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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

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Showing 61-90 of 436 questions

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61 Medium
61 of 436
Analyzing +4 | -1
Statement I: Combustion of 16 g of CH416 \mathrm{~g} {\text { of }} \mathrm{CH}_4 gives 18 g18 \mathrm{~g} of H2O\mathrm{H}_2 \mathrm{O}.

Statement II: In the combustion of CH4,H2O\mathrm{CH}_4, \mathrm{H}_2 \mathrm{O} is one of the products.
View answer and explanation
Correct option: D
Statement I is false; Statement II is true
The correct answer is D (Statement I is false; Statement II is true) because the combustion of 16g of CH4 produces 36g of H2O, and H2O is indeed a product.
62 Easy
62 of 436
Remembering +4 | -1
Dulong and Petit’s law is valid only for
View answer and explanation
Correct option: D
Solid elements
Option D (Solid elements) is correct because Dulong and Petit's law, which relates atomic mass and specific heat, is generally applicable to solid elements at room temperature.
63 Medium
63 of 436
Application +4 | -1
In the following change, 3Fe+4H2O→Fe3O4+4H23\mathrm{Fe} + 4{\mathrm{H}_2}\mathrm{O} \to \mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4} + 4{\mathrm{H}_2}. If the atomic weight of iron is 56, then its equivalent weight will be :
View answer and explanation
Correct option: B
21
Option A (64000 amu{6 4 0 0 0} \mathrm{~a m u}) is correct because it is the molecular weight calculated from the percentage of Fe and the number of Fe atoms in haemoglobin.
64 Medium
64 of 436
Application +4 | -1
Number of Zn\mathrm{Zn} atoms present in 122 g122 \mathrm{~g} of ZnO\mathrm{ZnO} is (Atomic wt. of Zn=65\mathrm{Zn}=65)
View answer and explanation
Correct option: C
3 NA2\frac{3 \mathrm{~N}_{\mathrm{A}}}{2}
Option C (3 NA2\frac{{3 \mathrm{~N}_{\mathrm{A}}}}{{2}}) is correct because 122 g of ZnO corresponds to 1.5 moles, hence 1.5 N_A atoms of Zn.
65 Medium
65 of 436
Analyzing +4 | -1
Statement I: SO2\mathrm{SO}_2 formed by the oxidation of sulphur or hydrogen sulphide contains sulphur and oxygen in 1:21: 2 mass ratio.

Statement II: Different samples of a pure chemical substance always contain the same proportions of elements by mass.
View answer and explanation
Correct option: D
Statement I is false; Statement II is true
Option D (Statement I is false; Statement II is true) is correct because the mass ratio of S and O in SO2SO_2 is not 1:2, but the law of definite proportions (Statement II) is always true.
66 Hard
66 of 436
Application +4 | -1
2.4 kg of carbon is made to react with 1.35 kg of aluminium to form Al4C3.Al_4C_3. The maximum amount (in kg) of aluminium carbide formed is
View answer and explanation
Correct option: D
1.8
The correct answer is 1.8 kg because Aluminium is the limiting reagent. The maximum amount of aluminium carbide formed is determined by the limiting reactant.
67 Medium
67 of 436
Application +4 | -1
20.0 kg of H2(g)H_2(g) and 32 kg of O2(g)O_2(g) are reacted to produce H2O(l).H_2O(l). The amount of H2O(l)H_2O(l) formed after completion of reaction is
View answer and explanation
Correct option: D
72 kg
Option D (72 kg) is correct because 4 kg of H2H_2 reacts with 32 kg of O2O_2 to produce 36 kg of H2OH_2O, so 20 kg of H2H_2 reacts with 32 kg of O2O_2 to produce 72 kg of H2OH_2O.
68 Hard
68 of 436
Application +4 | -1
The amount of BaSO4{BaSO}_4 formed upon mixing 100 mL100 {~mL} of 20.8% BaCl220.8 \% \,{BaCl}_2 solution with 50 mL50 {~mL} of 9.8%H2SO49.8 \% {H}_2 {SO}_4 solution with 50 mL50 {~mL} of 9.8% 9.8 \%\, H2SO4{H}_2 {SO}_4 solution will be:

