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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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Showing 271-300 of 436 questions

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271 Medium
271 of 436
Application +4 | -1
The mass of 2.5 g.atom of oxygen is
View answer and explanation
Correct option: A
40 g
Option A (40 g) is correct because 1 g.atom is equal to the gram atomic weight, and for oxygen, it's 16 g/g.atom. Therefore, 2.5 g.atom is 2.5 * 16 = 40 g.
272 Medium
272 of 436
Analyzing +4 | -1
Statement - I: For reaction: 2A(g)+3B(g)→4C(g)+D(g),2A(g) + 3B(g) \rightarrow 4C(g) + D(g), the vapour density remain
constant throughout the progress of reaction.

Statement - II: In all the gaseous chemical reactions, the vapour density remains constant.
View answer and explanation
Correct option: C
If Statement-I is true but Statement-II is false
Option C is correct because Statement-I is true (moles remain constant, so vapor density remains constant), but Statement-II is false (vapor density doesn't always remain constant in gaseous reactions).
273 Easy
273 of 436
Application +4 | -1
Which has minimum number of atoms of oxygen ?
View answer and explanation
Correct option: D
12.044×102212.044 \times 10^{22} molecules of CO2\mathrm{CO}_2
The correct answer is D (12.044×102212.044 \times 10^{22} molecules of CO2CO_2) because it contains the minimum number of oxygen atoms compared to the other options.
274 Hard
274 of 436
Analyzing +4 | -1
Rearrange the following (I to IV) in the order of increasing masses and choose the correct answer from (1), (2), (3) and (4) (Atomic mass: N=14,O=16,Cu=63N=14, O=16, C u=63).

I. 1 molecule of oxygen

II. 1 atom of nitrogen

III. 1×10−10 g1 \times 10^{-10} \mathrm{~g} molecular weight of oxygen

IV. 1×10−10 g1 \times 10^{-10} \mathrm{~g} atomic weight of copper
View answer and explanation
Correct option: A
II < I < III < IV
Option A (II < I < III < IV) is correct because it represents the increasing order of masses calculated for each species.
275 Medium
275 of 436
Understanding +4 | -1
Assertion (A): A one molal solution prepared at 20∘C20^{\circ} \mathrm{C} will retain the same molality at 100∘C,100^{\circ} \mathrm{C}, provided there is no loss of solute or solvent on heating.

