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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

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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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331 Medium
331 of 436
Remembering +4 | -1
The product of atomic weight and specific heat of a metal is approximately 6.4. This was given by:
View answer and explanation
Correct option: D
DulongPetit’s law
Option D (DulongPetit’s law) is correct because Dulong-Petit's law states that the product of atomic weight and specific heat of a metal is approximately 6.4.
332 Easy
332 of 436
Application +4 | -1
The volume occupied by one molecule of water (density 1  g  cm−31\;g\;\mathrm{c}{\mathrm{m}^-3} ) is:
View answer and explanation
Correct option: D
3.0×10−23  cm33.0 \times {10^{ - 23}}\;\mathrm{c}{\mathrm{m}^3}
Option D (3.0×10−23  cm33.0 \times {10^{ - 23}}\;\mathrm{c}{\mathrm{m}^3}) is correct because it is derived from dividing the molar volume of water by Avogadro's number.
333 Medium
333 of 436
Application +4 | -1
In a rocket motor filled with butane, how many kg\mathrm{kg} of liquid oxygen should be provided for each kg\mathrm{kg} of butane for its complete combustion?
C4H10+13/2O2→4CO2+5H2O\mathrm{C}_4 \mathrm{H}_{10}+13 / 2 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}
View answer and explanation
Correct option: A
4 kg
The correct answer is 4 kg because it represents the amount of liquid oxygen needed for complete combustion of 1 kg of butane based on the stoichiometry of the balanced chemical equation.
334 Easy
334 of 436
Application +4 | -1
1 mole of methyl amine on reaction with nitrous acid gives at NTP
View answer and explanation
Correct option: B
22.4 L of nitrogen
Option B is correct because 1 mole of methyl amine reacts with nitrous acid to produce 1 mole of nitrogen gas, which occupies 22.4 L at NTP.
335 Easy
335 of 436
Remembering +4 | -1
One mole of a gas is defined as:
View answer and explanation
Correct option: D
The number of molecules in 22.422.4 litre of a gas at STP
Option D (The number of molecules in 22.422.4 litre of a gas at STP) is correct because one mole of any gas occupies 22.4 liters at Standard Temperature and Pressure (STP).
336 Easy
336 of 436
Analyzing +4 | -1
Assertion : Equivalent weight of Cu{Cu} in CuO{CuO} is 63.6 and in Cu2O 31.8{Cu}_2 {O} \,31.8.

Reason : Equivalent weight of an element = Atomic weight of the element  Valency of the element =\frac{\text { Atomic weight of the element }}{\text { Valency of the element }}
View answer and explanation
Correct option: D
If assertion is false but reason is true.
Option D (If assertion is false but reason is true) is correct because the assertion incorrectly states the equivalent weight of Cu in both CuO and Cu2O, while the reason correctly defines equivalent weight.
337 Medium
337 of 436
Application +4 | -1
0.20 g of a hydrocarbon on combustion gave 0.66 g CO2.C{O_2}. The percentage of hydrogen in the hydrocarbon is about:
View answer and explanation
Correct option: C
10
Option C (10) is correct because the percentage of hydrogen is calculated by subtracting the percentage of carbon from 100.
338 Hard
338 of 436
Application +4 | -1
A mixture contains 1 mole each of Fe+2\mathrm{Fe}^{+2} and Fe+3\mathrm{Fe}^{+3} ions volume of 1 M KMnO41 \mathrm{~M} \mathrm{~KMnO}_4 required by mxiture in the oxidation of the given mixture is
View answer and explanation
Correct option: B
200 ml200 \mathrm{~ml}
Option B (200 ml200 \mathrm{~ml}) is correct because it is the volume of 1 M KMnO41 \mathrm{~M} \mathrm{~KMnO}_4 required to oxidize the mixture of Fe+2\mathrm{Fe}^{+2} and Fe+3\mathrm{Fe}^{+3} ions.
339 Medium
339 of 436
Application +4 | -1
A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2.C{O_2}. The empirical formula of the hydrocarbon is
View answer and explanation
Correct option: D
C7H8{C_7}{H_8}
Option D (C7H8{C_7}{H_8}) is correct because the ratio of moles of Carbon to Hydrogen is 7:8, which is determined from the combustion products.
340 Medium
340 of 436
Application +4 | -1
7.5 gs7.5 \mathrm{~gs} of a gas occupies 5.8 L5.8 \mathrm{~L} of volume at STP. The gas is:
View answer and explanation
Correct option: A
NO\mathrm{NO}
Option A (NO\mathrm{NO}) is correct because the molar mass calculated from the given volume and mass at STP is approximately 30 g/mol, which corresponds to nitric oxide (NO).
