JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 16

Practice MCQs for JEE Chemistry Some Basic Concepts Of Chemistry

Review structured MCQ sets for JEE Chemistry Some Basic Concepts Of Chemistry. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Access Some Basic Concepts Of Chemistry Questions and Solutions

Access the complete set of multiple-choice questions for Some Basic Concepts Of Chemistry below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The molecular masses of \( \ce{H2O} \), \( \ce{CO2} \) and \( \ce{CH4} \) (H = 1, C = 12, O = 16) are respectively
(a) 18, 28 and 16
(b) 16, 44 and 18
(c) 18, 44 and 12
(d) 18, 44 and 16
Show Answer & Explanation

Answer: (d) 18, 44 and 16

Explanation:
1. \( \ce{H2O} \) = 2 + 16 = 18.
2. \( \ce{CO2} \) = 12 + 32 = 44.
3. \( \ce{CH4} \) = 12 + 4 = 16.

Teacher's Note:
1. Molecular mass is the sum of the atomic masses of all atoms in the molecule.

Question 2: The mass percentages of Na, S and O in sodium sulphate, \( \ce{Na2SO4} \), are (Na = 23, S = 32, O = 16)
(a) 16.2%, 22.5% and 61.3%
(b) 32.4%, 22.5% and 45.1%
(c) 32.4%, 45.1% and 22.5%
(d) 23.0%, 32.0% and 45.0%
Show Answer & Explanation

Answer: (b) 32.4%, 22.5% and 45.1%

Explanation:
1. Molar mass = 46 + 32 + 64 = 142 g/mol.
2. Na = 46/142 = 32.4%, S = 32/142 = 22.5%, O = 64/142 = 45.1%.

Teacher's Note:
1. Percentage of an element = (mass of that element in one mole / molar mass) x 100.

Question 3: An oxide of iron contains 69.9% iron and 30.1% oxygen by mass. Its empirical formula is (Fe = 56)
(a) FeO
(b) \( \ce{Fe2O3} \)
(c) \( \ce{Fe3O4} \)
(d) \( \ce{Fe2O} \)
Show Answer & Explanation

Answer: (b) \( \ce{Fe2O3} \)

Explanation:
1. Moles: Fe = 69.9/56 = 1.248; O = 30.1/16 = 1.881.
2. Ratio Fe : O = 1 : 1.5 = 2 : 3.
3. Empirical formula = \( \ce{Fe2O3} \).

Teacher's Note:
1. Divide percentages by atomic masses, then divide by the smallest and clear fractions.

Question 4: The mass of copper that can be obtained from 100 g of copper sulphate, \( \ce{CuSO4} \), is (Cu = 63.5, S = 32, O = 16)
(a) 63.5 g
(b) 159.5 g
(c) 19.9 g
(d) 39.8 g
Show Answer & Explanation

Answer: (d) 39.8 g

Explanation:
1. Molar mass of \( \ce{CuSO4} \) = 159.5 g/mol.
2. Cu per 100 g = 63.5/159.5 x 100 = 39.8 g.

Teacher's Note:
1. Mass of an element from a compound = mass percentage x mass of compound.

Question 5: Rounded off to three significant figures, 34.216, 10.4107, 0.04597 and 2808 become respectively
(a) 34.2, 10.4, 0.0460 and 2810
(b) 34.2, 10.4, 0.046 and 2808
(c) 34.3, 10.5, 0.0459 and 2800
(d) 34.2, 10.4, 0.0459 and 2810
Show Answer & Explanation

Answer: (a) 34.2, 10.4, 0.0460 and 2810

Explanation:
1. 34.216 gives 34.2. 10.4107 gives 10.4.
2. 0.04597 gives 0.0460 (the last zero is significant). 2808 gives 2810.

Teacher's Note:
1. Look at the fourth significant digit to round and keep the trailing zero in 0.0460.

