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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 391-420 of 436 questions

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391 Hard
391 of 436
Analyzing +4 | -1
A quantity of 2.3 g2.3 \mathrm{~g} of a mixture of NO2\mathrm{NO}_2 and N2O4\mathrm{N}_2 \mathrm{O}_4 has a pressure of 0.82 atm0.82 \mathrm{~atm}, at temperature T K\text{T K} in a container of volume V\text{V} litres such that the ratio, T : V\text{T : V} is 300:1300: 1 in magnitude. What is the degree of dissociation of N2O4?\mathrm{N}_2 \mathrm{O}_4?
View answer and explanation
Correct option: B
0.33
Option B is correct because the degree of dissociation is calculated using the ideal gas law and the molar mass of the mixture.
392 Medium
392 of 436
Analyzing +4 | -1
Statement I : Values of R can be used to convert one energy parameter into another
Statement II : R has the dimension of work mol - 1  K - 1{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\,\,{{\text{K}}^{{\text{ - 1}}}}.
Which statement(s) is/are correct?
View answer and explanation
Correct option: B
Only II
Option B (Only II) is correct because Statement II accurately describes the dimensions of the ideal gas constant R, while Statement I is too general and not always applicable.
393 Medium
393 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 dividing the mass of NaClO per liter by its equivalent weight.
394 Medium
394 of 436
Remembering +4 | -1
The prefix 1018{10^{18}} is
View answer and explanation
Correct option: C
exa
Option C (exa) is correct because the prefix exa represents 1018{10^{18}}.
395 Medium
395 of 436
Analyzing +4 | -1
Which of the following is not a disproportion atom reaction
View answer and explanation
Correct option: D
F2+OH−→F−+OF2+H2O{F_2} + O{H^ - } \to {F^ - } + O{F_2} + {H_2}O
Option D (F2+OH−→F−+OF2+H2O{F_2} + O{H^ - } \to {F^ - } + O{F_2} + {H_2}O) is correct because fluorine cannot exhibit a positive oxidation state, so it cannot be both oxidized and reduced in the same reaction.
396 Medium
396 of 436
Understanding +4 | -1
Assertion (A): Laboratory reagents are usually made up to a specific molarity rather than a given molality.

Reason (R): The volume of a liquid is more easily measured than its mass.
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 molarity is preferred due to the ease of volume measurement, which is the reason why laboratory reagents are made to specific molarities.
397 Medium
397 of 436
Application +4 | -1
If NAN_A is Avogadro's number then number of valence electrons in 4.2 g4.2 \mathrm{~g} of nitride ions (N3−)\left(\mathrm{N}^{3-}\right) is
View answer and explanation
Correct option: B
2.4 NA2.4 \mathrm{~N}_{\mathrm{A}}
Option B (2.4NA2.4 N_A) is correct because each nitride ion has 8 valence electrons, and the number of moles of nitride ions is 0.3.
398 Hard
398 of 436
Application +4 | -1
A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 mL of CO2C{O_2} at T = 298.15 K and p=1p = 1 bar. If molar volume of CO2C{O_2} is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet?
[Molar mass of NaHCO3= 84  g  mol−1]NaHC{O_3} = \,84\,\,g\,\,mo{l^{ - 1}}]
View answer and explanation
Correct option: C
0.84
Option C (0.84) is correct because it is calculated by determining the moles of CO2 released and then using stoichiometry to find the mass percentage of sodium bicarbonate.
399 Medium
399 of 436
Application +4 | -1
Number of significant figures in the answer of following will be:

