Assertion: Equivalent weight of $\mathrm{NH}_3$ in the reaction $\mathrm{N}_2 \rightarrow \mathrm{NH}_3$ is $17 / 3$ while that of $\mathrm{N}_2$ is $28/6.$ Reason: Equivalent weight = $\frac{\text {Molecular weight}}{\text {number of } e^- \text{ lost or gained }}$.

Assertion is true, Reason is true and Reason is not correct explanation for Assertion

Assertion is true, Reason is false

Assertion is false, Reason is true

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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 79 questions

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31 Easy

31 of 79

Application +4 | -1

4 g-atom of Ag contains:

View answer and explanation

Correct option: C

432 g

Option C (432 g) is correct because it is the mass of 4 gram-atoms of Ag, calculated using the molar mass of Ag.

32 Easy

32 of 79

Remembering +4 | -1

A molal solution is one that contains one mole of a solute in:

View answer and explanation

Correct option: A

1000 g of the solvent

Option A is correct because a molal solution is defined as one containing one mole of solute in 1000 g of the solvent.

33 Easy

33 of 79

Application +4 | -1

The weight of 0.1 mole of

Na_2CO_3

View answer and explanation

Correct option: B

10.6 g

Option B (10.6 g) is correct because the weight of 0.1 mole of

Na_2CO_3

is calculated by multiplying the number of moles by the molar mass.

34 Easy

34 of 79

Application +4 | -1

A gas is found to have formula

{\left[ {\mathrm{CO}} \right]_x}

. Its

\mathrm{VD}

is 70. The value of

x

must be:

View answer and explanation

Correct option: B

Option B (5) is correct because the value of x is calculated using the vapor density formula and the given molecular formula.

35 Easy

35 of 79

Application +4 | -1

If 250 mL of a solution contains 2.7 g of

{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}

, the normality of the solution is:

View answer and explanation

Correct option: B

0.33

Option B (

0.33

) is correct because it is the normality of the solution calculated using the given mass of

{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}

and volume of the solution.

36 Easy

36 of 79

Analyzing +4 | -1

Which of the following samples contain the least number of atoms

View answer and explanation

Correct option: A

1g\,\,of\,C{O_2}\left( g \right)

Option A (

1g\,\,of\,C{O_2}\left( g \right)

) is correct because it contains the least number of atoms when considering the molar mass and atomicity of each compound.

37 Easy

37 of 79

Remembering +4 | -1

The modern atomic weight scale is based on

View answer and explanation

Correct option: B

{C}^{12}

Option B (

{C}^{12}

) is correct because the modern atomic weight scale is based on the carbon-12 isotope, which is assigned an atomic mass of exactly 12 atomic mass units.

38 Easy

38 of 79

Application +4 | -1

Four faradays of electricity is passed through a solution of

{CuSO}_4

. The mass of copper deposited at the cathode is: (At. mass of

{Cu}=63.5 \mathrm{amu}

View answer and explanation

Correct option: C

127 \mathrm{~g}

Option C (

127 \mathrm{~g}

) is correct because it represents the mass of copper deposited when four faradays of electricity are passed through a

CuSO_4

solution.

39 Easy

39 of 79

Application +4 | -1

Vapour density of a volatile substance is

(\mathrm{C}{\mathrm{H}_4} = \mathrm{ }1\mathrm{ })

. Its molecular weight would be:

View answer and explanation

Correct option: C

Option C (64) is correct because the molecular weight is twice the vapor density, and the vapor density is given relative to methane (CH4 = 1).

40 Easy

40 of 79

Remembering +4 | -1

Molarity is expressed as:

View answer and explanation

Correct option: B

\mathrm{Mol\; litr}{\mathrm{e}^{ - 1}}

Option B (

\mathrm{Mol\; litr}{\mathrm{e}^{ - 1}}

) is correct because molarity is defined as moles of solute per liter of solution.