Question 6: The molarity of nitric acid in a sample of density 1.41 g/mL containing 69% \( \ce{HNO3} \) by mass is (H = 1, N = 14, O = 16)
(a) 10.8 M
(b) 22.3 M
(c) 15.4 M
(d) 1.54 M
Show Answer & Explanation

Answer: (c) 15.4 M

Explanation:
1. 1 L of solution weighs 1410 g and holds 0.69 x 1410 = 972.9 g of \( \ce{HNO3} \).
2. Moles = 972.9/63 = 15.44.
3. Molarity = 15.4 M.

Teacher's Note:
1. Molarity = 10 x density x mass percent / molar mass.

Question 7: The concentration of sugar (\( \ce{C12H22O11} \), 342 g/mol), in mol/L, when 20 g is dissolved in enough water to make 2 L is
(a) 0.0585
(b) 0.0292
(c) 0.292
(d) 0.0146
Show Answer & Explanation

Answer: (b) 0.0292

Explanation:
1. Moles = 20/342 = 0.0585 mol.
2. Molarity = 0.0585/2 = 0.0292 mol/L.

Teacher's Note:
1. Divide moles by the final volume of solution in litre.

Question 8: The number of atoms in (i) 52 mol of Ar, (ii) 52 u of He and (iii) 52 g of He are respectively \( (N_A = 6.022\times10^{23} \), He = 4, Ar = 40)
(a) 52, 13 and 13
(b) \( 3.13\times10^{25} \), 52 and \( 3.13\times10^{25} \)
(c) \( 3.13\times10^{25} \), 13 and \( 3.13\times10^{25} \)
(d) \( 3.13\times10^{25} \), 13 and \( 7.83\times10^{24} \)
Show Answer & Explanation

Answer: (d) \( 3.13\times10^{25} \), 13 and \( 7.83\times10^{24} \)

Explanation:
1. Ar is monatomic: 52 mol = 52 x \( 6.022\times10^{23} = 3.13\times10^{25} \) atoms.
2. 52 u of He = 52/4 = 13 atoms (one atom weighs 4 u).
3. 52 g of He = 13 mol = \( 13\times6.022\times10^{23} = 7.83\times10^{24} \) atoms.

Teacher's Note:
1. Distinguish u (mass of a few atoms) from g (mass of a mole-scale sample).

Question 9: From 64 g of oxygen (11.2 L of \( \ce{O2} \) at STP and \( 6.02\times10^{23} \) oxygen atoms) are removed. The mass of oxygen gas left is
(a) 16 g
(b) 32 g
(c) 48 g
(d) 24 g
Show Answer & Explanation

Answer: (b) 32 g

Explanation:
1. Initial = 64/32 = 2 mol of \( \ce{O2} \).
2. 11.2 L at STP = 0.5 mol of \( \ce{O2} \) removed. \( 6.02\times10^{23} \) atoms = 1 mol atoms = 0.5 mol of \( \ce{O2} \) removed.
3. Left = 2 - 0.5 - 0.5 = 1 mol = 32 g.

Teacher's Note:
1. Convert each removed quantity to moles of \( \ce{O2} \) before subtracting.

Question 10: The molality of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 is (C = 12, O = 16, H = 1)
(a) 2.31 mol/kg
(b) 0.040 mol/kg
(c) 2.22 mol/kg
(d) 1.11 mol/kg
Show Answer & Explanation

Answer: (a) 2.31 mol/kg

Explanation:
1. Take 1 mol of solution: ethanol = 0.04 mol, water = 0.96 mol = 17.28 g.
2. Molality = 0.04/0.01728 kg = 2.31 mol/kg.

Teacher's Note:
1. Pick a convenient total of 1 mole of solution and convert water moles to kilogram.

Question 11: From the isotopic data for argon: \( ^{36}Ar \) 35.96755 g/mol (0.337%), \( ^{38}Ar \) 37.96272 g/mol (0.063%), \( ^{40}Ar \) 39.9624 g/mol (99.600%), the molar mass of naturally occurring argon is
(a) 39.95 g/mol
(b) 39.96 g/mol
(c) 37.96 g/mol
(d) 38.99 g/mol
Show Answer & Explanation

Answer: (a) 39.95 g/mol

Explanation:
1. Average = 0.00337 x 35.96755 + 0.00063 x 37.96272 + 0.99600 x 39.9624.
2. = 0.1212 + 0.0239 + 39.8026 = 39.95 g/mol.

