Skip to main content
Back to SOME BASIC CONCEPTS OF CHEMISTRY

Your Progress

0 / 436 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →
Showing 211-240 of 436 questions

Preparing practice and saved progress…

211 Medium
211 of 436
Analyzing +4 | -1
Assertion (AA) Significant figures for 0.200 is 3 whereas for 200 it is 1. Reason (RR) Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.
View answer and explanation
Correct option: C
AA is true but RR is false
Option C (A is true but R is false) is correct because the assertion about significant figures is correct, but the reason incorrectly states the rule for trailing zeros.
212 Hard
212 of 436
Application +4 | -1
The number of moles of KMnO4KMn{O_4} that will be needed to react completely with one mole of oxalate ion in acidic solution is
View answer and explanation
Correct option: C
25\frac{2}{5}
The correct answer is C (2/5) because the stoichiometry of the redox reaction between KMnO4KMn{O_4} and oxalate ion in acidic solution requires 2 moles of KMnO4KMn{O_4} for every 5 moles of oxalate.
213 Easy
213 of 436
Application +4 | -1
Vapour density of a gas is 32. Volume occupied by 16 g16 \mathrm{~g} substance at S.T.P.
View answer and explanation
Correct option: C
5.6 L5.6 \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.
214 Medium
214 of 436
Application +4 | -1
At 373 K373 \mathrm{~K} and 1 atm,1 \mathrm{~atm}, if the density of liquid water is 1.0 g/ml1.0 \mathrm{~g} / \mathrm{ml} and that of water vapour is 0.0006 g/ml,0.0006 \mathrm{~g} / \mathrm{ml}, then the volume occupied by water molecules in 1 litre of steam at that temperature is
View answer and explanation
Correct option: C
0.6 ml0.6 \mathrm{~ml}
Option C (0.6 ml) is correct because the volume occupied by water molecules in steam is equal to the mass of water in steam divided by the density of liquid water.
215 Medium
215 of 436
Application +4 | -1
A sample of ammonium phosphate (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4} contains 6.36 moles of hydrogen atoms. The number of moles of oxygen atom in the sample is (atomic mass of N = 14.04, H = 1, P = 31, O = 16)
View answer and explanation
Correct option: C
2.12
Option C (2.12) is correct because it is the number of moles of oxygen atoms calculated based on the given number of moles of hydrogen atoms in the ammonium phosphate sample.
216 Medium
216 of 436
Analyzing +4 | -1
Statement-I: Equivalent weight of NH3,{NH}_3, in the reaction N2→NH3N_2 \rightarrow N H_3 is 17/317 / 3 while that of N2N_2 is 28/6.28 / 6.

Statement-II: Equivalent weight = Molecular weight  number of e− lost or gained/mole =\frac{\text { Molecular weight }}{\text { number of } e^- \text{ lost or gained/mole }}
View answer and explanation
Correct option: A
Both the statements are correct
Option A (Both the statements are correct) is correct because the equivalent weight is calculated correctly in both cases using the number of electrons transferred.
217 Medium
217 of 436
Application +4 | -1
A metal M{M} forms the sulphate M2(SO4)3\mathrm{M}_2\left({SO}_4\right)_3. A 0.6 g0.6 \mathrm{~g} sample of the metal sulphate reacts with excess BaCl2{BaCl}_2 to give 1.2 g BaSO41.2 \mathrm{~g}\, {BaSO}_4. What is the atomic weight of M{M} ?