(Ba=137,Cl=35.5, S=32,H=1 and O=16 ) ({Ba}=137, {Cl}=35.5, {~S}=32, {H}=1 \text { and } {O}=16 \text { ) }
View answer and explanation
Correct option: A
23.3 g23.3 {~g}
Option A (23.3 g23.3 {~g}) is correct because it is the mass of BaSO4{BaSO}_4 formed based on the stoichiometry of the reaction between BaCl2{BaCl}_2 and H2SO4{H}_2 {SO}_4.
69 Medium
69 of 436
Analyzing +4 | -1
Statement I: When 10 g10 \mathrm{~g} of CaCO3\mathrm{CaCO}_3 is decomposed 5.6 g5.6 \mathrm{~g} of CaO\mathrm{CaO} and 4.4 g4.4 \mathrm{~g} of CO2\mathrm{CO}_2 are formed.

Statement II: Law of conservation of mass is followed.
View answer and explanation
Correct option: A
Statement I is true; Statement II is true; Statement II is the correct explanation of Statement I
Option A is correct because the decomposition of CaCO3 follows the law of conservation of mass, where the total mass of reactants equals the total mass of products.
70 Medium
70 of 436
Application +4 | -1
The number of moles of CaCl2\mathrm{CaCl}_2 needed to react with excess of AgNO3\mathrm{AgNO}_3 to produce 4.31gm4.31 \mathrm{gm} of AgCl\mathrm{AgCl}
View answer and explanation
Correct option: B
0.0150.015
Option B (0.0150.015) is correct because the mole ratio between CaCl2\mathrm{CaCl}_2 and AgCl\mathrm{AgCl} is 1:2, so the moles of CaCl2\mathrm{CaCl}_2 are half the moles of AgCl\mathrm{AgCl}.
71 Medium
71 of 436
Application +4 | -1
The volume of air needed for complete combustion of 1 kg carbon at STP is:
View answer and explanation
Correct option: A
9333.33 litre
Option A (9333.33 litre) is correct because it accurately calculates the volume of air needed for complete combustion of 1 kg of carbon at STP, considering the oxygen content in air.
72 Hard
72 of 436
Analyzing +4 | -1
Assertion:- 44 g44 \mathrm{~g} of CO2,28 g\mathrm{CO}_2, 28 \mathrm{~g} of CO\mathrm{CO} have same volume at STP.

