JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 32

Practice MCQs for JEE Chemistry Some Basic Concepts Of Chemistry

Access targeted multiple-choice questions for Some Basic Concepts Of Chemistry designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Some Basic Concepts Of Chemistry Questions and Solutions

View or download the dedicated Some Basic Concepts Of Chemistry MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: H3PO4 (98 g mol-1) is 98% by mass of a solution of density 1.8 g mL-1. The molarity is
(a) 98 M
(b) 18 M
(c) 9 M
(d) 1.8 M
Show Answer & Explanation

Answer: (b) 18 M

Explanation:
1. 1 L of solution weighs 1800 g and contains 0.98 × 1800 = 1764 g of H3PO4.
2. Moles = 1764/98 = 18, so molarity = 18 M.

Teacher's Note:
1. M = 10 × d × (% by mass) / molar mass.
2. Here 10 × 1.8 × 98 / 98 = 18.

Question 2: The volume of 90% alcohol by weight (d = 0.8 g mL-1) needed to prepare 80 mL of 10% alcohol by weight (d = 0.9 g mL-1) is
(a) 40 mL
(b) 20 mL
(c) 5 mL
(d) 10 mL
Show Answer & Explanation

Answer: (d) 10 mL

Explanation:
1. Mass of the final solution = 80 × 0.9 = 72 g, containing 7.2 g alcohol.
2. Mass of 90% solution needed = 7.2 / 0.9 = 8 g.
3. Volume = 8 / 0.8 = 10 mL.

Teacher's Note:
1. The alcohol mass is conserved on dilution.
2. Convert between mass and volume using each density.

Question 3: A solution containing 0.1 mol of a metal chloride MClx requires 500 mL of 0.8 M AgNO3 for complete reaction, MClx + xAgNO3 → xAgCl + M(NO3)x. The value of x is
(a) 4
(b) 5
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. Moles of AgNO3 = 0.5 × 0.8 = 0.4 mol.
2. 0.1 mol MClx reacts with 0.1x mol AgNO3, so 0.1x = 0.4 and x = 4.

Teacher's Note:
1. Each chloride ion needs one silver ion.
2. Compare moles of the two reactants.

Question 4: Which of the following statements about 1.7 g of NH3 are correct? (A) It contains 0.3 mol of H atoms. (B) It contains 2.408 × 1023 atoms. (C) The mass percentage of hydrogen is 17.65%. (D) It contains 0.3 mol of N atoms.
(a) B, C and D
(b) A, B and C
(c) A and D
(d) A, B, C and D
Show Answer & Explanation

Answer: (b) A, B and C

Explanation:
1. 1.7 g = 0.1 mol NH3, containing 0.1 mol N and 0.3 mol H, so (A) is true and (D) is false.
2. Total atoms = 0.4 mol × 6.02 × 1023 = 2.408 × 1023, so (B) is true.
3. Mass % H = 3/17 × 100 = 17.65%, so (C) is true.

Teacher's Note:
1. Count moles of each atom from the formula.
2. Total atoms = moles of molecules × atoms per molecule × N_A.

Question 5: The density of air is 0.001293 g cm-3 at STP. Which of the following are correct? (A) Vapour density = 14.48 (B) Molecular mass = 28.96 (C) Vapour density = 0.001293 g cm-3 (D) Vapour density and molecular mass cannot be determined
(a) A and C
(b) B and D
(c) C and D
(d) A and B
Show Answer & Explanation

Answer: (d) A and B

Explanation:
1. Molar mass = 1.293 g L-1 × 22.4 L mol-1 = 28.96 g mol-1.
2. Vapour density = M/2 = 14.48, and it is a ratio with no units, so (C) and (D) are false.

Teacher's Note:
1. Vapour density is a dimensionless ratio of gas density to hydrogen density.
2. Molar mass = density at STP × 22.4.

Question 6: 106.0 g of Na2CO3 reacts with 109.5 g of HCl: Na2CO3 + 2HCl → 2NaCl + CO2 + H2O. Which are correct? (A) HCl is in excess. (B) 117.0 g of NaCl is formed. (C) The volume of CO2 at NTP is 22.4 L. (D) None of these.
(a) B and C only
(b) A, B and C
(c) A only
(d) D only
Show Answer & Explanation

Answer: (b) A, B and C

Explanation:
1. Na2CO3 = 1 mol; HCl = 109.5/36.5 = 3 mol. 1 mol Na2CO3 needs only 2 mol HCl, so HCl is in excess (A true).
2. NaCl formed = 2 mol = 117 g (B true).
3. CO2 formed = 1 mol = 22.4 L at NTP (C true), so (D) is false.