Reason (R): Molality is independent of temperature.
View answer and explanation
Correct option: A
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because molality is defined as moles of solute per kg of solvent, which is independent of temperature, and this explains why molality remains constant with temperature changes.
276 Medium
276 of 436
Application +4 | -1
A nitrogen containing organic compound contain 28.9% nitrogen by weight. If one molecule compound have 4 nitrogen atom then molecular weight of compound is:
View answer and explanation
Correct option: B
194
Option B (194) is correct because it is the molecular weight calculated using the percentage of nitrogen and the number of nitrogen atoms in the molecule.
277 Easy
277 of 436
Application +4 | -1
The mass of 112  cm3  CH4\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.
278 Hard
278 of 436
Application +4 | -1
How many litres of detonating gas may be produced at 0∘C0^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm} from the decomposition of 0.1 mole of water by an electric current?
View answer and explanation
Correct option: C
3.36 L3.36 \mathrm{~L}
Option C (3.36 L) is correct because the volume of detonating gas (H₂ and O₂) produced from the decomposition of 0.1 mole of water is calculated using stoichiometry and the ideal gas law.
279 Easy
279 of 436
Application +4 | -1
Which has the maximum number of atoms?
View answer and explanation
Correct option: B
1  g  H21\;\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.
280 Medium
280 of 436
Application +4 | -1
0.72 gm0.72 \mathrm{~gm} of an oxide of a metal M\mathrm{M} on reduction with H2\mathrm{H}_2 gave 0.64 g0.64 \mathrm{~g} of the metal. The atomic weight of the metal is 64. The empirical formula of the compound is
View answer and explanation
Correct option: B
M2O\mathrm{M}_2 \mathrm{O}
Option B (M₂O) is correct because the empirical formula is determined by calculating the mole ratio of the metal and oxygen, which simplifies to 2:1.
281 Easy
281 of 436
Application +4 | -1
Volume occupied by 3.01×1023    3.01 \times {10^{23}}\;\; molecules of acetylene at NTP is:
View answer and explanation
Correct option: B
11.2  litre11.2\,\,\mathrm{litre}
Option B (11.2  litre11.2\,\,\mathrm{litre}) is correct because 3.01×10233.01 \times {10^{23}} molecules is half a mole, and half a mole of any gas at NTP occupies 11.2 liters.
282 Medium
282 of 436
Understanding +4 | -1
Molecular weight of tribasic acid is W. Its equivalent weight will be:
View answer and explanation
Correct option: B
W3\frac{W}{3}
Option B (W3\frac{W}{3}) is correct because equivalent weight is calculated by dividing the molecular weight by the basicity of the acid.
283 Easy
283 of 436
Application +4 | -1
How many gram of NaOH\mathrm{NaOH} will be required to neutralize 12.2 g12.2 \mathrm{~g} benzoic acid?
View answer and explanation
Correct option: D
4 g4 \mathrm{~g}
Option D (4 g4 \text{ g}) is correct because the mass of NaOH required is calculated based on the stoichiometry of the neutralization reaction.
284 Easy
284 of 436
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
3
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.
285 Easy
285 of 436
Remembering +4 | -1
The significant figures in 3400 are
View answer and explanation
Correct option: A
2
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.
286 Medium
286 of 436
Understanding +4 | -1
Which of the following statements is correct about the reaction given below?