Reason:- Both CO2\mathrm{CO}_2 and CO\mathrm{CO} are formed by C\mathrm{C} and oxygen.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct. The assertion is true because equal moles of gases occupy the same volume at STP. The reason is also true but doesn't explain the assertion.
73 Easy
73 of 436
Understanding +4 | -1
One mole of molecules of dioxygen represents
View answer and explanation
Correct option: A
6.02×10236.02 \times {10^{23}} molecules of oxygen
Option A is correct because one mole of any substance contains Avogadro's number of molecules, which is approximately 6.02×10236.02 \times 10^{23}.
74 Easy
74 of 436
Remembering +4 | -1
Which property of an element is not variable?
View answer and explanation
Correct option: B
At. wt.
Option B (At. wt.) is correct because the atomic weight of an element is a constant and not variable, unlike valence and equivalent weight.
75 Medium
75 of 436
Application +4 | -1
6.90 grams of metal carbonate was completely decomposed by 50 ml of 2N HCl. The equivalent weight of metal is
View answer and explanation
Correct option: A
39
Option A (39) is correct because the equivalent weight of the metal is calculated by subtracting the equivalent weight of carbonate from the equivalent weight of the metal carbonate.
76 Medium
76 of 436
Analyzing +4 | -1
Choose the correct option based on following statements. Here ‘T’ stands for true and ‘F’ stands for false statement.
(i) Molecular mass of cane sugar (C12H22O11)(C_{12}H_{22}O_{11}) is 182 amu.
(ii) 1 mole of cane sugar contains 6.022×10236.022 \times 10^{23} molecules of cane sugar.
(iii) 34.20 g of cane sugar contains 6.022×10216.022 \times 10^{21} molecules of cane sugar.
View answer and explanation
Correct option: C
FTF
Option C (FTF) is correct because the molecular mass of cane sugar is 342 amu, 1 mole contains 6.022 × 1023 molecules, and 34.20 g contains 6.022 × 1022 molecules.
77 Hard
77 of 436
Application +4 | -1
Calculate the molality of 1 litre solution of 93% H2SO493 \% \, \mathrm{H}_2 \mathrm{SO}_4 (weight/volume). The density of the solution is 1.84 g/mL1.84 \mathrm{~g} / \mathrm{mL}.
View answer and explanation
Correct option: C
10.42 m
Option C (10.42 m) is correct because the molality is calculated using the mass of solute and solvent in the solution.
78 Hard
78 of 436
Application +4 | -1
Complete combustion of 0.858 g0.858 \mathrm{~g} of compound XX gives 2.64 g2.64 \mathrm{~g} of CO2\mathrm{CO}_2 and 1.26 gof H2O1.26 \mathrm{~g}_{\text {of }} \mathrm{H}_2 \mathrm{O}. The lowest molecular mass XX can have :
View answer and explanation
Correct option: A
43 g43 \mathrm{~g}
Option B (21) is correct because it is the equivalent weight calculated based on the change in oxidation state of iron.
79 Medium
79 of 436
Analyzing +4 | -1
The following sequence of reaction occurs in commercial production of aqueous nitric acid.

4NH3( g)+5O2( g)⟶4NO(g)+6H2O(l)ΔH=−904 kJ.........(i)2NO(g)+O2( g)⟶2NO2( g)ΔH=−112 kJ.........(ii)3NO2( g)+H2O(l)⟶2HNO3(aq)+NO(g)ΔH=−140 kJ.........(iii)\begin{array}{cc} 4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) & \Delta \mathrm{H}=-904 \mathrm{~kJ}.........(i) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g}) & \Delta \mathrm{H}=-112 \mathrm{~kJ}.........(ii) \\ 3 \mathrm{NO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_3(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) & \Delta \mathrm{H}=-140 \mathrm{~kJ}.........(iii) \end{array}