41 Easy

41 of 79

Application +4 | -1

The number of oxygen atoms in

24.9 \mathrm{~g}

\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}

is (molar mass of

\mathrm{Cu}=63 \mathrm{~g} \mathrm{~mol}^{-1}

)

View answer and explanation

Correct option: C

5.42 \times 10^{23}

The correct answer is

5.42 \times 10^{23}

because the number of oxygen atoms is calculated from the moles of the compound and the number of oxygen atoms per molecule.

42 Easy

42 of 79

Application +4 | -1

Active mass of

0.64 \mathrm{~g}\, {SO}_2

in 10 lit vessel is

View answer and explanation

Correct option: B

10^{-3}\, \mathrm{M}

Option B (

10^{-3}

M) is correct because the active mass is calculated by dividing the number of moles by the volume in liters.

43 Easy

43 of 79

Application +4 | -1

Vapour density of a gas is 32. Volume occupied by

16 \mathrm{~g}

substance at S.T.P.

View answer and explanation

Correct option: C

5.6 \mathrm{~L}

Option C (5.6 L) is correct because it is the volume occupied by 16g of a gas with a vapor density of 32 at STP.

44 Easy

44 of 79

Application +4 | -1

14 g of element X combine with 16 g of oxygen. On the basis of this information, which of the followings is a correct statement?

View answer and explanation

Correct option: C

The element X could have an atomic weight of 7 and its oxide is

{\mathrm{X}_2}\mathrm{O}

Option D (80%) is correct because it is the calculated percentage of association of the solute based on the observed freezing point depression and the van't Hoff factor.

45 Easy

45 of 79

Analyzing +4 | -1

Assertion: Equivalent weight of

\mathrm{NH}_3

in the reaction

\mathrm{N}_2 \rightarrow \mathrm{NH}_3

17 / 3

while that of

\mathrm{N}_2

28/6.

Reason: Equivalent weight =

\frac{\text {Molecular weight}}{\text {number of } e^- \text{ lost or gained }}

View answer and explanation

Correct option: A

Assertion is true, Reason is true and Reason is correct explanation for Assertion

Option A is correct because both the assertion and reason are true, and the reason correctly explains how to calculate the equivalent weight based on the number of electrons gained or lost.

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P-BLOCK ELEMENTS 736 MCQs Equilibrium 696 MCQs STRUCTURE OF ATOM 634 MCQs THERMODYNAMICS 536 MCQs CHEMICAL BONDING & MOLECULAR STRUCTURE 514 MCQs SOLUTIONS 480 MCQs

46 Easy

46 of 79

Application +4 | -1

The weight of lime obtained by heating

200 \mathrm{~kg}

95 \%

pure lime stone is:

View answer and explanation

Correct option: B

106.4 \mathrm{~kg}

Option B (106.4 kg) is correct because it is the calculated weight of lime (CaO) obtained from heating 200 kg of 95% pure limestone (CaCO3).

47 Easy

47 of 79

Understanding +4 | -1

The law of definite proportions is not applicable to nitrogen oxide because

View answer and explanation

Correct option: C

Nitrogen equivalent weight is variable

Option C (Nitrogen equivalent weight is variable) is correct because nitrogen exhibits variable valency in its oxides, leading to different equivalent weights.

48 Easy

48 of 79

Application +4 | -1

At NTP,

5.6\,\,\mathrm{litre}

of a gas weighs

8\mathrm{g}

. The vapour density of the gas is

View answer and explanation

Correct option: C

Option C (16) is correct because the vapor density is half of the molar mass, which is calculated from the given mass and volume at NTP.

49 Easy

49 of 79

Application +4 | -1

One litre of

{\mathrm{C}{\mathrm{O}_2}}

is passed over hot coke. The volume becomes 1.4 litre. The per cent composition of products is:

View answer and explanation

Correct option: C

0.6 litre

{\mathrm{C}{\mathrm{O}_2}}

and 0.8 litre CO

Option C (0.6 litre

{\mathrm{C}{\mathrm{O}_2}}

and 0.8 litre CO) is correct because it accurately represents the product composition after the reaction of CO2 with hot coke.