341 Medium
341 of 436
Application +4 | -1
Versene, a chelating agent having chemical formula C2H4N2(C2H2O2Na)4{\mathrm{C}_2}{\mathrm{H}_4}{\mathrm{N}_2}{\left( {{\mathrm{C}_2}{\mathrm{H}_2}{\mathrm{O}_2}\mathrm{Na}} \right)_4}. If each mole of this compound could bind 1 mole of Ca2+\mathrm{C}{\mathrm{a}^{2 + }},then the rating of pure versene expressed as mg of CaCO3\mathrm{CaC}{\mathrm{O}_3} bound per g of chelating agent is:
View answer and explanation
Correct option: D
263 mg
The correct answer is D (263 mg) because the rating of pure versene is calculated based on the molar mass of CaCO3 and the chelating agent.
342 Medium
342 of 436
Application +4 | -1
A solution is prepared by dissolving 24.5 g24.5 {~g} of sodium hydroxide in distilled water to give 1 L1 {~L} solution. The molarity of NaOH{NaOH} in the solution is
(Given that molar mass of NaOH=40.0 g mol−1{NaOH}=40.0 {~g} {~mol}^{-1} )
View answer and explanation
Correct option: B
0.6125M0.6125 {M}
Option B (0.6125M0.6125 {M}) is correct because it is the molarity of NaOH calculated using the given mass and volume.
343 Medium
343 of 436
Application +4 | -1
Weight of FeSO4(mol.wt.=152)\mathrm{FeS}{\mathrm{O}_4}\left( {\mathrm{mol}\mathrm{.wt}\mathrm{.} = 152} \right) oxidized by 200 mL of 1 NN KMnO4\mathrm{KMn}{\mathrm{O}_4} solution is :
View answer and explanation
Correct option: A
30.4 g
Option A (30.4 g) is correct because it is calculated using the concept of gram equivalents and the balanced redox reaction.
344 Medium
344 of 436
Application +4 | -1
1 gram of sodium hydroxide was treated with 25 mL25 \mathrm{~mL} of 0.75 M HCl0.75 \mathrm{~M} \, {HCl} solution, the mass of sodium hydroxide left unreacted is equal to
View answer and explanation
Correct option: A
250 mg250 \mathrm{~mg}
Option A (250 mg) is correct because after reacting 1 g of NaOH with 25 mL of 0.75 M HCl, 250 mg of NaOH remains unreacted.
345 Easy
345 of 436
Analyzing +4 | -1
Assertion : A certain element XX, forms three binary compounds with chlorine containing 59.68%,68.95%59.68 \%, 68.95 \% and 74.75%74.75 \% chlorine respectively. These data illustrate the law of multiple proportions.

Reason : According to law of multiple proportions, the relative amounts of an element combining with some fixed amount of a second element in a series of compounds are the ratios of small whole numbers.
View answer and explanation
Correct option: A
If both assertion and reason are true and the reason is the correct explanation of the assertion.
The correct answer is A (If both assertion and reason are true and the reason is the correct explanation of the assertion) because the law of multiple proportions explains the fixed ratios in which elements combine.
346 Easy
346 of 436
Application +4 | -1
The number of moles of Hydrogen molecules required to produce 20 moles of NH3N{H_3} through Haber’s process
View answer and explanation
Correct option: D
30
Option D (30) is correct because the balanced Haber's process equation shows that 3 moles of hydrogen are required for every 2 moles of ammonia produced.
347 Medium
347 of 436
Application +4 | -1
Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is
View answer and explanation
Correct option: D
6
Option D (6) is correct because 6 moles of Mohr's salt are required to reduce 1 mole of dichromate, based on the balanced redox reaction.
348 Easy
348 of 436
Application +4 | -1
A compound having the empirical formula, C3H4O\mathrm{C}_3 \mathrm{H}_4 \mathrm{O} has a molecular weight of 170±5.170 \pm 5. The molecular formula of the compound is
View answer and explanation
Correct option: D
C9H12O3\mathrm{C}_9 \mathrm{H}_{12} \mathrm{O}_3
Option D (C9H12O3\mathrm{C}_9 \mathrm{H}_{12} \mathrm{O}_3) is correct because the molecular formula is a multiple of the empirical formula, with the multiplier determined by the molecular weight.
349 Medium
349 of 436
Application +4 | -1
Calculate the mass of lime (CaO)(\mathrm{CaO}) obtained by heating 400 g400 \mathrm{~g} of 80%80 \% pure lime stone (CaCO3)\left(\mathrm{CaCO}_3\right)
View answer and explanation
Correct option: A
179.2 g179.2 \mathrm{~g}
Option A (179.2 g179.2 \mathrm{~g}) is correct because it is calculated by first finding the mass of pure CaCO3CaCO_3 and then using stoichiometry to determine the mass of CaO produced.
350 Medium
350 of 436
Application +4 | -1
When HNO3{HNO}_3 is converted into NH3,NH_3, the equivalent weight of HNO3{HNO}_3 will be:

(M=\left({M}=\right. molecular weight of HNO3)\left.{HNO}_3\right)
View answer and explanation
Correct option: D
M/8{M} / 8
Option D (M/8) is correct because the equivalent weight is calculated by dividing the molecular weight by the change in oxidation number, which is 8 in this conversion.