(Atomic weights : S=32,Ba=137{S}=32, {Ba}=137 )
View answer and explanation
Correct option: A
30.75
Option A (30.75) is correct because it is calculated by equating the equivalents of the metal sulfate and barium sulfate.
218 Hard
218 of 436
Application +4 | -1
An unknown compound A\mathrm{A} dissociates at 500∘C500^{\circ} \mathrm{C} to give products as follows A(g)⇄B(g)+C(g)+D(g)A(g) \rightleftarrows B(g)+C(g)+D(g) Vapour density of the equilibrium mixture is 50 when it dissociates to the extent to 10%10 \%. What will be the molecular weight of compound A-
View answer and explanation
Correct option: A
120
Option A (120) is correct because the molecular weight of compound A is calculated using the vapour density at equilibrium and the degree of dissociation.
219 Medium
219 of 436
Application +4 | -1
Which contains the same number of molecules as 16 g16\,\mathrm{g} of oxygen?
View answer and explanation
Correct option: C
32 g SO232\,\mathrm{g}\,\mathrm{S}{\mathrm{O}_2}
Option C (32 g SO232\,\mathrm{g}\,\mathrm{S}{\mathrm{O}_2}) is correct because it contains the same number of moles as 16 g16\,\mathrm{g} of oxygen, and therefore the same number of molecules.
220 Medium
220 of 436
Application +4 | -1
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?
(Atomic weight: Mg=24Mg = 24)
View answer and explanation
Correct option: B
84
Option B (84) is correct because the percentage purity is calculated by comparing the actual yield of MgO to the theoretical yield based on the decomposition of MgCO3.
221 Medium
221 of 436
Application +4 | -1
3.92 g/lit3.92 \mathrm{~g} / \mathrm{lit} of a sample of ferrous ammonium sulphate reacts completely with 50mlN10KMnO450 \mathrm{ml} \frac{\mathrm{N}}{10} \mathrm{KMnO}_4 solution. The percentage purity of the sample is
View answer and explanation
Correct option: A
50
Option A (50) is correct because the percentage purity is calculated by comparing the strength of the pure salt to the strength of the sample.
222 Medium
222 of 436
Application +4 | -1
Haemoglobin contains 4.6%4.6 \% of FeFe by weight, the approximate minimum molecular mass of haemoglobin will be
View answer and explanation
Correct option: B
1200
Option C (1400) is correct because the minimum molecular mass is calculated assuming one iron atom per hemoglobin molecule.
223 Easy
223 of 436
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 X2O{\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.
224 Medium
224 of 436
Understanding +4 | -1
When 100 g100 \mathrm{~g} of ethylene polymerizes to polyethylene according to equation of nCH2=CH2→−(CH2−CH2)n−.n \mathrm{CH}_2=\mathrm{CH}_2 \rightarrow-\left(\mathrm{CH}_2-\mathrm{CH}_2\right)_n-. The weight of polyethylene produced will be
View answer and explanation
Correct option: B
100 g100 \mathrm{~g}
Option B (100 g) is correct because the mass is conserved during polymerization; the mass of the ethylene monomer equals the mass of the polyethylene polymer.
225 Medium
225 of 436
Remembering +4 | -1
Chemistry was not studied for its own sake, rather it came up as a result of search for
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because early alchemists sought the Philosopher's stone and the Elixir of life, and Paras is another name for the Philosopher's stone.
226 Medium
226 of 436
Application +4 | -1
D5W refers to one of the solutions used as an intravenous fluid. It is 5%5 \% by mass solution of dextrose, C6H12O6\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 in water. The density of D5W is 1.08 g/ml.1.08 \mathrm{~g} / \mathrm{ml}. The molarity of the solution is
View answer and explanation
Correct option: A
0.3 M0.3 \mathrm{~M}
Option B (11.3) is correct because it is calculated by finding the excess hydroxide ions and then calculating the pOH and pH.
227 Medium
227 of 436
Application +4 | -1
245 g245 \mathrm{~g} impure sample of KClO3( s){KClO}_3(\mathrm{~s}) on heating gives 12 g  O2( g)12 \mathrm{~g}\,\, {O}_2(\mathrm{~g}) according to

2KClO3(s)→2KCl(s)+3O2(g)2 {KClO}_3(\mathrm{s}) \rightarrow 2 {KCl}(\mathrm{s})+3 {O}_2(\mathrm{g}) Calculate %\% purity of sample?
View answer and explanation
Correct option: A
12.5
Option A (12.5) is correct because it is calculated using stoichiometry to relate the mass of oxygen produced to the mass of KClO3 in the impure sample.
228 Medium
228 of 436
Analyzing +4 | -1
In the following reaction, which choice has value twice that of the equivalent mass of the oxidizing agent?