Teacher's Note:
1. Na2CO3 is the limiting reagent here.
2. Use the 1 : 2 : 2 : 1 mole ratio.

Question 7: Starting with 1 mol K4[Fe(CN)6] and 5 mol H2SO4 (with enough water), the steps are (i) K4Fe(CN)6 + 3H2SO4 → 2K2SO4 + FeSO4 + 6HCN; (ii) 6HCN + 12H2O → 6HCOOH + 6NH3; (iii) 6NH3 + 3H2SO4 → 3(NH4)2SO4 and 6HCOOH → 6CO + 6H2O (conc. H2SO4 as catalyst in the last step). Which statements are correct?
(a) The limiting reagent of step (i) is H2SO4; 6 mol CO and 2 mol (NH4)2SO4 form.
(b) The limiting reagent of step (i) is K4Fe(CN)6; 6 mol CO and 2 mol (NH4)2SO4 form.
(c) The limiting reagent of step (i) is K4Fe(CN)6; 5 mol CO and 2.5 mol (NH4)2SO4 form.
(d) The limiting reagent of step (i) is H2SO4; 5 mol CO and 2.5 mol (NH4)2SO4 form.
Show Answer & Explanation

Answer: (b) The limiting reagent of step (i) is K4Fe(CN)6; 6 mol CO and 2 mol (NH4)2SO4 form.

Explanation:
1. Step (i) needs 3 mol H2SO4 per mol of complex; 5 mol is available, so K4Fe(CN)6 is limiting and 6 mol HCN forms.
2. 6 mol HCN gives 6 mol HCOOH and 6 mol NH3, and the HCOOH gives 6 mol CO.
3. After step (i), H2SO4 left = 5 - 3 = 2 mol, which can react with only 4 mol NH3 to give 2 mol (NH4)2SO4.

Teacher's Note:
1. Track the sulphuric acid used in step (i) before step (iii).
2. Atom conservation: 6 C from the cyanide ends up in CO.

Question 8: A + B → A3B2 and A3B2 + C → A3B2C2 (both unbalanced) are carried out with 3 mol A, 3 mol B and 1 mol C. Which statements are correct? (A) 1 mol A3B2C2 is formed. (B) 1/2 mol A3B2C2 is formed. (C) 1/2 mol A3B2 is formed. (D) 1/2 mol A3B2 is left finally.
(a) B and D
(b) A and D
(c) A and C
(d) B and C
Show Answer & Explanation

Answer: (a) B and D

Explanation:
1. Balanced: 3A + 2B → A3B2 and A3B2 + 2C → A3B2C2.
2. 3 mol A gives 1 mol A3B2; B (3 mol) is in excess.
3. 1 mol C reacts with 1/2 mol A3B2 to give 1/2 mol A3B2C2, leaving 1/2 mol A3B2.
4. So (B) and (D) are true.

Teacher's Note:
1. Balance each step first.
2. The last reagent, C, limits the second step.

Question 9: 1.0 g of a mixture of CaCO3 and glass beads gives 0.22 g of CO2 with excess HCl (glass does not react). Which are correct? (A) The mass of CaCO3 is 0.5 g. (B) The mass of calcium is 0.2 g. (C) The mass percent of Ca is 40%. (D) The mass percent of Ca is 20%.
(a) A, B and C
(b) A, B and D
(c) A and C
(d) B and D only
Show Answer & Explanation

Answer: (b) A, B and D

Explanation:
1. CO2 = 0.22/44 = 0.005 mol, so CaCO3 = 0.005 mol = 0.5 g (A true).
2. Ca = 0.005 × 40 = 0.2 g (B true), which is 0.2/1.0 = 20% of the mixture, so (D) is true and (C) is false.

Teacher's Note:
1. One mole of CaCO3 gives one mole of CO2.
2. Percentage is based on the 1.0 g mixture.

Question 10: A 21.2 g sample of impure Na2CO3 is reacted with CaCl2 solution and gives 10.0 g of CaCO3. Which are correct? (A) The purity of Na2CO3 is 50%. (B) The purity is 60%. (C) Moles of Na2CO3 = moles of CaCO3 = 0.1. (D) Moles of NaCl formed = 0.1.
(a) B and C
(b) A and C
(c) A and D
(d) B and D
Show Answer & Explanation

Answer: (b) A and C

Explanation:
1. Na2CO3 + CaCl2 → CaCO3 + 2NaCl; CaCO3 = 10/100 = 0.1 mol, so Na2CO3 = 0.1 mol (C true).
2. Pure Na2CO3 = 0.1 × 106 = 10.6 g, purity = 10.6/21.2 = 50% (A true).
3. NaCl formed = 0.2 mol, so (D) is false.

