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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

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Showing 31-60 of 436 questions
31 Medium
31 of 436
Analyzing +4 | -1
Assertion (A): On changing volume of the solution by 20%20 \%, molarity of solution also changes by 20%20 \%.

Reason (R): Molar concentration or molarity of solution is not affected on dilution.
View answer and explanation
Correct option: D
Both A and R are false
Option D (Both A and R are false) is correct because molarity changes inversely with volume, and the percentage change in molarity is not equal to the percentage change in volume. Also, molarity *is* affected by dilution.
32 Medium
32 of 436
Application +4 | -1
Some bottles of colourless liquids were being labelled when the technicians accidentally mixed them up and lost track of their contents. A 15.0 ml15.0 \mathrm{~ml} sample withdrawn from one bottle weighed 22.3 g.22.3 \mathrm{~g}. The technicians knew that the liquid was either acetone, benzene, chloroform or carbon tetrachloride (which have densities of 0.792 g/cm3,0.899 g/cm3,1.489 g/cm30.792 \mathrm{~g} /\mathrm{cm}^3, 0.899 \mathrm{~g} / \mathrm{cm}^3, 1.489 \mathrm{~g} / \mathrm{cm}^3, and 1.595 g/cm31.595 \mathrm{~g} / \mathrm{cm}^3, respectively). What was the identity of the liquid?
View answer and explanation
Correct option: C
Chloroform
Option C (Chloroform) is correct because the calculated density of the liquid matches the density of chloroform.
33 Easy
33 of 436
Application +4 | -1
The number of atoms in 4.25 g of NH3N{H_3} is
View answer and explanation
Correct option: D
6.023×10236.023 \times {10^{23}}
The correct answer is 6.023×10236.023 \times {10^{23}} because the number of atoms is calculated by multiplying the number of moles by Avogadro's number and the atomicity of the molecule.
34 Easy
34 of 436
Application +4 | -1
The ratio of the masses of methane and ethane in a gas mixture is 4:54: 5. The ratio of number of their molecules in the mixture is
View answer and explanation
Correct option: B
3:23: 2
Option B (3:23: 2) is correct because the ratio of the number of molecules is determined by the ratio of moles, which depends on the masses and molar masses of the gases.
35 Medium
35 of 436
Application +4 | -1
A sample of ammonium phosphate, (NH4)3PO4{\left( {N{H_4}} \right)_3}P{O_4}, contain 6 mole of hydrogen atoms. The number of moles of oxygen atoms in the sample is
View answer and explanation
Correct option: B
2
Option B (2) is correct because in one mole of ammonium phosphate, 12 moles of hydrogen atoms correspond to 4 moles of oxygen atoms, thus 6 moles of hydrogen correspond to 2 moles of oxygen.
36 Medium
36 of 436
Application +4 | -1
In a chemical reaction, A+B→ABA + B \to AB if B is acting as a limiting reagent then choose the correct option
View answer and explanation
Correct option: C
A, 50 atoms; B, 30 atoms
Option C (A, 50 atoms; B, 30 atoms) is correct because B is the limiting reagent, meaning there are fewer atoms of B than required to react with all of A.
37 Easy
37 of 436
Understanding +4 | -1
Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false.
(i) Gay-Lussac’s law of gaseous volumes is actually the law of definite proportion by volume.
(ii) Law of conservation of mass is true for physical change, but not for chemical change.
(iii) The percentage of oxygen in H2O2H_2O_2 is different from that in H2O.H_2O. Hence, it violates law of definite proportions.
(iv) Fixed mass of A reacts with two different masses of B (say x and y), then the ratio of x/y can be any positive integer.
(v) At STP, 5 mL of N2N_2 and H2H_2 have different no. of molecules.
View answer and explanation
Correct option: C
TFFTF
Option C (TFFTF) is correct because it accurately reflects the truth or falsehood of each statement regarding gas laws and chemical proportions.