4Fe(s)+3O2( g)⟶2Fe2O3( g)4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~g})
View answer and explanation
Correct option: A
Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore, it follows law of conservation of mass.
Option A is correct because the total mass of reactants equals the total mass of products, illustrating the law of conservation of mass.
287 Medium
287 of 436
Application +4 | -1
Number of atoms in 558.5 g of Fe is (at . wt = 55.85)
View answer and explanation
Correct option: A
Twice that in 60 g carbon
Option A (Twice that in 60 g carbon) is correct because 558.5 g of Fe is 10 moles, and 60 g of carbon is 5 moles, so the number of atoms in Fe is twice that in carbon.
288 Medium
288 of 436
Analyzing +4 | -1
Assertion: Density is expressed as g mL−1\mathrm{g} \mathrm{~mL}^{-1} whereas specific gravity is dimensionless.
Reason: Specific gravity is ratio of the masses of solution and solvent.
View answer and explanation
Correct option: C
Both Assertion and Reason are correct and Reason is correct explanation of Assertion
The correct answer is C because specific gravity is the ratio of densities (or masses) of a substance to a reference substance, making it dimensionless, and the reason correctly explains this.
289 Hard
289 of 436
Application +4 | -1
25 ml25 \mathrm{~ml} solution containing (6.35 g/L\left(6.35 \mathrm{~g} / \mathrm{L}\right. of KH3(C2O4)2⋅2H2O\mathrm{K} \mathrm{H}_3\left(\mathrm{C}_2 \mathrm{O}_4\right)_2 \cdot 2 \mathrm{H}_2 \mathrm{O} required V1ml\mathrm{V}_1 \mathrm{ml} of 0.1 N0.1 \mathrm{~N} NaOH\mathrm{NaOH} and V2ml\mathrm{V}_2 \mathrm{ml} of 0.1 N KMnO40.1 \mathrm{~N} \mathrm{~KMnO}_4 in two separate titrations. Calculate V1\mathrm{V}_1 and V2\mathrm{V}_2.
KH3(C2O4)2⋅2H2O;M0=254 gm/mol\mathrm{KH}_3\left(\mathrm{C}_2 \mathrm{O}_4\right)_2 \cdot 2 \mathrm{H}_2 \mathrm{O} ; \mathrm{M}_0=254 \mathrm{~gm} / \mathrm{mol}
View answer and explanation
Correct option: A
18.75 ml18.75 \mathrm{~ml} , 25 ml25 \mathrm{~ml}
Option A (18.75 ml , 25 ml) is correct because it calculates the volumes of NaOH and KMnO4 required based on the given normality and the amount of the compound.
290 Medium
290 of 436
Application +4 | -1
Sodium bicarbonate NaHCO3\mathrm{NaHCO}_3 can be purified by dissolving it in hot water (60∘C),\left(60^{\circ} \mathrm{C}\right), filtering to remove insoluble impurities, cooling to 0∘C0^{\circ} \mathrm{C} to precipitate solid NaHCO3,\mathrm{NaHCO}_3, and then filtering to remove the solid, leaving soluble impurities in solution. Any NaHCO3\mathrm{NaHCO}_3 that remains in the solution is not recovered. The solubility of NaHCO3\mathrm{NaHCO}_3 in hot water at 60∘C60^{\circ} \mathrm{C} is 164 g/164 \mathrm{~g} / litre and is 69 g/litre69 \mathrm{~g/litre} in cold water at 0∘C.0^{\circ} \mathrm{C}. What is the percentage yield of NaHCO3,\mathrm{NaHCO}_3, when it is purified by this method?
View answer and explanation
Correct option: A
57.93%57.93 \%
Option A (57.93%57.93 \%) is correct because the percentage yield is calculated by finding the difference in solubility between hot and cold water, then dividing by the solubility in hot water.
291 Hard
291 of 436
Application +4 | -1
One g\mathrm{g} of impure Na2CO3\mathrm{Na}_2 \mathrm{CO}_3 is dissolved in water and the solution is made upto 250 mL250 \mathrm{~mL}. To 50 mL50 \mathrm{~mL} of this made up solution, 50 mL50 \mathrm{~mL} of 0.1 N HCl0.1 \mathrm{~N} \mathrm{~HCl} is added and the mixture after shaking well required 10 mL10 \mathrm{~mL} of 0.16 N0.16 \mathrm{~N} sodium hydroxide solution for complete neutralization. Calculate the per cent purity of the sample of Na2CO3\mathrm{Na}_2 \mathrm{CO}_3.
View answer and explanation
Correct option: A
90%90 \%
Option A (90%90 \%) is correct because the purity is calculated based on the amount of Na2CO3Na_2CO_3 that reacts with HCl and NaOH.
292 Medium
292 of 436
Application +4 | -1
NaClO\mathrm{NaClO} solution reacts with H2SO3,as,NaClO+H2SO3→NaCl+H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3},as,\mathrm{NaClO} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3} \to \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} A solution of NaClO\mathrm{NaClO} used in the above reaction contained 15 g of NaClO\mathrm{NaClO} per litre. The normality of the solution would be :
