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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 91-120 of 436 questions

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91 Medium
91 of 436
Analyzing +4 | -1
Arrange the following aq. dilute solution in the increasing order of vapour pressure
(a) 10 % (w / v) glucose

(b) 10 % (w / v) sucrose

(c) 10 % (w / v) urea
View answer and explanation
Correct option: D
(c) << (a) << (b)
Option A ((a) < (b) < (c)) is the correct order because vapor pressure is inversely proportional to the molecular weight of the solute.
92 Medium
92 of 436
Application +4 | -1
Hydrazine reacts with KIO3\mathrm{KIO}_3 in presence of HCl\mathrm{HCl} as

N2H4+IO3−+2H++Cl−⟶ICl+N2+3H2O\mathrm{N}_2 \mathrm{H}_4+\mathrm{IO}_3^{-}+2 \mathrm{H}^{+}+\mathrm{Cl}^{-} \longrightarrow \mathrm{ICl}+\mathrm{N}_2+3 \mathrm{H}_2 \mathrm{O}

The equivalent masses of N2H4\mathrm{N}_2 \mathrm{H}_4 and KIO3\mathrm{KIO}_3 respectively are:
View answer and explanation
Correct option: A
8 and 53.5
Option A (8 and 53.5) is correct because the equivalent mass of N2H4N_2H_4 is calculated based on the change in oxidation number of nitrogen, and the equivalent mass of KIO3KIO_3 is calculated based on the change in oxidation number of iodine.
93 Easy
93 of 436
Analyzing +4 | -1
Moon takes 27.3 days to complete one orbit around the Earth. Now read the following statements and choose the correct code. Here T is for true statement and F is for ‘False statement’.
(i) Moon takes 655.2 hours to complete one orbit around the Earth.
(ii) Moon takes 39312 seconds to complete one orbit around the earth.
(iii) Moon takes 1638 minutes to complete one orbit around the Earth.
View answer and explanation
Correct option: C
T F F
Option C (T F F) is correct because only the conversion to hours is accurate; the conversions to seconds and minutes are incorrect.
94 Medium
94 of 436
Application +4 | -1
If potassium chlorate is 80%80 \% pure, then 48 g48 \mathrm{~g} of oxygen would be produced from (atomic mass of K=39\mathrm{K}=39 )
View answer and explanation
Correct option: A
153.12 gofKClO3153.12 \mathrm{~g} \mathrm{of} \mathrm{KClO}_3
Option A (153.12 gofKClO3153.12 \mathrm{~g} \mathrm{of} \mathrm{KClO}_3) is correct because it accounts for the purity of potassium chlorate to produce 48g of oxygen.
95 Easy
95 of 436
Application +4 | -1
One mole of P4{\mathrm{P}_4} molecules contain:
View answer and explanation
Correct option: D
24.088×1023  atoms24.088 \times {10^{23}}\;\mathrm{atoms}
Option D (24.088×1023  atoms24.088 \times {10^{23}}\;\mathrm{atoms}) is correct because one mole of P4 contains Avogadro's number of P4 molecules, and each P4 molecule contains 4 P atoms.
96 Medium
96 of 436
Application +4 | -1
The density of 3 M3 \mathrm{~M} solution of Na2S2O3\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 is 1.25 g mL−1.1.25 \mathrm{~g~mL}^{-1}. Calculate, the mole fraction of Na2SO3.\mathrm{Na}_2 \mathrm{SO}_3.
View answer and explanation
Correct option: A
0.065
Option A (0.065) is correct because the mole fraction is calculated by dividing moles of Na2S2O3 by the total moles of solute and solvent.
97 Medium
97 of 436
Application +4 | -1
The number of electron in 3.1 mg NO3−3.1 \mathrm{~mg} \mathrm{~NO}_3^{-} is:-
View answer and explanation
Correct option: C
9.6×10209.6 \times 10^{20}