(29.2−20.2)(1.79×105)1.39\frac{(29.2-20.2)\left(1.79 \times 10^5\right)}{1.39}
View answer and explanation
Correct option: B
3
Option B (3) is correct because the result of the calculation has three significant figures, determined by the least precise number in the calculation.
400 Medium
400 of 436
Application +4 | -1
How much Ca(NO3)2,\mathrm{Ca}\left(\mathrm{NO}_3\right)_2, in mg,\mathrm{mg}, must be present in 50 ml50 \mathrm{~ml} of a solution with 2.35 ppm2.35 \mathrm{~ppm} of Ca?\mathrm{Ca}?
View answer and explanation
Correct option: D
0.48
Option D (0.48) is correct because it represents the calculated mass of Ca(NO3)2\mathrm{Ca(NO_3)_2} in mg, present in the given solution based on the ppm of Ca.
401 Hard
401 of 436
Application +4 | -1
A volume of 500 ml500 \mathrm{~ml} of a 0.1 M0.1 \mathrm{~M} solution of AgNO3{AgNO}_3 is added to 500 ml500 \mathrm{~ml} of 0.1 M0.1 \mathrm{~M} solution of KCl.{KCl}. The concentration of nitrate ion in the resulting solution is
View answer and explanation
Correct option: A
0.05 M0.05 \mathrm{~M}
Option A (0.05 M0.05 \mathrm{~M}) is correct because the concentration of nitrate ions is calculated by dividing the moles of nitrate by the total volume.
402 Hard
402 of 436
Application +4 | -1
Common salt obtained from sea-water contains 8.775%NaCl8.775 \% \mathrm{NaCl} by mass. The number of formula units of NaCl\mathrm{NaCl} present in 25 g25 \mathrm{~g} of this salt is :
View answer and explanation
Correct option: B
2.258×10222.258 \times 10^{22} formula units
Option B (2.258×10222.258 \times 10^{22} formula units) is correct because it is calculated by finding the mass of NaCl in the sample, converting it to moles, and then multiplying by Avogadro's number.
403 Medium
403 of 436
Application +4 | -1
What volume of 0.40 MM Na2S2O3\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3} would be required to react with the I2{\mathrm{I}_2} liberated by adding excess of KI to 50 mL of 0.20 MM CuSO4\mathrm{CuS}{\mathrm{O}_4} solution?
View answer and explanation
Correct option: B
25 mL
Option B (25 mL) is correct because the milliequivalents of Na2S2O3 equals the milliequivalents of CuSO4.
404 Easy
404 of 436
Application +4 | -1
The chloride of a metal contains 52.8%52.8 \% chlorine by mass. The equivalent weight of metal is
View answer and explanation
Correct option: A
31.7
Option A (31.7) is correct because the equivalent weight of the metal can be calculated using the percentage of chlorine in the metal chloride.
405 Medium
405 of 436
Application +4 | -1
What is the percentage of 'free SO3\mathrm{SO}_3' in a sample of oleum labelled as 104.5%?104.5 \%?
View answer and explanation
Correct option: A
20%20 \%
Option A (20%) is correct because the percentage of free SO3 is calculated based on the excess percentage of oleum.
406 Hard
406 of 436
Application +4 | -1
Hydrazine N2H4{N}_2 {H}_4 (used as a fuel in rocket system) can be produced according to the following reaction.

