Its molar mass is 180 g/mol. What is the molecular formula? (NEET 2018)

(A) $C_2H_4O_2$

(B) $C_3H_6O_3$

(C) $C_6H_{12}O_6$

(D) $C_4H_8O_4$

Correct Answer: (C)

Explanation

Empirical formula mass ($CH_2O$) = $12 + 2 + 16 = 30$ g/mol. n = $180 ÷ 30 = 6$. Molecular formula = $C_6H_{12}O_6$ (Glucose).

Ques 4: The volume occupied by 4.4 g of $CO_2$ at STP is: (Molar mass of $CO_2$ = 44 g/mol) (NEET 2021)

(A) 2.24 L

(B) 4.48 L

(C) 22.4 L

(D) 1.12 L

Correct Answer: (A)

Explanation

Moles of $CO_2$ = $4.4 ÷ 44 = 0.1$ mol. Volume at STP = $0.1 \times 22.4 = 2.24$ L.

Ques 5: The percentage of nitrogen in urea [$CO(NH_2)_2$] is approximately: (Molar mass = 60 g/mol) (NEET 2019)

(A) 26.67%

(B) 20%

(C) 46.67%

(D) 33.33%

Correct Answer: (C)

Explanation:

Urea has 2 N atoms. Mass of N = $2 \times 14 = 28$ g. % N = $(28 / 60) \times 100 = 46.67%$.

Ques 6: How many total atoms are present in 0.5 mol of $H_2SO_4$? (NEET 2022)

(A) $1.806 \times 10^{23}$

(B) $3.011 \times 10^{23}$

(C) $2.108 \times 10^{24}$

(D) $4.216 \times 10^{24}$

Correct Answer: (C)

Explanation

$H_2SO_4$ has $2+1+4 = 7$ atoms per molecule. Total = $0.5 \times 7 \times 6.022 \times 10^{23} = 2.108 \times 10^{24}$ atoms.

Ques 7: In the reaction $N_2 + 3H_2 \rightarrow 2NH_3$, if 28 g of $N_2$ reacts with 6 g of $H_2$, which is the limiting reagent? (NEET 2020)

(A) $H_2$

(B) $N_2$

(C) Both are equally limiting

(D) Neither is limiting

Correct Answer: (C)

Explanation

Moles of $N_2$ = $28/28 = 1$ mol → ratio = $1/1 = 1.0$. Moles of $H_2$ = $6/2 = 3$ mol → ratio = $3/3 = 1.0$. Both ratios are equal; both are consumed simultaneously.

Ques 8: The highest number of helium atoms is present in which of the following? (Molar mass of He = 4 g/mol) (NEET 2024)

(A) 4 mol of helium

(B) 4 u of helium

(C) 4 g of helium

(D) 2.271 L of helium at STP

Correct Answer: (A)

Explanation

4 mol → $4 \times 6.022 \times 10^{23} = 2.409 \times 10^{24}$ atoms. 4 u = only 1 atom. 4 g = 1 mol = $6.022 \times 10^{23}$ atoms. 2.271 L ≈ 0.101 mol = $6.1 \times 10^{22}$ atoms. Option A is the highest.

Ques 9: 1 g of NaOH is treated with 25 mL of 0.75 M HCl. The mass of NaOH left unreacted is: (Molar mass of NaOH = 40 g/mol; $NaOH + HCl \rightarrow NaCl + H_2O$) (NEET 2024)

(A) 750 mg

(B) 250 mg

(C) 0 mg

(D) 200 mg

Correct Answer: (B)

Explanation

Moles of NaOH = $1/40 = 0.025$ mol. Moles of HCl = $0.75 \times 0.025 = 0.01875$ mol. HCl is limiting. Excess NaOH = $0.025 − 0.01875 = 0.00625$ mol → $0.00625 \times 40 = 0.25$ g = 250 mg.

