Basic Concepts of Chemistry Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Basic Concepts of Chemistry — mole concept, empirical formula, and stoichiometry — with step-by-step solutions.
Solved JEE Main 2026 previous year questions from the Basic Concepts of Chemistry chapter, each with a full step-by-step solution.
Solutions are AI-generated and pending review.
Questions
Solution
Given: Volume of $\text{SO}_2$ at STP $= 1.4187\ \text{L}$; molar volume $V_m = 22.4\ \text{L mol}^{-1}$.
Step 1 — Moles:
$$n = \frac{V}{V_m} = \frac{1.4187}{22.4} \approx 0.0633\ \text{mol}$$Step 2 — Number of molecules:
$$N = n \times N_A = 0.0633 \times 6.022 \times 10^{23} \approx 3.812 \times 10^{22}$$Answer: B
Solution
Step 1 — Mass of each element in 100 g:
- Fe $= 69.9\ \text{g}$, O $= 100 - 69.9 = 30.1\ \text{g}$
Step 2 — Convert to moles:
$$n_\text{Fe} = \frac{69.9}{56} \approx 1.248, \qquad n_\text{O} = \frac{30.1}{16} \approx 1.881$$Step 3 — Simplest ratio (divide by smallest):
$$\text{Fe} : \text{O} = \frac{1.248}{1.248} : \frac{1.881}{1.248} = 1 : 1.507 \approx 2 : 3$$Empirical formula $= \text{Fe}_2\text{O}_3$ (check: %Fe $= \tfrac{112}{160} = 70.0\%$).
Answer: B
Solution
Reaction:
$$3\,\text{Fe} + 4\,\text{H}_2\text{O} \longrightarrow \text{Fe}_3\text{O}_4 + 4\,\text{H}_2$$Step 1 — Moles of steam:
$$n_{\text{H}_2\text{O}} = \frac{18}{18} = 1\ \text{mol}$$Step 2 — Moles of Fe (ratio $3:4$ with water):
$$n_{\text{Fe}} = \frac{3}{4} \times 1 = 0.75\ \text{mol}$$Step 3 — Mass of Fe:
$$m_{\text{Fe}} = 0.75 \times 56 = 42\ \text{g}$$Answer: D
Solution
Express each total atom count in units of $N_A$.
(A) Cyclohexane $\text{C}_6\text{H}_{12}$: $18$ atoms per molecule.
$$2 \times 18 = 36$$(B) Sucrose $\text{C}_{12}\text{H}_{22}\text{O}_{11}$, $M = 342$: $45$ atoms per molecule.
$$n = \frac{684}{342} = 2\ \text{mol} \Rightarrow 2 \times 45 = 90$$(C) Dihydrogen at STP ($V_m = 22.7\ \text{L mol}^{-1}$): $2$ atoms per molecule.
$$n = \frac{90.8}{22.7} = 4\ \text{mol} \Rightarrow 4 \times 2 = 8$$Order: $B(90) > A(36) > C(8)$, i.e. $B > A > C$.
Answer: D
Solution
Step 1 — Mole ratio of atoms from mass ratio $12:1$:
$$\text{C} : \text{H} = \frac{12}{12} : \frac{1}{1} = 1 : 1$$Empirical formula $\text{CH}$; with two carbons the molecule is $\text{C}_2\text{H}_2$, $M = 26$.
Step 2 — Moles of hydrocarbon:
$$n = \frac{3.38}{26} = 0.13\ \text{mol}$$Step 3 — Moles of $\text{CO}_2$ ($2$ carbons per molecule):
$$n_{\text{CO}_2} = 2 \times 0.13 = 0.26\ \text{mol}$$Step 4 — Mass:
$$m_{\text{CO}_2} = 0.26 \times 44 = 11.44\ \text{g}$$Answer: B
Solution
Total atoms (in mol) $=$ moles $\times$ atoms per molecule, with $V_m = 22.4\ \text{L mol}^{-1}$.
$$\text{A: } \frac{2}{32}\times 2 = 0.0625 \times 2 = 0.125$$$$\text{B: } \frac{4}{64}\times 3 = 0.0625 \times 3 = 0.1875$$$$\text{C: } \frac{1.4}{22.4}\times 2 = 0.0625 \times 2 = 0.125$$$$\text{D: } \frac{0.05}{22.4}\times 1 = 0.00223$$$$\text{E: } 0.0625 \times 2 = 0.125$$A, C and E each give $0.125$ mol of atoms — the same number.
Answer: D
Solution
Reactions — the final residue is metallic silver:
$$\text{Ag}_2\text{CO}_3 \xrightarrow{\Delta} \text{Ag}_2\text{O} + \text{CO}_2, \qquad 2\,\text{Ag}_2\text{O} \xrightarrow{\Delta} 4\,\text{Ag} + \text{O}_2$$Step 1 — Molar mass of $\text{Ag}_2\text{CO}_3$:
$$2(108) + 12 + 3(16) = 276\ \text{g mol}^{-1}$$Step 2 — Moles:
$$n = \frac{2.76}{276} = 0.01\ \text{mol}$$Step 3 — Mass of Ag ($2$ mol Ag per mol $\text{Ag}_2\text{CO}_3$):
$$m_{\text{Ag}} = (0.01 \times 2) \times 108 = 2.16\ \text{g}$$Answer: B
Solution
Reaction:
$$\text{Zn} + \text{H}_2\text{SO}_4 \longrightarrow \text{ZnSO}_4 + \text{H}_2$$Step 1 — Moles of $\text{H}_2\text{SO}_4$: mass of solution $= 50 \times 1.3 = 65\ \text{g}$; pure acid $= 50\% = 32.5\ \text{g}$.
$$n_{\text{H}_2\text{SO}_4} = \frac{32.5}{98} = 0.332\ \text{mol}$$Step 2 — Moles of Zn:
$$n_{\text{Zn}} = \frac{20}{65} = 0.308\ \text{mol}$$Step 3 — Limiting reagent: ratio $1:1$, so Zn limits ($0.308 < 0.332$). Hence $n_{\text{H}_2} = 0.308\ \text{mol}$.
Step 4 — Volume at STP ($V_m = 22.4\ \text{L mol}^{-1}$):
$$V_{\text{H}_2} = 0.308 \times 22.4 = 6.892\ \text{L}$$Answer: C
Solution
For each, atoms $=$ moles $\times$ atoms per formula unit (in units of $N_A$).
$$\text{A. } \text{H}_2\text{O}\ (M=18): \frac{1.8\times 10^{-3}}{18}\times 3 = 3\times 10^{-4}\,N_A \to \text{III}$$$$\text{B. } \text{H}_2\text{SO}_4\ (M=98): \frac{9.8\times 10^{-3}}{98}\times 7 = 7\times 10^{-4}\,N_A \to \text{IV}$$$$\text{C. C}\ (M=12): \frac{1.8\times 10^{-3}}{12}\times 1 = 1.5\times 10^{-4}\,N_A \to \text{II}$$$$\text{D. NaCl}\ (M=58.5): \frac{5.85\times 10^{-3}}{58.5}\times 2 = 2\times 10^{-4}\,N_A \to \text{I}$$Thus A-III, B-IV, C-II, D-I.
Answer: C