351 Medium
351 of 436
Application +4 | -1
If NA{N_A} is Avogadro’s number, then number of valence electrons in 4.2 g of nitride ions (N3−{N^{3 - }}) is
View answer and explanation
Correct option: A
2.4 NA 2.4\,{N_A}\,
Option A (2.4 NA 2.4\,{N_A}\,) is correct because the number of valence electrons in 4.2 g of nitride ions is calculated based on Avogadro's number and the number of valence electrons per ion.
352 Medium
352 of 436
Analyzing +4 | -1
In the reaction, 2Al(s)+6HCl(aq)→2Al3+(aq)+6Cl−(aq)+3H2(g)2 \mathrm{Al(s})+6 \mathrm{HCl(a q}) \rightarrow 2 \mathrm{Al^{3+}(a q})+6 \mathrm{Cl^{-}(a q})+3 \mathrm{H_2(g})
View answer and explanation
Correct option: D
11.2 L H2(g)11.2 \mathrm{~L} \mathrm{~H}_2(\mathrm{g}) at STP is produced for every mole HCl(aq)\mathrm{HCl(a q}) consumed
Option D (11.2 L H2(g)11.2 \mathrm{~L} \mathrm{~H}_2(\mathrm{g}) at STP is produced for every mole HCl(aq)\mathrm{HCl(a q}) consumed) is correct because the stoichiometry of the reaction dictates this relationship.
353 Medium
353 of 436
Application +4 | -1
When 3.0 litre solution of normality N\mathrm{N} is mixed with 5.0 litre of 4MHCl4 \mathrm{M} \mathrm{HCl}, then the resultant solution has the normality 10 . Find the value of N\mathrm{N}.
View answer and explanation
Correct option: B
20
Option B (20) is correct because it accurately calculates the normality N using the mixing equation N1V1 + N2V2 = NV.
354 Medium
354 of 436
Application +4 | -1
A sample of clay contains 50%50 \% silica and 10%10 \% water. The sample is partially dried by which it loses 8 g8 \mathrm{~g} water. If the percentage of silica in the partially dried clay is 52, then what is the percentage of water in the partially dried clay?
View answer and explanation
Correct option: B
6.4%6.4 \%
Option B (6.4%) is correct because the percentage of water is calculated based on the conservation of silica mass before and after drying.
355 Hard
355 of 436
Analyzing +4 | -1
Assume that the atomic mass of oxygen is 7.7. A sample of 11 g11 \mathrm{~g} of an oxide of uranium contains 10 g10 \mathrm{~g} of uranium. Which of the following formula for the oxide is compatible with the data?
View answer and explanation
Correct option: A
Uranium oxide is UO\text{UO} and the atomic mass of U\mathrm{U} is 70
Option A is correct because the ratio of Uranium to Oxygen is 1:1, which corresponds to the formula UO, and the atomic mass of U is 70.
356 Medium
356 of 436
Understanding +4 | -1
A mixture contains two or more substances in ________ which are called its ________.
View answer and explanation
Correct option: C
any ratio, components
Option C (any ratio, components) is correct because a mixture is defined as a combination of two or more substances (components) in any proportion, where each substance retains its individual properties.
357 Hard
357 of 436
Application +4 | -1
In the mixture of (NaHCO3+Na2CO3)\left(\mathrm{NaHCO}_3+\mathrm{Na}_2 \mathrm{CO}_3\right), volume of HCl\mathrm{HCl} required is x mLx \mathrm{~mL} with phenolphthalein indicator and y mLy \mathrm{~mL}, with methyl orange indicator in the same titration. Hence, volume of HCl\mathrm{HCl} for complete reaction of Na2CO3\mathrm{Na}_2 \mathrm{CO}_3 is:
View answer and explanation
Correct option: A
2x2 x
Option A (2x2x) is correct because the volume of HCl required for complete reaction of Na2CO3 is twice the volume required to convert it to NaHCO3 using phenolphthalein.
358 Medium
358 of 436
Application +4 | -1
34 g of hydrogen peroxide is present in 1120 mL of solution. This solution is called:
View answer and explanation
Correct option: A
10 vol solution
Option A (10 vol solution) is correct because 34 g of H2O2 in 1120 mL of solution produces 10 mL of O2 per mL of solution.
359 Medium
359 of 436
Understanding +4 | -1
Assertion: NaNO3\mathrm{NaNO}_3 has no definite molecule.
Reason: Its formula mass is 85.
View answer and explanation
Correct option: D
Both Assertion and Reason are correct but Reason is not correct explanation of Assertion
Option D is correct because while both the assertion and reason are true, the reason (formula mass) does not explain why NaNO3 lacks a definite molecule (lattice structure).
360 Medium
360 of 436
Application +4 | -1
10 g of metal oxide contains 4 g of oxygen. The equivalent weight of the metal carbonate is
View answer and explanation
Correct option: A
42
Option A (42) is correct because the equivalent weight of the metal carbonate is calculated by first finding the equivalent weight of the metal and then adding the equivalent weight of the carbonate ion.
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