JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 07

Practice MCQs for JEE Chemistry Some Basic Concepts Of Chemistry

Explore reliable objective questions for Some Basic Concepts Of Chemistry tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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 vapour density of a gas is 16. The ratio of specific heats at constant pressure and constant volume is 1.4. Then its atomic weight is
(a) 8
(b) 16
(c) 24
(d) 32
Show Answer & Explanation

Answer: (b) 16

Explanation:
1. Molar mass = 2 × 16 = 32.
2. A ratio of specific heats of 1.4 shows a diatomic gas, so atomic weight = 32/2 = 16.

Teacher's Note:
1. The ratio 1.67 means monatomic and 1.4 means diatomic.
2. Divide the molar mass by the atomicity.

Question 2: The weight of the substance that displaces 22.4 L of air at NTP is
(a) Molecular weight
(b) Atomic weight
(c) Equivalent weight
(d) All
Show Answer & Explanation

Answer: (a) Molecular weight

Explanation:
1. 22.4 L of vapour at NTP is one mole, and its mass is the molecular weight in grams.

Teacher's Note:
1. This is the principle of Victor Meyer's method.
2. The volume of vapour equals the volume of air displaced.

Question 3: 0.39 g of a liquid on vapourisation gave 112 mL of vapour at STP. Its molecular weight is
(a) 39
(b) 18.5
(c) 78
(d) 112
Show Answer & Explanation

Answer: (c) 78

Explanation:
1. Molar mass = 0.39 × 22400/112 = 78.

Teacher's Note:
1. Mass of 22400 mL gives the molar mass.
2. This is benzene, with molar mass 78.

Question 4: In Victor Meyer's method, 0.2 g of a volatile compound on volatilisation gave 56 mL of vapour at STP. Its molecular weight is
(a) 40
(b) 60
(c) 80
(d) 120
Show Answer & Explanation

Answer: (c) 80

Explanation:
1. Molar mass = 0.2 × 22400/56 = 80.

Teacher's Note:
1. The air displaced equals the vapour volume at STP.
2. Scale to 22400 mL.

Question 5: 510 mg of a liquid on vapourisation in Victor Meyer's apparatus displaces 67.2 cc of dry air (at NTP). The molecular weight of the liquid is
(a) 130
(b) 17
(c) 1700
(d) 170
Show Answer & Explanation

Answer: (d) 170

Explanation:
1. Mass = 0.510 g.
2. Molar mass = 0.510 × 22400/67.2 = 170.

Teacher's Note:
1. Convert mg to g first.
2. The result is 170 g/mol.

Question 6: 5 L of a gas at STP weighs 6.25 g. What is its gram molecular weight?
(a) 1.25
(b) 14
(c) 28
(d) 56
Show Answer & Explanation

Answer: (c) 28

Explanation:
1. Molar mass = 6.25 × 22.4/5 = 28 g/mol.

Teacher's Note:
1. Mass of 22.4 L is the gram molecular weight.
2. This is the molar mass of N2 or CO.

Question 7: 0.44 g of a colourless oxide of nitrogen occupies 224 mL at STP. The compound is
(a) N2O
(b) NO
(c) N2O4
(d) NO2
Show Answer & Explanation

Answer: (a) N2O

Explanation:
1. Molar mass = 0.44 × 22400/224 = 44 g/mol.
2. N2O has molar mass 44 and is colourless; NO2 is brown.

Teacher's Note:
1. Compute the molar mass and match the oxide.
2. N2O is laughing gas.

Question 8: One litre of a certain gas weighs 1.16 g at STP. The gas may possibly be
(a) C2H2
(b) CO
(c) O2
(d) NH3
Show Answer & Explanation

Answer: (a) C2H2

Explanation:
1. Molar mass = 1.16 × 22.4 = 26 g/mol.
2. C2H2 has molar mass 26.

Teacher's Note:
1. Density × 22.4 gives molar mass.
2. CO is 28, O2 is 32 and NH3 is 17.

Question 9: The equivalent weight of a bivalent metal is 32.7. The molecular weight of its chloride is
(a) 68.2
(b) 103.7
(c) 136.4
(d) 166.3
Show Answer & Explanation

Answer: (c) 136.4

Explanation:
1. Atomic weight = 2 × 32.7 = 65.4.
2. MCl2 = 65.4 + 71 = 136.4.

Teacher's Note:
1. Valency is 2, so the chloride is MCl2.
2. Add 2 × 35.5 = 71.

Question 10: The oxide of an element has the molecular formula M2O3. If the equivalent mass of the metal is 9, the molecular mass of the oxide will be
(a) 27
(b) 75
(c) 102
(d) 18
Show Answer & Explanation

Answer: (c) 102

Explanation:
1. Valency = 3, so atomic mass of M = 9 × 3 = 27.
2. Molar mass of M2O3 = 2 × 27 + 48 = 102.