Determine the total heat (in kJ/mole\mathrm{kJ} / \mathrm{mole} ) liberated at constant pressure for the production of exactly 1 mole of aqueous nitric acid by this process.
View answer and explanation
Correct option: A
493
Option A (493) is correct because the total heat liberated is calculated by summing the enthalpy changes for the reactions required to produce 1 mole of aqueous nitric acid.
80 Medium
80 of 436
Application +4 | -1
The mass of 11.2 L of ammonia gas at STP is
View answer and explanation
Correct option: A
8.5 g
Option A (8.5 g) is correct because it is calculated by using the molar mass of ammonia and the volume at STP.
81 Easy
81 of 436
Application +4 | -1
The mole fraction of solute in one molal aqueous solution is:
View answer and explanation
Correct option: B
0.018
Option B (0.018) is correct because the mole fraction is calculated by dividing the moles of solute by the total moles of solute and solvent, considering that one molal solution contains 1 mole of solute in 1000g of water.
82 Easy
82 of 436
Application +4 | -1
40 milligram diatomic volatile substance (X2)\left(X_2\right) is converted to vapour that displaced 4.92 mL4.92 \mathrm{~mL} of air at 1 atm1 \mathrm{~atm} and 300 K300 \mathrm{~K}. Atomic weight of element XX is nearly:
View answer and explanation
Correct option: D
100
Option C (200) is correct because the atomic weight of element X is calculated using the ideal gas law and the given conditions.
83 Easy
83 of 436
Application +4 | -1
Max. no. of gram molecules are present in:
View answer and explanation
Correct option: B
36 g36 \mathrm{~g} of H2O\mathrm{H}_2 \mathrm{O}
Option B (36 g of H2O) is correct because it contains the maximum number of moles (2 moles) compared to the other options.
84 Medium
84 of 436
Understanding +4 | -1
45.4 L45.4 \mathrm{~L} of dinitrogen reacted with 22.7 L22.7 \mathrm{~L} of dioxygen and 45.4 L45.4 \mathrm{~L} of nitrous oxide was formed. The reaction is given below.
2 N2(g)+O2(g)⟶2 N2O(g)2 \mathrm{~N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g})
Which law is being obeyed in this experiment?
View answer and explanation
Correct option: D
Gay Lussac's law of gaseous volumes
Option D (Gay Lussac's law of gaseous volumes) is correct because the volumes of the reacting gases and the product bear a simple whole-number ratio to each other, which is consistent with Gay-Lussac's Law.
85 Medium
85 of 436
Application +4 | -1
10 g of hydrogen and 64 of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:
View answer and explanation
Correct option: D
4 moles
Option D (4 moles) is correct because oxygen is the limiting reactant, and the stoichiometry of the reaction indicates that 1 mole of oxygen produces 2 moles of water.
86 Medium
86 of 436
Analyzing +4 | -1
Assertion: 1 equivalent of K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 has 1 equivalent of K,Cr\mathrm{K}, \mathrm{Cr} and O\mathrm{O} each.
Reason: Equivalent and milliequivalent reacts in equal number to give same equivalent of product.
View answer and explanation
Correct option: C
Both Assertion and Reason are correct and Reason is correct explanation of Assertion
Option C is correct because both the assertion and reason are true, and the reason correctly explains why equivalents react in equal numbers.
87 Medium
87 of 436
Analyzing +4 | -1
Maximum number of atoms are present in
View answer and explanation
Correct option: D
1 L H2O(l)1 \mathrm{~L} \mathrm{~H}_2 \mathrm{O}(\mathrm{l})
The correct answer is D (1 L H2O(l)1 \mathrm{~L} \mathrm{~H}_2 \mathrm{O}(\mathrm{l})) because it contains the largest number of moles of water, and thus the largest number of atoms.
88 Hard
88 of 436
Application +4 | -1
A 4 g4 \mathrm{~g} mixture of Na2O\mathrm{Na}_2 \mathrm{O} and NaCl\mathrm{NaCl} was dissolved in 400 mL400 \mathrm{~mL} of 0.3 M HCl0.3 \mathrm{~M} \mathrm{~HCl}. After complete reaction, the resulting solution required 200 mL200 \mathrm{~mL} of 0.2 M NaOH0.2 \mathrm{~M} \mathrm{~NaOH} for complete neutralization of the excess HCl\mathrm{HCl}. What mass %\% of Na2O\mathrm{Na}_2 \mathrm{O} is present in the mixture?
View answer and explanation
Correct option: C
62%62 \%
Option C (62%) is correct because it is calculated by finding the moles of Na2O that reacted with HCl and then calculating the mass percent.
89 Medium
89 of 436
Application +4 | -1
2 g of metal carbonate is neutralised completely by 100mL of 0.1 (N) HCl. The equivalent weight of metal carbonate is
View answer and explanation
Correct option: D
200
The correct answer is D (200) because the equivalent weight of the metal carbonate is calculated based on the neutralization reaction with HCl.
90 Medium
90 of 436
Analyzing +4 | -1
Out of 1.0 g1.0 \mathrm{~g} dioxygen, 1.0 g1.0 \mathrm{~g} atomic oxygen and 1.0 g1.0 \mathrm{~g} ozone, the maximum number of oxygen atoms are contained in
View answer and explanation
Correct option: D
All contain the same number of atoms
The correct answer is D (All contain the same number of atoms) because equal masses of the same element, regardless of their molecular form, contain the same number of atoms.
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