Teacher's Note:
1. Purity = mass of pure compound / mass of sample.
2. Note the 1 : 2 ratio of Na2CO3 to NaCl.

Question 11: 100 g of clay containing 19% water, 40% silica and the rest inert matter is partly dried until it contains 10% water. Which statements are correct? (A) Silica is 44.4% of the dried clay. (B) The dried clay weighs 90.0 g. (C) Inert matter is 45.6% of the dried clay. (D) The mass of water evaporated is 10.0 g.
(a) A and B only
(b) A, B and C only
(c) A, B, C and D
(d) B, C and D only
Show Answer & Explanation

Answer: (c) A, B, C and D

Explanation:
1. Inert = 100 - 19 - 40 = 41 g. Dry solids = 81 g, which must be 90% of the partly dried clay, so its mass = 81/0.9 = 90 g (B true).
2. Silica = 40/90 = 44.4% (A true); inert = 41/90 = 45.6% (C true).
3. Water left = 9 g, so water lost = 19 - 9 = 10 g (D true).

Teacher's Note:
1. The mass of non-volatile solids does not change on drying.
2. Use the new fraction of water to find the new mass.

Question 12: Which of the following solutions have the same molar concentration? (A) 166 g KI per litre (B) 33.0 g (NH4)2SO4 in 200 mL (C) 25.0 g CuSO4·5H2O in 100 mL (D) 27.0 mg Al3+ per mL (K = 39, I = 127, Al = 27, Cu = 63.5)
(a) A, B, C and D
(b) A, C and D
(c) A and B
(d) B, C and D
Show Answer & Explanation

Answer: (b) A, C and D

Explanation:
1. (A) 166/166 = 1.0 M.
2. (B) 33/132 = 0.25 mol in 0.2 L = 1.25 M.
3. (C) 25/249.5 = 0.100 mol in 0.1 L = 1.0 M.
4. (D) 27 mg per mL = 27 g per L = 1.0 M.
5. So (A), (C) and (D) are 1.0 M.

Teacher's Note:
1. Convert each to mol L-1.
2. Mg per mL is the same number as g per L.

Question 13: Which of the following solutions contain 23 g of HCOOH (molar mass 46 g mol-1)? (A) 46 g of 70% w/v HCOOH (d = 1.40 g mL-1) (B) 50 g of 10 M HCOOH (d = 1 g mL-1) (C) 50 g of 25% w/w HCOOH (D) 46 g of 5 M HCOOH (d = 1 g mL-1)
(a) B and C
(b) C and D
(c) A and D
(d) A and B
Show Answer & Explanation

Answer: (d) A and B

Explanation:
1. (A) Volume = 46/1.4 = 32.86 mL, containing 0.70 × 32.86 = 23.0 g. True.
2. (B) 50 g is 50 mL = 0.05 L, so moles = 0.5 mol = 23 g. True.
3. (C) 25% of 50 g = 12.5 g. False.
4. (D) 46 mL × 5 M = 0.23 mol = 10.6 g. False.

Teacher's Note:
1. w/v percent is grams per 100 mL of solution.
2. Find the volume of each sample first.

Question 14: 100 mL of 1 M H2SO4 is mixed with 100 mL of 9.8% (w/w) H2SO4 (d = 1 g mL-1). Which are correct? (A) The concentration stays the same. (B) The volume becomes 200 mL. (C) The mass of H2SO4 is 98 g. (D) The mass of H2SO4 is 19.6 g.
(a) A and C
(b) B and C
(c) A, B and D
(d) A, B and C
Show Answer & Explanation

Answer: (c) A, B and D

Explanation:
1. The second solution: 9.8 g per 100 g (100 mL) = 0.1 mol per 100 mL = 1 M, the same as the first, so (A) is true.
2. Total volume = 200 mL (B true), and the acid is 0.2 mol = 19.6 g (D true; C false).

Teacher's Note:
1. Convert the percentage to molarity using the density.
2. Mixing equal concentrations leaves it unchanged.