38 Medium
38 of 436
Application +4 | -1
1 mole of chlorine combines with a certain weight of a metal giving 111 g of its chloride. The atomic weight of the metal (assuming its valency to be 2) is :
View answer and explanation
Correct option: C
80
Option C (80) is correct because the equivalent weight of the metal is 40, and since the valency is 2, the atomic weight is 80.
39 Medium
39 of 436
Application +4 | -1
The molarity of 4.9%4.9 \% (w / w) H2SO4\mathrm{H}_2 \mathrm{SO}_4 solution having density 1.98 g/ml1.98 \mathrm{~g} / \mathrm{ml} will be
View answer and explanation
Correct option: A
0.99 M0.99 \mathrm{~M}
The correct answer is A (0.99 M) because it is calculated using the formula Molarity = (% by weight * density * 10) / Molar mass.
40 Medium
40 of 436
Application +4 | -1
How many moles of P4O6\mathrm{P}_4 \mathrm{O}_6 and P4O10\mathrm{P}_4 \mathrm{O}_{10} will be produced by the combustion of 12.4 g12.4 \mathrm{~g} of phosphorous (atomic mass 31) in 12.8 g12.8 \mathrm{~g} of oxygen, leaving no P4\mathrm{P}_4 or O2\mathrm{O}_2 ?
View answer and explanation
Correct option: C
0.05 mol0.05 \mathrm{~mol} each
Option C (0.05 mol0.05 \mathrm{~mol} each) is correct because the reaction stoichiometry and limiting reactant determine the moles of P4O6\mathrm{P}_4 \mathrm{O}_6 and P4O10\mathrm{P}_4 \mathrm{O}_{10} produced.
41 Easy
41 of 436
Analyzing +4 | -1
A: Number of ions in 9 gram of NH4+{NH}_4^{+}is equal to Avogadro's number (NA)\left({N}_{{A}}\right).

R: Number of ions is equal to number of atoms.
View answer and explanation
Correct option: D
Both the A and R are incorrect.
Option D is correct because the number of ions in 9 grams of NH4+NH_4^+ is half of Avogadro's number, and the number of ions is not always equal to the number of atoms.
42 Medium
42 of 436
Application +4 | -1
How much AgCl{AgCl} will be formed by adding 200 mL200 \mathrm{~mL} of 5 N5 {~N} HCl{HCl} to a solution containing 1.7 g AgNO31.7 \mathrm{~g} \, {AgNO}_3 (Ag=108)?({Ag}=108)?
View answer and explanation
Correct option: B
1.435 g1.435 \mathrm{~g}
Option B (1.435 g) is correct because it is calculated based on the limiting reagent and the stoichiometry of the reaction.
43 Medium
43 of 436
Application +4 | -1
If 65.5 g of a metal chloride contains 35.5 g chlorine, then equivalent mass of metal is
View answer and explanation
Correct option: A
30
Option A (30) is correct because the equivalent mass of the metal is calculated using the ratio of masses and equivalent masses of the metal and chlorine.
44 Easy
44 of 436
Analyzing +4 | -1
Equivalent weight of FeS2\mathrm{FeS}_2 in the half reaction, FeS2⟶Fe2O3+SO2\mathrm{FeS}_2 \longrightarrow \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{SO}_2 is :
View answer and explanation
Correct option: B
M/11M / 11
The correct answer is B (M/11M / 11) because the equivalent weight is calculated by dividing the molar mass by the n-factor, which is 11 in this reaction.
45 Medium
45 of 436
Remembering +4 | -1
One gram mole of a gas at NTP occupies 22.4 L. This fact was derived from
View answer and explanation
Correct option: B
Avogadro’s hypothesis
Option B (Avogadro’s hypothesis) is correct because Avogadro's hypothesis states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules, which leads to the molar volume concept.
46 Medium
46 of 436
Application +4 | -1
Solid NaHCO3{NaHCO}_3 will be neutralized by 40.0 mL40.0 \mathrm{~mL} of 0.1 MH2SO40.1\, {MH}_2 {SO}_4 solution. What would be the weight of solid NaHCO3{NaHCO}_3 in gram?