Teacher's Note:
1. Valency in M2O3 is 3.
2. Add the mass of three oxygen atoms.

Question 11: The law of multiple proportions was proposed by
(a) Lavoisier
(b) Dalton
(c) Proust
(d) Gay-Lussac
Show Answer & Explanation

Answer: (b) Dalton

Explanation:
1. John Dalton proposed the law of multiple proportions in 1803.

Teacher's Note:
1. Lavoisier: conservation of mass. Proust: definite proportions.
2. Gay-Lussac: combining volumes.

Question 12: Which one of the following pairs of compounds illustrates the law of multiple proportions?
(a) H2O, Na2O
(b) MgO, Na2O
(c) Na2O, BaO
(d) SnCl2, SnCl4
Show Answer & Explanation

Answer: (d) SnCl2, SnCl4

Explanation:
1. The law applies to two compounds made of the same two elements.
2. SnCl2 and SnCl4 contain Sn and Cl, with Cl masses in the ratio 1 : 2 for a fixed mass of Sn.

Teacher's Note:
1. Look for the same pair of elements forming two compounds.
2. The ratio of the second element's masses is a simple whole number.

Question 13: In the reaction N2 + 3H2 → 2NH3, the ratio by volume of N2, H2 and NH3 is 1 : 3 : 2. This illustrates the law of
(a) Definite proportion
(b) Multiple proportion
(c) Conservation of mass
(d) Gaseous volumes
Show Answer & Explanation

Answer: (d) Gaseous volumes

Explanation:
1. Gases combine in simple whole number volume ratios at the same T and P.
2. This is Gay-Lussac's law of gaseous volumes.

Teacher's Note:
1. Volume ratio of gases means Gay-Lussac's law.
2. The ratio is the same as the mole ratio.

Question 14: Different proportions of oxygen in the various oxides of nitrogen prove the law of
(a) Equivalent proportion
(b) Multiple proportion
(c) Constant proportion
(d) Conservation of matter
Show Answer & Explanation

Answer: (b) Multiple proportion

Explanation:
1. Nitrogen and oxygen form several oxides with oxygen masses in simple whole number ratios for fixed nitrogen mass.
2. This is the law of multiple proportions.

Teacher's Note:
1. Same two elements, different compounds: multiple proportions.
2. N2O, NO, N2O3, NO2 and N2O5 are examples.

Question 15: Oxygen combines with two isotopes of carbon, 12C and 14C, to form two samples of carbon dioxide. The data illustrates
(a) Law of conservation of mass
(b) Law of multiple proportions
(c) Law of gaseous volume
(d) None of these
Show Answer & Explanation

Answer: (d) None of these

Explanation:
1. The two samples have different masses of carbon per mole, so they are not two different compounds of the same elements in a simple ratio.
2. None of the given laws is illustrated.

Teacher's Note:
1. Isotopes have different masses but identical chemistry.
2. The laws of chemical combination apply to natural samples.

Question 16: The law of conservation of mass holds good for all of the following except
(a) All chemical reactions
(b) Nuclear reactions
(c) Endothermic reactions
(d) Exothermic reactions
Show Answer & Explanation

Answer: (b) Nuclear reactions

Explanation:
1. In nuclear reactions, a small part of mass is converted into energy.
2. Mass is conserved in all chemical reactions, including endothermic and exothermic ones.

Teacher's Note:
1. Einstein's mass-energy relation explains nuclear mass defect.
2. Chemical energy changes are too small to measure as a mass change.

Question 17: The number of molecules in 100 mL each of O2, NH3 and CO2 at STP is
(a) In the order CO2 < O2 < NH3
(b) In the order NH3 < O2 < CO2
(c) The same
(d) NH3 = CO2 < O2
Show Answer & Explanation

Answer: (c) The same

Explanation:
1. Equal volumes of gases at the same T and P contain equal numbers of molecules (Avogadro's law).

Teacher's Note:
1. The identity of the gas does not matter.
2. This holds for ideal gases at the same T and P.

Question 18: The empirical formula of an organic compound containing carbon and hydrogen is CH2. The mass of one litre of this organic gas is exactly equal to that of one litre of N2 at the same temperature and pressure. The molecular formula of the organic gas is
(a) C2H4
(b) C3H6
(c) C6H12
(d) C4H8
Show Answer & Explanation

Answer: (a) C2H4

Explanation:
1. Equal volumes of gases at the same T and P have the same number of molecules, so equal mass means equal molar mass: 28.
2. Mass of CH2 = 14, so n = 28/14 = 2 and the formula is C2H4.