Ques 10: The mass of $CO_2$ produced by heating 20 g of 20% pure limestone: (M: $CaCO_3 = 100$, $CO_2 = 44$ g/mol; $CaCO_3 \rightarrow CaO + CO_2$) (NEET 2023)

(A) 1.32 g

(B) 1.12 g

(C) 1.76 g

(D) 2.64 g

Correct Answer: (C)

Explanation

Pure $CaCO_3$ = 20% of 20 g = 4 g. Moles = $4/100 = 0.04$ mol. Ratio 1:1 → moles of $CO_2$ = 0.04 mol. Mass = $0.04 \times 44 = 1.76$ g.

Ques 11: A compound X contains 32% of A, 20% of B, and the rest is C. Its empirical formula is: (Atomic masses: A = 64 u, B = 40 u, C = 32 u) (NEET 2024)

(A) $A_2BC_2$

(B) $ABC_3$

(C) $AB_2C_2$

(D) $ABC_4$

Correct Answer: (B)

Explanation:

% C = $100 − 32 − 20 = 48%$. Moles: A = $32/64 = 0.5$, B = $20/40 = 0.5$, C = $48/32 = 1.5$. Ratio = $0.5 : 0.5 : 1.5$ → divide by 0.5 → $1 : 1 : 3$. Empirical formula = $ABC_3$.

Ques 12: Which of the following set contains an equal number of total atoms? ($N_A = 6.022 \times 10^{23}$) (NEET 2025)

A = 212 g of $Na_2CO_3$ (M=106), B = 248 g of $Na_2O$ (M=62), C = 240 g of NaOH (M=40), D = 12 g of $H_2$ (M=2)

(A) B, C, and D only

(B) B, D, and E only

(C) A, B, and C only

(D) A, B, and D only

Correct Answer: (D)

Explanation

Total atoms = moles × atoms per molecule. A: $(212/106)\times6 = 12$ mol atoms. B: $(248/62)\times3 = 12$ mol atoms. C: $(240/40)\times3 = 18$ mol atoms. D: $(12/2)\times2 = 12$ mol atoms. A, B, and D each give $12 \times N_A$ total atoms.

Ques 13: What is the molarity of a solution containing 5.85 g of NaCl dissolved in 500 mL of solution? (Molar mass of NaCl = 58.5 g/mol) (NEET 2022)

(A) 0.1 M

(B) 0.2 M

(C) 0.5 M

(D) 1.0 M

Correct Answer: (B)

Explanation

Moles of NaCl = $5.85 / 58.5 = 0.1$ mol. Volume = 500 mL = 0.5 L. Molarity = $0.1 / 0.5 = 0.2$ M.

Ques 14: What is the normality of 0.5 M $H_2SO_4$ solution? ($H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$) (NEET 2021)

(A) 0.5 N

(B) 2 N

(C) 1 N

(D) 4 N

Correct Answer: (C)

Explanation

$H_2SO_4$ furnishes $2H^+$ ions per molecule, so its n-factor = 2. Normality = Molarity × n-factor = $0.5 \times 2 = 1$ N.

Ques 15: What volume (at STP) is occupied by 2.5 mol of an ideal gas? (NEET 2023)

(A) 22.4 L

(B) 44.8 L

(C) 56.0 L

(D) 11.2 L

Correct Answer: (C)

Explanation

At STP, 1 mol of an ideal gas occupies 22.4 L. Volume = $2.5 \times 22.4 = 56.0$ L.

Short Tricks to Solve Mole Concept Questions Quickly

Here are some quick tricks and shortcuts to solve mole concept questions faster in NEET, helping you save time and improve accuracy in numerical problems.

1. Mole Fraction Always Sums to 1: The sum of all mole fractions in any mixture = 1. Use this to cross-check answers quickly or to find the mole fraction of the last component without extra calculation.

2. Always count atoms in the formula: 1 mole of CH₄ = 1C + 4H = 5 atoms per molecule $=5 \times 6.022 \times 10^{23}$ atoms total. Write out the atom count before computing.

3. Number of Molecules in ANY Gas at STP: Equal volumes of any gas at the same T and P contain equal molecules (Avogadro's Law). So, 22.4 L of any gas at STP → $6.022 \times 10^{23}$molecules. This is a direct question asked in NEET.