50 Easy

50 of 79

Understanding +4 | -1

S.I. unit of molarity will be?

View answer and explanation

Correct option: C

\mathrm{mol} / \mathrm{m}^3

Option C (

\mathrm{mol} / \mathrm{m}^3

) is correct because the S.I. unit for volume is

m^3

, and molarity is defined as moles per liter.

51 Easy

51 of 79

Analyzing +4 | -1

In the reaction

\mathrm{Br}_2+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{NaBr}+\mathrm{NaBrO}_3+\mathrm{CO}_2

The equivalent weight of

\mathrm{NaBrO}_3

View answer and explanation

Correct option: C

\frac{\text {Molecular weight}}{5}

The correct answer is C (Molecular weight/5) because the equivalent weight of NaBrO3 is its molecular weight divided by the change in oxidation state of Br, which is 5.

52 Easy

52 of 79

Application +4 | -1

The total number of electrons in 18 ml of water (density = 1/gml

View answer and explanation

Correct option: B

6.023\,\, \times \,\,{10^{^{24}}}

Option B (

6.023 \times 10^{24}

) is correct because it is calculated by finding the number of moles of water and multiplying by the number of electrons per water molecule and Avogadro's number.

53 Easy

53 of 79

Application +4 | -1

How many moles of

\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}

would be in 50 g of the substance?

View answer and explanation

Correct option: D

0.140 mol

Option D (0.140 mol) is correct because the number of moles is calculated by dividing the given mass by the molar mass of the substance.

54 Easy

54 of 79

Application +4 | -1

Which has minimum number of atoms of oxygen ?

View answer and explanation

Correct option: D

12.044 \times 10^{22}

molecules of

\mathrm{CO}_2

The correct answer is D (

12.044 \times 10^{22}

molecules of

CO_2

) because it contains the minimum number of oxygen atoms compared to the other options.

55 Easy

55 of 79

Application +4 | -1

The mass of

\mathrm{112}\;\mathrm{c}{\mathrm{m}^3}\;\mathrm{C}{\mathrm{H}_4}

gas a STP is

View answer and explanation

Correct option: C

0.08 g

Option C (0.08 g) is correct because it is the mass of methane calculated using the molar volume at STP.

56 Easy

56 of 79

Application +4 | -1

Which has the maximum number of atoms?

View answer and explanation

Correct option: B

1\;\mathrm{g}\;{\mathrm{H}_2}

Option B is correct because 1 g of H₂ contains the maximum number of moles, and therefore the maximum number of atoms.

57 Easy

57 of 79

Application +4 | -1

Volume occupied by

3.01 \times {10^{23}}\;\;

molecules of acetylene at NTP is:

View answer and explanation

Correct option: B

11.2\,\,\mathrm{litre}

Option B (

11.2\,\,\mathrm{litre}

) is correct because

3.01 \times {10^{23}}

molecules is half a mole, and half a mole of any gas at NTP occupies 11.2 liters.

58 Easy

58 of 79

Application +4 | -1

How many gram of

\mathrm{NaOH}

will be required to neutralize

12.2 \mathrm{~g}

benzoic acid?

View answer and explanation

Correct option: D

4 \mathrm{~g}

Option D (

4 \text{ g}

) is correct because the mass of NaOH required is calculated based on the stoichiometry of the neutralization reaction.

59 Easy

59 of 79

Application +4 | -1

The correctly reported answer of addition of 29.4406, 3.2 and 2.25 will have significant figures

View answer and explanation

Correct option: A

The correct answer is A (3) because the result of the addition should be rounded to the least precise decimal place, which is one decimal place in this case.

60 Easy

60 of 79

Remembering +4 | -1

The significant figures in 3400 are

View answer and explanation

Correct option: A

Option A (2) is correct because in the number 3400, only the digits 3 and 4 are significant; the trailing zeros are not significant unless indicated otherwise.

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