ClNH2+2NH3→N2H4+NH4Cl{ClNH}_2+2 {NH}_3 \rightarrow {N}_2 {H}_4+{NH}_4 {Cl}

When 1.0 kg ClNH21.0 \mathrm{~kg} \, {ClNH}_2 is reacted with excess of NH3,{NH}_3, 473 g473 \mathrm{~g} of N2H4{N}_2 {H}_4 is produced. What is the percentage yield?
View answer and explanation
Correct option: A
76.12
Option A (76.12) is correct because the percentage yield is calculated by dividing the actual yield (473 g) by the theoretical yield (621.36 g) and multiplying by 100.
407 Medium
407 of 436
Analyzing +4 | -1
A closed vessel contains equal numbers of O2{O_2} and H2{H_2} molecules at same TT. Which of the following is not true?
View answer and explanation
Correct option: D
The weight of H2{H_2} is the same as the weight of oxygen
Option D (The weight of H2{H_2} is the same as the weight of oxygen) is correct because if the number of molecules is equal, the weights are not the same due to different molar masses.
408 Medium
408 of 436
Application +4 | -1
Volume of O2\mathrm{O}_2 required at STP to completely burn 3 g3 \mathrm{~g} of ethane is
View answer and explanation
Correct option: B
7.84 L7.84 \mathrm{~L}
The correct answer is D (2-methylprop-1-ene) because it correctly identifies the longest carbon chain containing the double bond and the position of the methyl substituent.
409 Medium
409 of 436
Application +4 | -1
When BrO3−\mathrm{BrO}_3^{-}ion reacts with Br−\mathrm{Br}^{-}in acid medium, Br2\mathrm{Br}_2 is liberated. The equivalent weight of Br2\mathrm{Br}_2 in this reaction is:
View answer and explanation
Correct option: C
3M5\frac{3 M}{5}
Option C (3M5\frac{3 M}{5}) is correct because the equivalent weight of Br2 is calculated based on the change in oxidation state during the reaction.
410 Medium
410 of 436
Application +4 | -1
30 mL30 \mathrm{~mL} of 0.2N BaCl0.2 \mathrm{N} \mathrm{~BaC}{l} is mixed with 40 mL40 \mathrm{~mL} of 0.3N Al2(SO4)30.3 \mathrm{N} \mathrm{~A}{l}_2\left(\mathrm{SO}_4\right)_3. How many g\mathrm{g} of BaSO4\mathrm{BaSO}_4 are formed?
View answer and explanation
Correct option: D
0.70 g0.70 \mathrm{~g}
Option D (0.70 g0.70 \mathrm{~g}) is correct because the amount of BaSO4 formed is determined by the limiting reactant, which in this case is BaCl2.
411 Medium
411 of 436
Application +4 | -1
The weight of 1×10221 \times 10^{22} molecules of CuSO4⋅5H2O{CuSO}_4 \cdot 5 {H}_2 {O} is
View answer and explanation
Correct option: C
4.159 g4.159 \mathrm{~g}
Option C (4.159 g4.159 \mathrm{~g}) is correct because it is calculated by dividing the number of molecules by Avogadro's number and multiplying by the molar mass.
412 Easy
412 of 436
Understanding +4 | -1
The quantity which changes with temperature is 
View answer and explanation
Correct option: C
Molarity
Option C is correct because molarity is defined as moles of solute per liter of solution, and the volume of a solution changes with temperature, thus affecting molarity.
413 Medium
413 of 436
Application +4 | -1
Volume of N2\mathrm{N}_2 required at STP to produce 10 moles of NH3\mathrm{NH}_3 through Haber's process is
View answer and explanation
Correct option: A
112 L112 \mathrm{~L}
The correct answer is A (112 L112 \mathrm{~L}) because 5 moles of N2 are required to produce 10 moles of NH3, and the volume of 5 moles of any gas at STP is 5 * 22.4 L = 112 L.
414 Medium
414 of 436
Application +4 | -1
Volume of one molecule of oxygen gas at S.T.P. is
View answer and explanation
Correct option: A
3.7×10−20 mL3.7 \times 10^{-20} {~mL}
Option A (3.7×10−20 mL3.7 \times 10^{-20} {~mL}) is correct because it's calculated by dividing the molar volume at STP by Avogadro's number.
415 Medium
415 of 436
Analyzing +4 | -1
Select the option having maximum number of carbon atoms among the following
View answer and explanation
Correct option: D
0.5  g C5H100.5 \,{~g}\, {C}_5 {H}_{10}
Option D (0.5 g C5H10) is correct because it contains the maximum number of carbon atoms compared to the other options when calculated based on their molar masses.
416 Hard
416 of 436
Application +4 | -1
A mixture contains FeSO4\mathrm{FeSO}_4 and Fe2(SO4)3.\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3. If both FeSO4\mathrm{FeSO}_4 and Fe2(SO4)3\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 provide equal number of sulphate ions then, the ratio of Fe2+\mathrm{Fe}^{2+} and Fe3+\mathrm{Fe}^{3+} ions in mixture is
View answer and explanation
Correct option: D
3:23: 2
Option D (3:23: 2) is correct because the ratio of Fe²⁺ to Fe³⁺ ions is determined by the equal number of sulfate ions provided by FeSO₄ and Fe₂(SO₄)₃.
417 Medium
417 of 436
Application +4 | -1
If 6.3 g of NaHCO3NaHC{O_3} are added to 15.0 g CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} solution, the residue is found to weight 18.0 g. what is the mass of CO2C{O_2} released in the reaction?
View answer and explanation
Correct option: B
3.3 g
Option B (3.3 g) is correct because the mass of CO2 released is calculated using the law of conservation of mass.
418 Easy
418 of 436
Application +4 | -1
What is the percentage of carbon in calcium carbide (CaC2)(Ca{C_2})
View answer and explanation
Correct option: A
37.5
Option A (37.5) is correct because the percentage of carbon in calcium carbide is calculated by dividing the mass of carbon by the molecular mass of calcium carbide and multiplying by 100.
419 Medium
419 of 436
Analyzing +4 | -1
Statement-I: Molarity of a solution does not depend upon temperature whereas molality depends.

Statement-II: Molarity and molality both depend only on the number of moles of solute particles.
View answer and explanation
Correct option: D
Statement II is correct but statement I is incorrect
Option D is correct because molarity depends on volume, which changes with temperature, while molality depends on mass, which is independent of temperature. Also, both depend on the number of moles of solute.
420 Hard
420 of 436
Application +4 | -1
If air contains N2{\mathrm{N}_2} and O2{\mathrm{O}_2} in volume ratio 4:14:1. The average vapour density of air is:
View answer and explanation
Correct option: C
14.4
Option C (14.4) is correct because the average vapor density is calculated from the average molecular weight of the air mixture, considering the volume ratio of nitrogen and oxygen.
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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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