SO2+H2O→3S+2H2O\mathrm{S}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O} \to 3\mathrm{S} + 2{\mathrm{H}_2}\mathrm{O}
View answer and explanation
Correct option: B
32
Option B (32) is correct because the equivalent weight of SO₂ is 16, and twice that value is 32.
229 Medium
229 of 436
Application +4 | -1
A gaseous mixture contains oxygen and nitrogen in the ratio of 1:41:4 by weight. Therefore, the ratio of their number of molecules is:
View answer and explanation
Correct option: C
7:87:8
Option C (7:8) is correct because the ratio of the number of molecules is calculated based on their molar masses and the given weight ratio.
230 Medium
230 of 436
Understanding +4 | -1
Assertion (A): Molarity of a solution in liquid state depends upon temperature.
Reason (R): Volume of a solution is temperature independent entity.
View answer and explanation
Correct option: C
Assertion (A) is true, but Reason (R) is false
Option C (Assertion (A) is true, but Reason (R) is false) is correct because molarity is temperature-dependent, and the volume of a solution is also temperature-dependent.
231 Medium
231 of 436
Application +4 | -1
What will be the volume of CO2C{O_2} at NTP obtained on heating 10 grams of (90 % pure) limestone
View answer and explanation
Correct option: B
2.016 L
Option C (CH3CHClCOOH\mathrm{C}{\mathrm{H}_3}\mathrm{CHClCOOH}) is correct because the reaction of propanoic acid with Cl2 in the presence of red phosphorus leads to alpha-halogenation (Hell-Volhard-Zelinsky reaction).
232 Hard
232 of 436
Analyzing +4 | -1
A mixture of NaHC2O4\mathrm{NaHC}_2 \mathrm{O}_4 and KHC2O4⋅H2C2O4\mathrm{KHC}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 required equal volumes of 0.2 N KMnO40.2 \mathrm{~N} \mathrm{~KMnO}_4 and 0.12 NNaOH0.12 \mathrm{~N} \mathrm{NaOH} separately. What is the molar ratio of NaHC2O4\mathrm{NaHC}_2 \mathrm{O}_4 and KHC2O4⋅H2C2O4\mathrm{KHC}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 in the mixture?
View answer and explanation
Correct option: D
3:13: 1
Option D (3:1) is correct because by setting up equations based on the stoichiometry of the reactions with KMnO4 and NaOH, we can solve for the molar ratio of NaHC2O4 and KHC2O4·H2C2O4.
233 Medium
233 of 436
Application +4 | -1
17 g17 \mathrm{~g} of H2O2\mathrm{H}_2 \mathrm{O}_2 is present in 100ml100 \mathrm{ml} an aqueous solution 10ml10 \mathrm{ml} of this on decomposition at STP will produce mℓm \ell of O2\mathrm{O}_2 at STP?
View answer and explanation
Correct option: B
560
Option B (560) is correct because the volume of O2 produced at STP is calculated based on the decomposition of H2O2, using stoichiometry and the molar volume of a gas at STP.
234 Medium
234 of 436
Application +4 | -1
The equivalent weight of KIO3\mathrm{KI}{\mathrm{O}_3} in the reaction, 2Cr(OH)3+OH−+KIO3→2CrO42−+5H2O+KI2\mathrm{Cr}{\left( {\mathrm{OH}} \right)_3} + \mathrm{O}{\mathrm{H}^ - } + \mathrm{KI}{\mathrm{O}_3} \to 2\mathrm{CrO}_4^{2 - } + 5{\mathrm{H}_2}\mathrm{O} + \mathrm{KI} is
View answer and explanation
Correct option: C
Mol. wt./6
Option C (Mol. wt./6) is correct because the change in oxidation number of iodine in KIO3 is 6, and the equivalent weight is the molecular weight divided by this change.
235 Medium
235 of 436
Application +4 | -1
One mole of N2H4\mathrm{N}_2 \mathrm{H}_4 loses 10 moles of electrons to form a new compound y. Assuming that all nitrogen appear in the new compound, what is the oxidation state of nitrogen in y (There is no change in the oxidation state of hydrogen)
View answer and explanation
Correct option: C
+3+3
Option C (+3) is correct because the oxidation state of nitrogen changes from -2 in N₂H₄ to +3 in the new compound after losing 10 electrons.
236 Medium
236 of 436
Application +4 | -1
At constant temperature for 1 mole of an ideal gas, if its pressure is decreased by 20%20 \% then its volume will be increased by
View answer and explanation
Correct option: C
25%25 \%
Option C (25%25 \%) is correct because when the pressure is decreased by 20%, the volume increases by 25% according to Boyle's Law.
237 Medium
237 of 436
Application +4 | -1
Which one of the following sets of compounds correctly illustrate the law of reciprocal proportions?
View answer and explanation
Correct option: A
P2O3,PH3,H2OP_2O_3, PH_3, H_2O
Option A (P2O3,PH3,H2OP_2O_3, PH_3, H_2O) is correct because the ratios of elements combining with a fixed mass of a third element are related to the ratio in which they combine with each other.
238 Medium
238 of 436
Application +4 | -1
Oxidation number of carbon in carbon suboxide is :
View answer and explanation
Correct option: B
+43+ \frac{4}{3}
Option B (+4/3) is correct because it is calculated by setting up an equation where the sum of oxidation states equals zero.
239 Medium
239 of 436
Application +4 | -1
1.5276 g1.5276 \mathrm{~g} of CdCl2\mathrm{CdCl}_2 was found to contain 0.9367 g0.9367 \mathrm{~g} of Cd\mathrm{Cd}. The atomic mass of Cd\mathrm{Cd} is (to the nearest integer):
View answer and explanation
Correct option: A
113 g
Option A (113 g) is correct because the atomic mass of Cd is calculated using the equivalent mass concept and the given masses of Cd and Cl in CdCl2.
240 Medium
240 of 436
Application +4 | -1
In an experiment 50 mL of 0.1 MM solution of a salt reacted with 25 mL of 0.1 MM solution of sodium sulphite. The half equation for the oxidation of SO32−(aq)+H2O(l)→SO42−(aq)+2H+(aq)+2e−\mathrm{SO}_3^{2 - }\left( {aq} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right) \to \mathrm{SO}_4^{2 - }\left( {aq} \right) + 2{\mathrm{H}^ + }\left( {aq} \right) + 2{e^ - } If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal?
View answer and explanation
Correct option: C
2
Option C (2) is correct because the change in oxidation number of the metal is 1, resulting in a new oxidation number of 2.
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.

Showing 211-240 of 436 questions

Your practice

0 / 436 answered · 436 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.