Teacher's Note:
1. Weight each isotopic mass by its fractional abundance.

Question 12: The amounts of carbon dioxide produced when (i) 1 mol of carbon is burnt in air, (ii) 1 mol of carbon is burnt in 16 g of dioxygen and (iii) 2 mol of carbon are burnt in 16 g of dioxygen are respectively
(a) 44 g, 22 g and 22 g
(b) 44 g, 44 g and 44 g
(c) 44 g, 22 g and 44 g
(d) 22 g, 22 g and 44 g
Show Answer & Explanation

Answer: (a) 44 g, 22 g and 22 g

Explanation:
1. \( \ce{C + O2 -> CO2} \). (i) Air has excess oxygen, so 1 mol \( \ce{CO2} \) = 44 g.
2. (ii) 16 g \( \ce{O2} \) = 0.5 mol limiting, so \( \ce{CO2} \) = 0.5 mol = 22 g.
3. (iii) Again \( \ce{O2} \) is limiting (0.5 mol), so 22 g.

Teacher's Note:
1. Find the limiting reagent each time. Excess carbon does not make more product.

Question 13: A drinking water sample is contaminated with chloroform (\( \ce{CHCl3} \), 119.5 g/mol) at 15 ppm by mass. The contamination as percent by mass and as molality are respectively
(a) \( 1.5\times10^{-3} \)% and \( 1.26\times10^{-4} \) m
(b) \( 1.5\times10^{-5} \)% and \( 1.26\times10^{-4} \) m
(c) \( 1.5\times10^{-3} \)% and \( 1.26\times10^{-2} \) m
(d) 15% and \( 1.26\times10^{-4} \) m
Show Answer & Explanation

Answer: (a) \( 1.5\times10^{-3} \)% and \( 1.26\times10^{-4} \) m

Explanation:
1. 15 ppm = 15 g in \( 10^6 \) g of solution = \( 15/10^6\times100 = 1.5\times10^{-3} \)%.
2. Moles of \( \ce{CHCl3} \) = 15/119.5 = 0.1255 mol in about 1000 kg of water.
3. Molality = \( 0.1255/1000 = 1.26\times10^{-4} \) mol/kg.

Teacher's Note:
1. ppm is mass per million parts of solution. Convert to percent by multiplying by \( 10^{-4} \).

Question 14: \( \ce{N2 + 3H2 -> 2NH3} \). When 2.00 x \( 10^{3} \) g of \( \ce{N2} \) reacts with 1.00 x \( 10^{3} \) g of \( \ce{H2} \), the mass of ammonia produced and the mass of the reactant left unreacted are (N = 14)
(a) 2429 g of \( \ce{NH3} \); 1000 g of \( \ce{N2} \) left
(b) 3000 g of \( \ce{NH3} \); none left
(c) 1214 g of \( \ce{NH3} \); 571 g of \( \ce{H2} \) left
(d) 2429 g of \( \ce{NH3} \); 571 g of \( \ce{H2} \) left
Show Answer & Explanation

Answer: (d) 2429 g of \( \ce{NH3} \); 571 g of \( \ce{H2} \) left

Explanation:
1. Moles: \( \ce{N2} \) = 2000/28 = 71.4; \( \ce{H2} \) = 1000/2 = 500.
2. \( \ce{H2} \) needed for all the nitrogen = 3 x 71.4 = 214.3 mol, so \( \ce{N2} \) is limiting.
3. \( \ce{NH3} \) = 2 x 71.4 = 142.9 mol = 2429 g.
4. \( \ce{H2} \) left = 500 - 214.3 = 285.7 mol = 571 g.

Teacher's Note:
1. Compare moles divided by coefficients. The leftover is the excess reagent minus what is used.