Teacher's Note:
1. Same volume and same mass means same molar mass.
2. Molecular formula = n × empirical formula.

Question 19: Four one-litre flasks are separately filled with the gases hydrogen, helium, oxygen and ozone at the same room temperature and pressure. The ratio of the total number of atoms of these gases present in the different flasks would be
(a) 1 : 1 : 1 : 1
(b) 1 : 2 : 2 : 3
(c) 2 : 1 : 2 : 3
(d) 2 : 1 : 3 : 2
Show Answer & Explanation

Answer: (c) 2 : 1 : 2 : 3

Explanation:
1. Equal volumes have equal numbers of molecules, say N.
2. Atoms: H2 = 2N, He = N, O2 = 2N, O3 = 3N.
3. Ratio = 2 : 1 : 2 : 3.

Teacher's Note:
1. Multiply the common molecule count by atomicity.
2. Helium is monatomic.

Question 20: A container of volume V contains 0.28 g of N2 gas. If the same volume of an unknown gas under similar conditions of temperature and pressure weighs 0.44 g, the molecular mass of the gas is
(a) 22
(b) 44
(c) 66
(d) 88
Show Answer & Explanation

Answer: (b) 44

Explanation:
1. Same volume means the same number of moles.
2. 0.28/28 = 0.01 mol, so the molar mass of the unknown gas = 0.44/0.01 = 44.

Teacher's Note:
1. Equal moles in the same volume at the same T and P.
2. Use the ratio of masses times 28.

Question 21: A and B are two identical vessels. A contains 15 g of ethane at 1 atm and 298 K. The vessel B contains 75 g of a gas X2 at the same temperature and pressure. The vapour density of X2 is
(a) 75
(b) 150
(c) 37.5
(d) 45
Show Answer & Explanation

Answer: (a) 75

Explanation:
1. 15 g of ethane (30 g/mol) = 0.5 mol.
2. The same moles of X2 weigh 75 g, so molar mass = 150 and vapour density = 75.

Teacher's Note:
1. Same vessel, T and P means the same number of moles.
2. V.D. = molar mass ÷ 2.

Question 22: When 100 g of ethylene polymerises to polyethylene according to the equation nCH2=CH2 → −(CH2−CH2)_n−, the weight of polyethylene produced will be
(a) \( \frac{n}{2} \) g
(b) 100 g
(c) \( \frac{100}{n} \) g
(d) 100n g
Show Answer & Explanation

Answer: (b) 100 g

Explanation:
1. Polymerisation only joins molecules, so no mass is lost or gained.
2. Mass of polymer = 100 g.

Teacher's Note:
1. The law of conservation of mass applies.
2. No by-product is formed in addition polymerisation.

Question 23: If the law of conservation of mass is to hold true, then 20.8 g of BaCl2 on reaction with 9.8 g of H2SO4 will produce 7.3 g of HCl. Determine the weight of BaSO4 produced.
(a) 11.65 g
(b) 23.3 g
(c) 25.5 g
(d) 30.6 g
Show Answer & Explanation

Answer: (b) 23.3 g

Explanation:
1. Total mass of reactants = 20.8 + 9.8 = 30.6 g.
2. Mass of BaSO4 = 30.6 − 7.3 = 23.3 g.

Teacher's Note:
1. Total mass of reactants equals total mass of products.
2. Check: 0.1 mol BaSO4 = 23.3 g.

Question 24: A chemical equation is balanced according to the law of
(a) Multiple proportion
(b) Constant composition
(c) Gaseous volume
(d) Conservation of mass
Show Answer & Explanation

Answer: (d) Conservation of mass

Explanation:
1. Balancing makes the number of atoms of each element equal on both sides, which follows the law of conservation of mass.

Teacher's Note:
1. Balanced equation means mass is conserved.
2. Other laws deal with ratios of combining masses or volumes.

Question 25: Two flasks A and B of equal capacity contain NH3 and SO2 gas respectively under similar conditions. Which flask has more number of moles?
(a) A
(b) B
(c) Both have the same moles
(d) None
Show Answer & Explanation

Answer: (c) Both have the same moles

Explanation:
1. Equal volumes at the same T and P have equal moles (Avogadro's law).

Teacher's Note:
1. The nature of the gas is irrelevant.
2. Same capacity, T and P means same moles.

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)

Test your conceptual understanding of Some Basic Concepts Of Chemistry with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Chemistry, these problem sets build accuracy and prepare students for objective exams.

Concept Clarification for Some Basic Concepts Of Chemistry

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

Additional Study Resources for JEE Chemistry

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 07?

You can get most exhaustive JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 07 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 07 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 07, 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 07?

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