Question 15: A compound contains C, H and O only, with H : O atoms in the ratio 2 : 1, 40% C by mass and molecular mass 180 g. The percentage by mass of oxygen is
(a) 53.33%
(b) 88.88%
(c) 33.33%
(d) none of these
Show Answer & Explanation

Answer: (a) 53.33%

Explanation:
1. Mass of C = 0.40 × 180 = 72 g (6 atoms).
2. Remaining 108 g is H and O with 2n H atoms and n O atoms: 2n + 16n = 108, so n = 6.
3. Mass of O = 96 g, so % O = 96/180 × 100 = 53.33%.

Teacher's Note:
1. Use the atom ratio to write the mass of H and O in terms of n.
2. Mass percentages must add up to 100.

Question 16: A compound contains C, H and O only, with H : O atoms in the ratio 2 : 1, 40% C by mass and molecular mass 180 g. Its empirical formula is
(a) CH3O2
(b) CH3O
(c) C2H2O
(d) CH2O
Show Answer & Explanation

Answer: (d) CH2O

Explanation:
1. C = 72 g = 6 mol, H = 12 g = 12 mol, O = 96 g = 6 mol.
2. Ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH2O.

Teacher's Note:
1. Convert each mass to moles and take the simplest ratio.
2. This is the empirical formula of glucose.

Question 17: A compound contains C, H and O only, with H : O atoms in the ratio 2 : 1, 40% C by mass and molecular mass 180 g. Which could be its molecular formula?
(a) C6H14O6
(b) C6H12O6
(c) C6H14O12
(d) C6H6O6
Show Answer & Explanation

Answer: (b) C6H12O6

Explanation:
1. Empirical formula CH2O has mass 30 g; n = 180/30 = 6.
2. Molecular formula = C6H12O6. The others do not have H : O = 2 : 1 or a mass of 180 g.

Teacher's Note:
1. Molecular formula = n × empirical formula.
2. Check the mass: 72 + 12 + 96 = 180.

Question 18: Air is taken as 80% N2 and 20% O2 by volume. Treating air as a solution of O2 in N2, the percentage by mass (w/w) of oxygen is
(a) 200/9
(b) 10/9
(c) 700/9
(d) 350/9
Show Answer & Explanation

Answer: (a) 200/9

Explanation:
1. For 1 mol of air: O2 = 0.2 × 32 = 6.4 g and N2 = 0.8 × 28 = 22.4 g; total 28.8 g.
2. % w/w of O2 = 6.4/28.8 × 100 = 22.2% = 200/9.

Teacher's Note:
1. Convert mole fractions to masses.
2. 200/9 = 22.2.

Question 19: Air is taken as 80% N2 and 20% O2 by volume. The density of air at NTP is
(a) cannot be determined
(b) 1 g L-1
(c) 2/7 g L-1
(d) 9/7 g L-1
Show Answer & Explanation

Answer: (d) 9/7 g L-1

Explanation:
1. Average molar mass of air = 0.8 × 28 + 0.2 × 32 = 28.8 g.
2. Density = 28.8/22.4 = 9/7 g L-1 (about 1.29).

Teacher's Note:
1. Density at NTP = molar mass / 22.4.
2. The mixture behaves like a single gas of average molar mass.

Question 20: 60 g of a solution containing 40% by mass NaCl is mixed with 100 g of a solution containing 15% by mass NaCl. The mass percent of NaCl in the final solution is
(a) 24.4%
(b) 78%
(c) 19.68%
(d) 48.8%
Show Answer & Explanation

Answer: (a) 24.4%

Explanation:
1. NaCl = 0.40 × 60 + 0.15 × 100 = 24 + 15 = 39 g.
2. Total mass = 160 g, so mass percent = 39/160 × 100 = 24.4%.

Teacher's Note:
1. Masses and mass of solute add on mixing.
2. Percent = solute / total solution × 100.

Question 21: 60 g of a 40% (w/w) NaCl solution is mixed with 100 g of a 15% (w/w) NaCl solution. If the final solution has density 1.6 g mL-1, its molarity is (NaCl = 58.5)
(a) 5.5 M
(b) 6.67 M
(c) 2.59 M
(d) none of these
Show Answer & Explanation

Answer: (b) 6.67 M

Explanation:
1. Volume of solution = 160/1.6 = 100 mL = 0.1 L.
2. Moles of NaCl = 39/58.5 = 0.667, so molarity = 0.667/0.1 = 6.67 M.

Teacher's Note:
1. Molarity uses the volume of solution, found from mass and density.
2. Moles of NaCl are the same as in the previous part.