Question 15: 28.7 pm, 15.15 microseconds and 25365 mg expressed in basic SI units are respectively
(a) \( 2.87\times10^{-9} \) m, \( 1.515\times10^{-3} \) s and 25.365 kg
(b) \( 2.87\times10^{-11} \) m, \( 1.515\times10^{-5} \) s and 0.025365 kg
(c) \( 2.87\times10^{-11} \) m, \( 1.515\times10^{-5} \) s and 25.365 g
(d) \( 2.87\times10^{-8} \) m, \( 1.515\times10^{-6} \) s and 0.25365 kg
Show Answer & Explanation

Answer: (b) \( 2.87\times10^{-11} \) m, \( 1.515\times10^{-5} \) s and 0.025365 kg

Explanation:
1. pico = \( 10^{-12} \): 28.7 pm = \( 2.87\times10^{-11} \) m.
2. micro = \( 10^{-6} \): 15.15 us = \( 1.515\times10^{-5} \) s.
3. 25365 mg = 25.365 g = 0.025365 kg (the SI unit of mass is kg).

Teacher's Note:
1. Mass in SI is kilogram, not gram. Multiply by the prefix power of ten.

Question 16: A welding gas contains only carbon and hydrogen. Burning a sample gives 3.38 g of \( \ce{CO2} \) and 0.690 g of water. 10.0 L of the gas at STP weighs 11.6 g. The empirical formula, molar mass and molecular formula are respectively
(a) \( \ce{CH2} \), 28 g/mol and \( \ce{C2H4} \)
(b) CH, 26 g/mol and \( \ce{C2H2} \)
(c) CH, 78 g/mol and \( \ce{C6H6} \)
(d) \( \ce{CH3} \), 30 g/mol and \( \ce{C2H6} \)
Show Answer & Explanation

Answer: (b) CH, 26 g/mol and \( \ce{C2H2} \)

Explanation:
1. C = 3.38/44 = 0.0768 mol. H = 2 x 0.690/18 = 0.0767 mol. Ratio 1 : 1, so empirical formula CH.
2. Molar mass = 11.6/10.0 x 22.4 = 26 g/mol.
3. 26/13 = 2, so molecular formula = \( \ce{C2H2} \).

Teacher's Note:
1. Moles of C and H come from \( \ce{CO2} \) and \( \ce{H2O} \). Molar mass from the STP data gives the multiplier.

Question 17: The numbers of H, S and O atoms in 5 mol of \( \ce{H2S2O8} \) are respectively (in terms of \( N_A \))
(a) 10 \( N_A \), 10 \( N_A \) and 40 \( N_A \)
(b) 5 \( N_A \), 5 \( N_A \) and 20 \( N_A \)
(c) 10 \( N_A \), 10 \( N_A \) and 8 \( N_A \)
(d) 2 \( N_A \), 2 \( N_A \) and 8 \( N_A \)
Show Answer & Explanation

Answer: (a) 10 \( N_A \), 10 \( N_A \) and 40 \( N_A \)

Explanation:
1. Each formula unit has 2 H, 2 S and 8 O.
2. For 5 mol: H = 10, S = 10, O = 40 mol of atoms, that is 10 \( N_A \), 10 \( N_A \) and 40 \( N_A \) atoms.

Teacher's Note:
1. Number of atoms = moles x atoms per formula unit x \( N_A \).

Question 18: 10 mol of \( \ce{NH3} \) and 5 mol of \( \ce{H2SO4} \) have all their H atoms removed to form \( \ce{H2} \) gas. The number of \( \ce{H2} \) molecules formed is \( (N_A = 6\times10^{23} \))
(a) \( 6\times10^{24} \)
(b) \( 2.4\times10^{25} \)
(c) \( 3.6\times10^{25} \)
(d) \( 1.2\times10^{25} \)
Show Answer & Explanation

Answer: (d) \( 1.2\times10^{25} \)

Explanation:
1. H atoms = 10 x 3 + 5 x 2 = 40 mol.
2. \( \ce{H2} \) = 40/2 = 20 mol.
3. Molecules = \( 20\times6\times10^{23} = 1.2\times10^{25} \).

Teacher's Note:
1. Two H atoms make one \( \ce{H2} \) molecule, so halve the number of hydrogen atoms.