View answer and explanation
Correct option: A
0.672gm0.672 {gm}
Option A (0.672 gm) is correct because the weight of solid NaHCO₃ is calculated based on the stoichiometry of the reaction with H₂SO₄.
47 Medium
47 of 436
Application +4 | -1
How many moles of H2SO4{H_2}S{O_4} are present in 4.9 g H2SO4{H_2}S{O_4}
View answer and explanation
Correct option: B
0.05
Option B (0.05) is correct because the number of moles is calculated by dividing the given mass by the molar mass of H2SO4.
48 Medium
48 of 436
Application +4 | -1
Which among the following has maximum number of atoms among the following?
View answer and explanation
Correct option: A
22gCO222\mathrm{ }g\mathrm{ }C{O_2}
Option A (22gCO222\mathrm{ }g\mathrm{ }C{O_2}) is correct because it contains the maximum number of atoms, calculated as 1.5 times Avogadro's number.
49 Hard
49 of 436
Application +4 | -1
If 6.256.25 is to be rounded off to AA and if 6.356.35 is to be rounded off to BB. Here, AA and BB refer to
View answer and explanation
Correct option: A
A→6.2,B→6.4A \rightarrow 6.2, B \rightarrow 6.4
Option A (A→6.2,B→6.4A \rightarrow 6.2, B \rightarrow 6.4) is correct because it follows the rounding rules where if the rightmost digit to be removed is 5, the preceding even number remains unchanged, and the preceding odd number is increased by one.
50 Medium
50 of 436
Analyzing +4 | -1
The no. of Pi bonds in increasing order is present in
View answer and explanation
Correct option: A
CrO4−2<MnO4−<Cr2O7−2Cr{O_4}^{ - 2} < Mn{O_4}^ - < C{r_2}{O_7}^{ - 2}
Option A (CrO4−2<MnO4−<Cr2O7−2Cr{O_4}^{ - 2} < Mn{O_4}^ - < C{r_2}{O_7}^{ - 2}) is correct because the number of pi bonds increases in the order: CrO4−2CrO_4^{-2} (2 π bonds), MnO4−MnO_4^- (3 π bonds), Cr2O7−2Cr_2O_7^{-2} (6 π bonds).
51 Medium
51 of 436
Application +4 | -1
How many g of KCl would have to be dissolved in 60 g H2O{H_2}O to give 20% by weight of solution?
View answer and explanation
Correct option: A
15 g
Option A (15 g) is correct because it is the mass of KCl required to make a 20% by weight solution when dissolved in 60 g of water.
52 Medium
52 of 436
Application +4 | -1
What is the number of moles of Fe(OH)3(s){\left( {\mathrm{OH}} \right)_3}\left( \mathrm{s} \right) that can be produced by allowing 1 mole of Fe2S3\mathrm{F}{\mathrm{e}_2}{\mathrm{S}_3}, 2 moles of H2O{\mathrm{H}_2}\mathrm{O} and 3 moles of O2{\mathrm{O}_2} to react as
2Fe2S3+6H2O+3O→4Fe(OH)3+6S?2\mathrm{F}{\mathrm{e}_2}{\mathrm{S}_3} + 6{\mathrm{H}_2}\mathrm{O} + 3\mathrm{O} \to 4\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3} + 6\mathrm{S}?
View answer and explanation
Correct option: C
1.34 mol
Option C (1.34 mol) is correct because the limiting reagent is H2O, and the stoichiometry of the reaction dictates the amount of Fe(OH)3 produced.
53 Easy
53 of 436
Application +4 | -1
Insulin contains 3.4% Sulphur. The minimum mol. weight of insulin is:
View answer and explanation
Correct option: A
941.176
Option A (941.176) is correct because it is calculated by assuming that insulin contains at least one atom of sulfur per molecule.
54 Medium
54 of 436
Application +4 | -1
2KMnO4+16HCl→2 \mathrm{KMnO}_4+16 \mathrm{HCl} \rightarrow 2KCl+2MnCl2+5Cl2+8H2O.2 \mathrm{KCl}+2 \mathrm{MnCl}_2+5 \mathrm{Cl}_2+8 \mathrm{H}_2 \mathrm{O}. Find nfactor n_{\text {factor }} of HCl\mathrm{HCl} in above reaction.