Question 22: A gaseous organic compound has C = 52.17%, H = 13.04% and O = 34.78% by mass and molar mass 46 g mol-1. Which of the following properties does it have? (p) 1 mol contains 4 N_A atoms of H; (q) empirical formula is the same as its molecular formula; (r) combustion gives more moles of CO2 than H2O; (s) 0.25 mol on combustion gives 11.2 L of CO2 at NTP.
(a) q and r only
(b) p, q and r only
(c) q and s only
(d) p and q only
Show Answer & Explanation

Answer: (c) q and s only

Explanation:
1. C : H : O = 4.35 : 13.04 : 2.17 = 2 : 6 : 1, so the formula is C2H6O (mass 46).
2. (p) false: 6 H atoms. (q) true: empirical mass = molar mass. (r) false: 2 mol CO2 and 3 mol H2O.
3. (s) true: 0.25 mol gives 0.5 mol CO2 = 11.2 L.

Teacher's Note:
1. Find the formula from the composition first.
2. Then test each property in turn.

Question 23: A hydrocarbon has 10.5 g of carbon per gram of hydrogen and vapour density 46. Which properties does it have? (p) 1 mol contains 4 N_A atoms of H; (q) empirical formula is the same as the molecular formula; (r) combustion gives more moles of CO2 than H2O; (s) 0.25 mol gives 11.2 L of CO2 at NTP.
(a) p, q and r only
(b) p and q only
(c) q and r only
(d) r and s only
Show Answer & Explanation

Answer: (c) q and r only

Explanation:
1. Molar mass = 2 × 46 = 92. C : H = 10.5/12 : 1 = 0.875 : 1 = 7 : 8, so the formula is C7H8 (mass 92).
2. (q) true. (r) true: 7 mol CO2 against 4 mol H2O.
3. (p) false: 8 H atoms. (s) false: 0.25 mol gives 1.75 mol CO2.

Teacher's Note:
1. The molar mass is twice the vapour density.
2. Compare the formula mass with the empirical mass.

Question 24: 0.3 g of an organic compound containing C, H and O gives 0.44 g of CO2 and 0.18 g of H2O on combustion, and has two O atoms per molecule. Which properties does it have? (p) 1 mol contains 4 N_A atoms of H; (q) empirical formula = molecular formula; (r) combustion gives more moles of CO2 than H2O; (s) 0.25 mol gives 11.2 L of CO2 at NTP.
(a) q and r only
(b) p and q only
(c) p and s only
(d) r and s only
Show Answer & Explanation

Answer: (c) p and s only

Explanation:
1. C = 0.44 × 12/44 = 0.12 g; H = 0.18 × 2/18 = 0.02 g; O = 0.30 - 0.14 = 0.16 g.
2. Moles C : H : O = 0.01 : 0.02 : 0.01, so the empirical formula is CH2O; with two O atoms, the formula is C2H4O2.
3. (p) true: 4 H. (q) false. (r) false: 2 mol CO2 and 2 mol H2O. (s) true: 0.25 mol gives 0.5 mol CO2 = 11.2 L.

Teacher's Note:
1. Find the mass of each element from the combustion products.
2. Two O atoms per molecule fixes the multiple of the empirical formula.

Question 25: 112.0 mL of NO2 at STP is liquefied; the density of the liquid is 1.15 g mL-1. The volume of the liquid and the number of molecules in it are (N = 14, O = 16)
(a) 0.40 mL and 6.02 × 1021
(b) 0.20 mL and 3.01 × 1021
(c) 0.20 mL and 6.02 × 1023
(d) 0.10 mL and 3.01 × 1022
Show Answer & Explanation

Answer: (b) 0.20 mL and 3.01 × 1021

Explanation:
1. Moles = 112 / 22 400 = 0.005 mol; mass = 0.005 × 46 = 0.23 g.
2. Volume of liquid = 0.23 / 1.15 = 0.20 mL.
3. Molecules = 0.005 × 6.02 × 1023 = 3.01 × 1021.

Teacher's Note:
1. Convert gas volume to moles, then to mass.
2. Liquid volume = mass / density.

Free study material for Chemistry

Practice MCQs for JEE Chemistry Some Basic Concepts Of Chemistry

Download Multiple Choice Questions: Some Basic Concepts Of Chemistry (JEE Chemistry)

Explore reliable practice questions for Some Basic Concepts Of Chemistry tailored for JEE Chemistry learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

Concept Clarification for Some Basic Concepts Of Chemistry

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Additional Study Resources for JEE Chemistry

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 32?

You can get most exhaustive JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 32 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 32 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 32, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 32?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 32 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.