Question 19: Two compounds of tin and oxygen: A has 78.77% Sn and 21.23% O, B has 88.12% Sn and 11.88% O. The ratio of the masses of oxygen combining with a fixed mass of tin in A and B is
(a) 1 : 2
(b) 3 : 2
(c) 4 : 3
(d) 2 : 1
Show Answer & Explanation

Answer: (d) 2 : 1

Explanation:
1. Oxygen per g of Sn: A = 21.23/78.77 = 0.2695; B = 11.88/88.12 = 0.1348.
2. Ratio = 0.2695 : 0.1348 = 2 : 1, a simple ratio (law of multiple proportions).

Teacher's Note:
1. Divide the oxygen mass by the tin mass in each compound and compare.

Question 20: The density of a gaseous sample at NTP is 1.20 g/L. The weight of 1 mol of the gas is (molar volume 22.4 L)
(a) 26.9 g
(b) 22.4 g
(c) 18.7 g
(d) 1.20 g
Show Answer & Explanation

Answer: (a) 26.9 g

Explanation:
1. Mass of 22.4 L = 1.20 x 22.4 = 26.88 g.
2. So the molar mass is about 26.9 g/mol.

Teacher's Note:
1. Molar mass = density at STP x 22.4.

Question 21: \( \ce{N2O} \), \( \ce{NO2} \) and \( \ce{N2O3} \) are mixed in the mole ratio 3 : 2 : 1. The average molar mass of the mixture is (N = 14, O = 16)
(a) 46 g/mol
(b) 55 g/mol
(c) 50 g/mol
(d) 44 g/mol
Show Answer & Explanation

Answer: (c) 50 g/mol

Explanation:
1. Molar masses: 44, 46, 76.
2. Average = (3 x 44 + 2 x 46 + 1 x 76)/6 = (132 + 92 + 76)/6 = 50 g/mol.

Teacher's Note:
1. Average molar mass is the mole-fraction-weighted mean.

Question 22: A gaseous mixture contains 2 mol of \( \ce{CO2} \), 1 mol of \( \ce{H2} \) and 2 mol of He. The mole percentage of \( \ce{CO2} \) is
(a) 20%
(b) 40%
(c) 50%
(d) 25%
Show Answer & Explanation

Answer: (b) 40%

Explanation:
1. Total moles = 2 + 1 + 2 = 5.
2. Mole percentage of \( \ce{CO2} \) = 2/5 x 100 = 40%.

Teacher's Note:
1. Mole percent = moles of one component / total moles x 100.

Question 23: A compound has a haemoglobin-like structure with one Fe atom per molecule and contains 4.6% Fe. Its approximate molar mass is (Fe = 56)
(a) 56 g/mol
(b) 258 g/mol
(c) 1217 g/mol
(d) 12170 g/mol
Show Answer & Explanation

Answer: (c) 1217 g/mol

Explanation:
1. One mole of the compound contains 56 g of Fe, which is 4.6% of the molar mass.
2. Molar mass = 56/0.046 = 1217 g/mol.

Teacher's Note:
1. Minimum molar mass = atomic mass of the element divided by its mass fraction.

Question 24: The residue left on strongly heating 2.76 g of \( \ce{Ag2CO3} \) (\( \ce{Ag2CO3 -> 2Ag + CO2 + 1/2 O2} \)) is (Ag = 108, C = 12, O = 16)
(a) 2.16 g
(b) 1.08 g
(c) 2.76 g
(d) 4.32 g
Show Answer & Explanation

Answer: (a) 2.16 g

Explanation:
1. Molar mass of \( \ce{Ag2CO3} \) = 276 g/mol, so 2.76 g = 0.01 mol.
2. Ag = 2 x 0.01 = 0.02 mol = 2.16 g (the only solid residue).

Teacher's Note:
1. Only silver is left behind. The carbon dioxide and oxygen escape as gases.

Free study material for Chemistry

Practice MCQs for JEE Chemistry Some Basic Concepts Of Chemistry

About Some Basic Concepts Of Chemistry MCQs for 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.

How to Verify Your MCQ Answers

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Chemistry to ensure absolute clarity across all sub-topics in this chapter.

Enhance Speed with Online MCQ Tests

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 16?

You can get most exhaustive JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 16 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 16 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 16, 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 16?

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 16 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.