View answer and explanation
Correct option: C
1016\frac{10}{16}
Option C (1016\frac{10}{16}) is correct because it represents the change in oxidation state of HCl divided by the number of moles of HCl in the balanced reaction.
55 Medium
55 of 436
Application +4 | -1
A mixture of salts was treated to determine the barium content. A 0.230 g\mathrm{A}\, 0.230 \mathrm{~g} sample was dissolved and treated with excess potassium chromate. The precipitate of BaCrO4\mathrm{BaCrO}_4 was dissolved in dil HCl\mathrm{HCl} to 2 convert BaCrO4\mathrm{BaCrO}_4 to Cr2O72−\mathrm{Cr}_2 \mathrm{O}_7^{2-} anion. The solution treated with excess sodium iodide and tri lodide produced was titrated with 21 mL21 \mathrm{~mL} of 0.095 M0.095 \mathrm{~M} sodium thiosulphate. Calculate the %\% of BaCl2.2H2O\mathrm{BaCl}_2 . 2 \mathrm{H}_2 \mathrm{O} in the sample. [given Molar Mass of BaCl2.2H2O=244\mathrm{BaCl}_2. 2 \mathrm{H}_2 \mathrm{O}=244]
View answer and explanation
Correct option: A
70.5%70.5 \%
Option A (70.5 %) is correct because the percentage of BaCl2.2H2O is calculated based on the stoichiometry of the reactions and the titration data.
56 Hard
56 of 436
Application +4 | -1
Density of dry air containing only N2\mathrm{N}_2 and O2\mathrm{O}_2 is 1.15 g/L1.15 \mathrm{~g} / \mathrm{L} at 740 mm740 \mathrm{~mm} of Hg\mathrm{Hg} and 300 K300 \mathrm{~K}. What is %\% composition of N2\mathrm{N}_2 by weight in the air?
View answer and explanation
Correct option: C
70.02%70.02 \%
Option C (70.02%70.02 \%) is correct because it is calculated using the ideal gas law and weighted average molar mass to find the percentage composition of N2 by weight.
57 Medium
57 of 436
Remembering +4 | -1
The equivalent weight of an element can be calculated from:
View answer and explanation
Correct option: A
6.4 divided by specific heat and valence
Option A (6.4 divided by specific heat and valence) is correct because the approximate atomic weight can be found by Dulong-Petit's law, and equivalent weight is related to atomic weight and valence.
58 Easy
58 of 436
Analyzing +4 | -1
Which one of the following is not a redox reaction
View answer and explanation
Correct option: A
CaCO3→CaO+CO2CaC{O_3} \to CaO + C{O_2}
Option A is correct because in CaCO3 -> CaO + CO2, the oxidation states of all elements remain unchanged, indicating it is not a redox reaction.
59 Medium
59 of 436
Application +4 | -1
BaCl2+Na2SO4→BaSO4+2NaCl\mathrm{BaCl}_2+\mathrm{Na}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4+2 \mathrm{NaCl}
How many moles of NaCl\mathrm{NaCl} are produced when initially there are 3 moles of Na2SO4\mathrm{Na}_2 \mathrm{SO}_4 and 1 mole BaCl2\mathrm{BaCl}_2 ?
View answer and explanation
Correct option: C
2 moles
Option C (2 moles) is correct because BaCl2 is the limiting reagent, and the stoichiometry of the reaction dictates the amount of NaCl produced.
60 Medium
60 of 436
Application +4 | -1
The isotopic abundance of C-12 and C-14 is 98% and 2% respectively. What would be the number of C-14 isotope in 12 g carbon sample?
View answer and explanation
Correct option: A
1.032×10221.032 \times {10^{22}}
Option A (1.032×10221.032 \times {10^{22}}) is correct because it accurately calculates the number of C-14 isotopes in the given carbon sample based on isotopic abundance.
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