Option C (9.6 × 1020) is correct because the number of electrons is calculated by finding the moles of NO₃⁻, then multiplying by Avogadro's number and the number of electrons in the ion.
98 Medium
98 of 436
Analyzing +4 | -1
N2\mathrm{N}_2 and H2\mathrm{H}_2 are mixed in 14:314:3 mass ratio. After certain time ammonia was found to be 40%40 \% by mol. The mole fraction of N2\text{N}_2 at that time in the mixture of N2,H2\mathrm{N}_2,\mathrm{H}_2 and NH3\text{NH}_3 is
View answer and explanation
Correct option: A
0.15
Option A (0.15) is correct because after calculating the moles of each component at equilibrium, the mole fraction of N2 is found to be 0.15.
99 Medium
99 of 436
Understanding +4 | -1
XX' is the amount of a substance that contains as many particles as there are atoms in exactly 12 g12 \mathrm{~g} of the YY isotope. XX and YY respectively are
View answer and explanation
Correct option: C
A→A \rightarrow one mole, B→12CB \rightarrow{ }^{12} \mathrm{C}
Option C is correct because one mole is defined as the amount of substance containing as many particles as there are atoms in exactly 12g of the carbon-12 isotope.
100 Easy
100 of 436
Application +4 | -1
Maximum number of atoms is present in
View answer and explanation
Correct option: B
0.2 mol C6H12O60.2 \mathrm{~mol}\, {C}_6 {H}_{12} {O}_6
Option B (0.2 mol C6H12O60.2 \mathrm{~mol}\, {C}_6 {H}_{12} {O}_6) is correct because it contains the maximum number of atoms (4.8 moles) compared to the other options.
101 Medium
101 of 436
Application +4 | -1
A mixture contains Na2C2O4{Na}_2 {C}_2 {O}_4 and KHC2O4{KHC}_2 {O}_4 in 1:1 molar ratio. Mixture is neutralised by 100 mL100 \mathrm{~mL} of 0.01 M KOH0.01 \, \mathrm{M} \, {KOH}. What volume of 0.01 M KMnO4?0.01 \, \mathrm{M} \, {KMnO}_4?
Thus, the same mixture is oxidised by
View answer and explanation
Correct option: D
80 mL80 \mathrm{~mL}
Option D (80 mL80 \mathrm{~mL}) is correct because it is calculated by finding the millimoles of C2O42−{C}_2 {O}_4^{2-} and equating it to the milliequivalents of KMnO4KMnO_4.
102 Medium
102 of 436
Application +4 | -1
A mixture of gases contains H2  and  O2{H_2}\,\,\mathrm{and}\,\,{O_2} gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture
View answer and explanation
Correct option: D
0.167361111
Option D (04:01) is correct because the molar ratio is calculated based on the given weight ratio and the molar masses of hydrogen and oxygen.
103 Medium
103 of 436
Application +4 | -1
The mineral haematite is Fe2O3.\mathrm{Fe}_2 \mathrm{O}_3. Haematite ore contains unwanted material called gangue in addition to Fe2O3.\mathrm{Fe}_2 \mathrm{O}_3. If 5 kg5 \mathrm{~kg} of ore contains 2.78 kg2.78 \mathrm{~kg} of Fe,\mathrm{Fe}, then what percentage of ore is gangue? (Fe=56)(\mathrm{Fe}=56)
View answer and explanation
Correct option: C
20.6%20.6 \%
Option D (20.6%) is correct because it is the percentage of gangue calculated from the mass of iron and the total mass of the ore.
104 Medium
104 of 436
Understanding +4 | -1
Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation states changes from
View answer and explanation
Correct option: B
#ERROR!
Option B (+6 to +3) is correct because during the oxidizing action of potassium dichromate in acidic solution, chromium's oxidation state changes from +6 in dichromate to +3 in the chromium(III) ion.
105 Easy
105 of 436
Analyzing +4 | -1
If the true value for an experimental result is 6.23 and the results reported by three students X, Y and Z are :

