Previous year questions: Some Basic Concepts of Chemistry

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Iron reacts with steam to form Fe3O4 and hydrogen gas. Excess iron is treated with 18 g of steam. Calculate the mass of iron that is converted into Fe3O4. (Molar masses: H = 1, O = 16, Fe = 56 g mol^-1)

Answer: The balanced equation is 3Fe + 4H2O giving Fe3O4 + 4H2. The molar mass of water is 18 g mol^-1, so 18 g of steam is 1.00 mol. Four moles of water consume three moles of iron, therefore one mole of water consumes 3/4 = 0.75 mol of iron. The mass of iron is 0.75 x 56 = 42 g. The mass of Fe3O4 formed would be 0.25 x 232 = 58 g, but the question asks for the iron consumed, so the answer is 42 g and the option is 42 g.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Find the number of moles and the number of molecules in 1.4187 L of sulphur dioxide measured at STP.

Answer: At STP the molar volume used by JEE papers is 22.4 L mol^-1, so the number of moles is 1.4187 divided by 22.4, which is 0.0633 mol. The number of molecules is 0.0633 x 6.022 x 10^23 = 3.812 x 10^22. Both figures appear in the second option. The rationalised NCERT text states the molar volume as 22.7 L mol^-1 at 1 bar, and with that value the moles would be 0.0625; the answer key of the paper uses 22.4 L, so the paper is following the 1 atm convention. State the convention you use in the answer.

Q3 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

An oxide of iron contains 69.9 percent iron by mass. Determine its empirical formula. (Molar masses: Fe = 56, O = 16 g mol^-1)

Answer: Take 100 g of the oxide, so the sample contains 69.9 g of iron and 30.1 g of oxygen. The moles of iron are 69.9 divided by 56, which is 1.248, and the moles of oxygen are 30.1 divided by 16, which is 1.881. Divide both by the smaller value: iron gives 1 and oxygen gives 1.507, which is close to 1.5. Multiply by two to reach whole numbers, giving 2 iron atoms for every 3 oxygen atoms. The empirical formula is Fe2O3. The formula Fe3O4 would need 72.4 percent iron, so it does not fit the data.

Q4 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

Which of the following samples contain the same number of atoms? (A) 2 g of O2 gas, (B) 4 g of SO2 gas, (C) 1400 mL of O2 at STP, (D) 0.05 L of He at STP, (E) 0.0625 mol of H2 gas. (Molar masses: H = 1, He = 4, O = 16, S = 32 g mol^-1)

Answer: Convert each sample into moles of atoms. Sample A: 2 g of O2 is 0.0625 mol of molecules, and each molecule has two atoms, so 0.125 mol of atoms. Sample C: 1400 mL is 1.4 L, which at 22.4 L mol^-1 is 0.0625 mol of O2, again 0.125 mol of atoms. Sample E: 0.0625 mol of H2 has two atoms per molecule, so 0.125 mol of atoms. Sample B: 4 g of SO2 is 4/64 = 0.0625 mol of molecules, and with three atoms per molecule it gives 0.1875 mol of atoms. Sample D: 0.05 L of helium at STP is about 0.0022 mol of atoms. Only A, C and E agree, so the correct option is the one listing A, C and E.

Q5 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

In a hydrocarbon the ratio of the mass percentages of carbon and hydrogen is 12 : 1, and one molecule contains two carbon atoms. What mass of carbon dioxide is formed when 3.38 g of this hydrocarbon is completely burnt in oxygen? (Molar masses: C = 12, H = 1, O = 16 g mol^-1)

Answer: With two carbon atoms the carbon mass in one molecule is 24 u, and the 12 : 1 mass ratio makes the hydrogen mass 2 u, so the molecule is C2H2 with a molar mass of 26 g mol^-1. The sample is 3.38/26 = 0.13 mol. Each molecule of C2H2 gives two molecules of CO2, so the CO2 formed is 0.26 mol. Its mass is 0.26 x 44 = 11.44 g. The option 11.44 g matches, and the check is that all the carbon, 0.26 mol, appears in the carbon dioxide.