View answer and explanation
Correct option: A
0.4
Option A (0.4) is correct because the normality is calculated by considering the number of equivalents of NaClO per liter of solution.
293 Medium
293 of 436
Understanding +4 | -1
In a gaseous reaction of the type aA+bB→cC+dD,aA + bB \to cC + dD, which is wrong?
View answer and explanation
Correct option: C
aa g of AA combines with bb g of BB to give CC and DD
Option C (aa g of AA combines with bb g of BB to give CC and DD) is correct because stoichiometry in chemical reactions represents mole ratios or volume ratios (for gases), but not directly mass ratios.
294 Medium
294 of 436
Analyzing +4 | -1
Which of the following statements illustrate the law of multiple proportions?
View answer and explanation
Correct option: C
An element forms two oxides, XO\mathrm{XO} and XO2\mathrm{XO}_2 containing 50%50 \% and 60%60 \% oxygen respectively. The ratio of masses of oxygen which combines with 1 g1 \mathrm{~g} of element is 2:32: 3
Option C is correct because it directly demonstrates the law of multiple proportions by showing that the ratio of oxygen masses combining with a fixed mass of the element is a simple whole number ratio (2:3).
295 Hard
295 of 436
Application +4 | -1
A 0.518 g sample of lime stone is dissolved in HClHCl and then the calcium is precipitated as CaC2O4\mathrm{Ca}{\mathrm{C}_2}{\mathrm{O}_4}. After filtering and washing the precipitate, it requires 40.0 mL of 0.250NKMnO4,0.250N\mathrm{KMn}{\mathrm{O}_4},, solution acidified with H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} to titrate is as, MnO4−+H++C2O42−→Mn2++CO2+2H2O\mathrm{MnO}_4^ - + {\mathrm{H}^ + } + {\mathrm{C}_2}\mathrm{O}_4^{2 - } \to \mathrm{M}{\mathrm{n}^{2 + }} + \mathrm{C}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}. The percentage of CaO\mathrm{CaO} in the sample is :
View answer and explanation
Correct option: A
54.0%54.0\mathrm{ }\%
Option A (54.0%54.0\%) is correct because the percentage of CaO is calculated by equating the meq of limestone, CaC2O4, KMnO4 and CaO.
296 Medium
296 of 436
Analyzing +4 | -1
In redox reaction, H2O2{H}_2 {O}_2 oxidises K4[Fe(CN)6]{K}_4\left[{Fe}({CN})_6\right] into K+,Fe+3,CO3−2{K}^{+}, {Fe}^{+3}, {CO}_3{ }^{-2} and NO3−{NO}_3^- ions in acidic medium, than how many moles of H2O2{H}_2 {O}_2 will react with 1mole of K4[Fe(CN)6]{K}_4\left[{Fe}({CN})_6\right]
View answer and explanation
Correct option: D
30.5 moles
Option D is correct because the n-factor of H2O2 is 2 and the n-factor of K4[Fe(CN)6] is 61. Therefore, 30.5 moles of H2O2 will react with 1 mole of K4[Fe(CN)6].
297 Medium
297 of 436
Remembering +4 | -1
Number of gas molecules present in 1 ml1 \mathrm{~ml} of gas at 0∘C0^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm} is called Loschmidt number. Its value is about
View answer and explanation
Correct option: A
2.7×10192.7 \times 10^{19}
Option A (2.7×10192.7 \times 10^{19}) is correct because the Loschmidt number, representing the number of gas molecules in 1 ml at standard conditions, is approximately 2.7×10192.7 \times 10^{19}.
298 Hard
298 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 based on the mass of solute and solvent in the solution.
299 Medium
299 of 436
Application +4 | -1
100 mL100 \mathrm{~mL} of 2.4 V H2O22.4 \mathrm{~V} {~H}_2 {O}_2 require how many grams of KMnO4{KMnO}_4 in acidic medium for complete reaction?
View answer and explanation
Correct option: D
1.35 g1.35 \mathrm{~g}
Option D (1.35 g1.35 \mathrm{~g}) is correct because it is calculated by equating the milliequivalents of H2O2 and KMnO4 in acidic medium.
300 Medium
300 of 436
Application +4 | -1
A sample of nitric acid is 69%69 \% by mass and it has a concentration of 15.44 moles per litre. Its density is
View answer and explanation
Correct option: B
1.41 g/cc1.41 \mathrm{~g} / \mathrm{cc}
Option B (1.41 g/cc) is correct because the density is calculated using the molarity, percentage by mass, and molar mass of nitric acid.
Revision notes and key takeaways

Review SOME BASIC CONCEPTS OF CHEMISTRY in your textbook or class notes, then use this set to check your understanding.

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