X : 6.18 and 6.28

Y : 6.20 and 6.023

Z : 6.22 and 6.24

Which of the following option is correct :
View answer and explanation
Correct option: D
Both X & Y neither precise nor accurate, Z both precise
and accurate.
Option D (Both X & Y neither precise nor accurate, Z both precise and accurate) is correct because X and Y's values are not close to each other (imprecise) or the true value (inaccurate), while Z's values are close to each other and the true value.
106 Hard
106 of 436
Application +4 | -1
A spherical ball of radius 7 cm7 \mathrm{~cm} contains 56%56 \% iron. If density is 1.4 g cm−3,1.4 \mathrm{~g} \mathrm{~cm}^{-3}, number of moles of Fe\text{Fe} present approximately is
View answer and explanation
Correct option: C
20
Option C (20) is correct because it is calculated based on the volume of the sphere, density, percentage of iron, and the molar mass of iron.
107 Medium
107 of 436
Application +4 | -1
A metal M of atomic mass 54.94 has a density of 7.42 g/cc7.42 \mathrm{~g} / \mathrm{cc}. Calculate the apparent volume occupied by one atom of the metal.
View answer and explanation
Correct option: A
1.23×10−23 cm31.23 \times 10^{-23} \mathrm{~cm}^3
Option A (1.23×10−23 cm31.23 \times 10^{-23} \mathrm{~cm}^3) is correct because it is the calculated apparent volume occupied by one atom of the metal, based on its density and atomic mass.
108 Medium
108 of 436
Application +4 | -1
Number of atoms in 3 mol of NH3N{H_3}
View answer and explanation
Correct option: A
7.22×10247.22 \times {10^{24}} atoms
Option A (7.22×10247.22 \times {10^{24}} atoms) is correct because it is the total number of atoms calculated by multiplying the number of moles, atomicity, and Avogadro's number.
109 Medium
109 of 436
Application +4 | -1
BrO3−{BrO}_3^{-}changes into Br2{Br}_2 in an acidic medium of a unbalanced equation. How many electron should be present on the balanced equation?
View answer and explanation
Correct option: A
10 electron in left
The correct answer is A (10 electron in left) because balancing the equation requires adding 10 electrons to the left side to balance the charge.
110 Easy
110 of 436
Analyzing +4 | -1
Assertion: Density == specific gravity ×0.99823\times 0.99823 at 20∘C20^{\circ} \mathrm{C}.
Reason: Density == specific gravity at 4∘C4^{\circ} \mathrm{C}.
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 both the assertion and the reason are true statements, but the reason does not explain the assertion.
111 Medium
111 of 436
Analyzing +4 | -1
The explosion of a mixture consisting of one volume of a gas being studied and one volume of H2\mathrm{H}_2 yielded one volume water vapour and one volume of N2.\mathrm{N}_2. The formula of gas being studied, is
View answer and explanation
Correct option: C
N2O\mathrm{N}_2 \mathrm{O}
Option C (N2O\mathrm{N}_2 \mathrm{O}) is correct because the reaction indicates that one volume of the gas and one volume of H2 produce one volume of N2 and one volume of H2O, which is consistent with the decomposition of N2O.
112 Medium
112 of 436
Application +4 | -1
Which of the following has largest number of molecules?
View answer and explanation
Correct option: B
56 g56 \mathrm{~g} of CO\mathrm{CO}
Option B (56 g56 \mathrm{~g} of CO\mathrm{CO}) is correct because it contains the largest number of moles, and thus the largest number of molecules.
113 Medium
113 of 436
Analyzing +4 | -1
Statement I : For H3PO2{H_3}P{O_2}, 1 M = 3 N
Statement II : H3PO2{H_3}P{O_2} has three H-atoms hence, it is a tribasic acid
Which statement(s) is/are correct ?
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because both statements are incorrect. H3PO2{H_3}P{O_2} is a monobasic acid, and therefore 1 M = 1 N.
114 Medium
114 of 436
Understanding +4 | -1
In order to determine the mole precisely, the mass of a carbon AA was determined by a BB and found to be equal to C gC \mathrm{~g}. A,BA, B and CC respectively are
View answer and explanation
Correct option: A
A→12A \rightarrow 12 atom, B→B \rightarrow mass spectrometer, C→1.992648×10−23C \rightarrow 1.992648 \times 10^{-23}
Option A (A→12A \rightarrow 12 atom, B→B \rightarrow mass spectrometer, C→1.992648×10−23C \rightarrow 1.992648 \times 10^{-23}) is correct because the mass of a carbon-12 atom is determined using a mass spectrometer and is approximately 1.992648×10−231.992648 \times 10^{-23} g.
115 Hard
115 of 436
Remembering +4 | -1
Which of the following is not described in Charaka Samhita?
View answer and explanation
Correct option: C
Preparation of perfumes and cosmetics
Option C (Preparation of perfumes and cosmetics) is correct because the preparation of perfumes and cosmetics was described in Brihat Samhita, not Charaka Samhita.
116 Medium
116 of 436
Application +4 | -1
A compound contains elements X\text{X} and Y\text{Y} in 1:41: 4 mass ratio. If the atomic masses of X\mathrm{X} and Y\mathrm{Y} are in 1:21: 2 ratio, the empirical formula of the compound should be
View answer and explanation
Correct option: A
XY2\mathrm{XY}_2
Option A (XY2\mathrm{XY}_2) is correct because the empirical formula is determined by the simplest whole number ratio of the elements, calculated from their mass ratio and atomic mass ratio.
117 Easy
117 of 436
Application +4 | -1
2.79 g of silver carbonate on being strongly heated yields a residue weighing:
View answer and explanation
Correct option: A
2.16 g
Option A (2.16 g) is correct because the decomposition of silver carbonate yields silver as a residue, and the mass can be calculated based on stoichiometry.
118 Medium
118 of 436
Analyzing +4 | -1
Consider the following redox reaction :

H2O2+O3→H2O+2O2{H}_2 {O}_2+{O}_3 \rightarrow {H}_2 {O}+2 {O}_2

Total loss or gain of electrons by one mole of ozone are
View answer and explanation
Correct option: A
Gain of 2 moles of electrons
Option D (Both A and R are correct but R is NOT the correct explanation of A) is correct because both the assertion about the size of the hydrogen nucleus and the reason that protons exist in combined form are true, but they are unrelated statements.
119 Medium
119 of 436
Understanding +4 | -1
Which of the following reactions is the metal displacement reaction? Choose the right option.
View answer and explanation
Correct option: D
Cr2O3+2Al→ΔAl2O3+2CrC{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr
Option D (Cr2O3+2Al→ΔAl2O3+2CrC{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr) is correct because aluminum displaces chromium from its oxide, which is a metal displacement reaction.
120 Hard
120 of 436
Application +4 | -1
A compound used in making nylon, is 43.8%43.8 \% oxygen. There are four oxygen atoms per molecule. What is the molecular weight of compound?
View answer and explanation
Correct option: D
146146
Option D (146146) is correct because it is the molecular weight calculated from the percentage of oxygen and the number of oxygen atoms per molecule.
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Review SOME BASIC CONCEPTS OF CHEMISTRY in your textbook or class notes, then use this set to check your understanding.

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