Q6 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

What is the mole fraction of water in a 10 percent by mass aqueous urea solution? (Molar masses: H = 1, C = 12, N = 14, O = 16 g mol^-1)

Answer: A 10 percent by mass solution has 10 g of urea and 90 g of water in every 100 g. Urea, NH2CONH2, has a molar mass of 60 g mol^-1, so 10 g is 0.1667 mol. Water has a molar mass of 18 g mol^-1, so 90 g is 5.00 mol. The mole fraction of water is 5.00 divided by 5.00 plus 0.1667, which is 5.00/5.1667 = 0.968. The option 0.967 is the answer, and it is close to one because water is present in a far greater amount.

Q7 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

What volume of hydrogen at STP is liberated when 20 g of zinc reacts with 50 mL of sulphuric acid of 50 percent purity and density 1.3 g mL^-1? (Molar masses: H = 1, O = 16, S = 32, Zn = 65 g mol^-1)

Answer: The mass of the acid solution is 50 x 1.3 = 65 g, and at 50 percent purity the mass of pure H2SO4 is 32.5 g, which is 32.5/98 = 0.3316 mol. Zinc is 20/65 = 0.3077 mol. The reaction Zn + H2SO4 giving ZnSO4 + H2 uses them in a 1 : 1 ratio, so zinc is the limiting reagent and the hydrogen formed is 0.3077 mol. At 22.4 L mol^-1 this is 0.3077 x 22.4 = 6.89 L. The answer is 6.892 L. Using the acid as the basis instead would overestimate the gas, which is why the limiting reagent test comes first.

Q8 (NCERT Exemplar, 3 marks)

A temperature measured on the Fahrenheit scale is 200 degrees Fahrenheit. What is the reading on the Celsius scale? (i) 40 degrees Celsius (ii) 94 degrees Celsius (iii) 93.3 degrees Celsius (iv) 30 degrees Celsius

Source: NCERT Exemplar Class 11 Chemistry, Unit 1, MCQ 2.

Answer: Use the conversion relation, Celsius equals (Fahrenheit minus 32) multiplied by 5/9. Substituting 200 gives (200 - 32) x 5/9, which is 168 x 5/9 = 93.33 degrees Celsius. The correct option is (iii) 93.3 degrees Celsius. The kelvin reading would be 93.3 + 273.15 = 366.5 K. A common error is to apply the 5/9 factor without subtracting 32 first.

Q9 (NCERT Exemplar, 3 marks)

If 500 mL of a 5 M solution is diluted to 1500 mL, what is the molarity of the solution obtained? (i) 1.5 M (ii) 1.66 M (iii) 0.017 M (iv) 1.59 M

Source: NCERT Exemplar Class 11 Chemistry, Unit 1, MCQ 4.

Answer: Dilution keeps the number of moles of solute unchanged, so M1V1 = M2V2. Here M1 is 5 M, V1 is 0.500 L and V2 is 1.500 L. The moles of solute are 5 x 0.500 = 2.5 mol, and the new molarity is 2.5 divided by 1.500, which is 1.67 M. The option given as 1.66 M is the intended answer, the small difference coming from rounding. Option (i) would be right only if the volume had tripled and the moles had also changed, which does not happen in dilution.

Q10 (NCERT Exemplar, 3 marks)

What is the molarity of a solution that contains 5.85 g of sodium chloride per 500 mL of solution? (Molar masses: Na = 23, Cl = 35.5 g mol^-1)

Source: NCERT Exemplar Class 11 Chemistry, Unit 1, MCQ 3.

Answer: The molar mass of sodium chloride is 23 + 35.5 = 58.5 g mol^-1, so 5.85 g is 5.85/58.5 = 0.100 mol. The volume of the solution is 500 mL, which is 0.500 L. Molarity is moles of solute divided by volume of solution in litres, so 0.100/0.500 = 0.200 mol L^-1, or 0.2 M. The answer is not 0.1 M, because the moles are spread over half a litre and the volume must be converted before dividing.