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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

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Showing 121-150 of 436 questions

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121 Easy
121 of 436
Remembering +4 | -1
The hardness of water is usually expressed in:
View answer and explanation
Correct option: A
ppm
Option A (ppm) is correct because hardness of water is commonly expressed in parts per million.
122 Medium
122 of 436
Application +4 | -1
7.5 gr of a gas occupy 5.8 L of volume at STP, the gas is
View answer and explanation
Correct option: A
NONO
Option A (NONO) is correct because the molar mass calculated from the given weight and volume at STP matches the molar mass of NO.
123 Medium
123 of 436
Application +4 | -1
Mixing up of equal volumes of 0.1  M  NaOH  and  0.1  M  CH3COOH0.1\;M\;\mathrm{NaOH}\;\mathrm{and}\;0.1\;M\;\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} yields a solution which is:
View answer and explanation
Correct option: A
Basic
Option A (Basic) is correct because NaOH is a strong base and CH3COOH is a weak acid, so the resulting solution will be basic due to the hydrolysis of CH3COONa.
124 Medium
124 of 436
Understanding +4 | -1
CH4(g)+2O2(g)⟶CO2(g)+2H2O(g)\mathrm{CH_4(g})+2 \mathrm{O_2(g}) \longrightarrow \mathrm{CO_2(g})+2 \mathrm{H}_2 \mathrm{O(g})
According to the above chemical reaction, following information have been given.

(i) One mole of CH4(g)\mathrm{CH}_4(\mathrm{g}) reacts with two moles of O2(g)\mathrm{O}_2(\mathrm{g}) to give one mole of CO2(g)\mathrm{CO}_2(\mathrm{g}) and two moles of H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{g})

(ii) One molecule of CH4(g)\mathrm{CH}_4(\mathrm{g}) reacts with 2 molecules of O2(g)\mathrm{O}_2(\mathrm{g}) to give one molecule of CO2(g)\mathrm{CO}_2(\mathrm{g}) and 2 molecules of H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{g})

(iii) 22.7 L22.7 \mathrm{~L} of CH4(g)\mathrm{CH}_4(\mathrm{g}) reacts with 45.4 L45.4 \mathrm{~L} of O2\mathrm{O}_2 to give 22.7 L22.7 \mathrm{~L} of CO2(g)\mathrm{CO}_2(\mathrm{g}) and 45.4 L45.4 \mathrm{~L} of H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{g})

(iv) 16 g16 \mathrm{~g} of CH4(g)\mathrm{CH}_4(\mathrm{g}) reacts with 2×32 g2 \times 32 \mathrm{~g} of O2(g)\mathrm{O}_2(\mathrm{g}) to give 44 g44 \mathrm{~g} of CO2(g)\mathrm{CO}_2(\mathrm{g}) and 2×18 g2 \times 18 \mathrm{~g} of H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{g})

From these information the correct interconversion should be
View answer and explanation
Correct option: B
Mass ⇌\rightleftharpoons moles ⇌\rightleftharpoons number of molecules
Option B (Mass ⇌ moles ⇌ number of molecules) is correct because mass, moles, and the number of molecules can be interconverted using molar mass and Avogadro's number.
125 Medium
125 of 436
Analyzing +4 | -1
Which of the following contains greatest number of oxygen atoms?
View answer and explanation
Correct option: D
All have the same number of atoms
Option D (All have the same number of atoms) is correct because the number of oxygen atoms is calculated by (moles x Avogadro's number x atomicity), which is the same for all options.
126 Medium
126 of 436
Understanding +4 | -1
The O18/O16\mathrm{O}^{18} / \mathrm{O}^{16} ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. The average mass of an atom of oxygen in these meteorites is _____ that of a terrestrial oxygen atom?
View answer and explanation
Correct option: B
greater than
The correct answer is B (greater than) because a higher O18/O16\mathrm{O}^{18} / \mathrm{O}^{16} ratio indicates a greater proportion of the heavier isotope, resulting in a higher average atomic mass.
127 Hard
127 of 436
Understanding +4 | -1
Assertion: Equivalence point is a theoretical value.
Reason: End point is an experimental value.
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 Reason are correct, but the Reason is not the correct explanation of the Assertion. The equivalence point is a theoretical value, and the endpoint is an experimental value, but the endpoint being experimental doesn't explain why the equivalence point is theoretical.
128 Medium
128 of 436
Application +4 | -1
Mass of CO2C{O_2} produced by heating 40 g of 20% pure lime stone
View answer and explanation
Correct option: B
3.52 g
Option B (3.52 g) is correct because it is the mass of CO2 produced from the decomposition of 40g of 20% pure limestone.
129 Easy
129 of 436
Application +4 | -1
Mole fraction of solvent in 10%(ww)10 \%\left(\frac{w}{w}\right) aqueous NaOH\mathrm{NaOH} is
View answer and explanation
Correct option: B
0.952
Option B (0.952) is correct because it is the mole fraction of the solvent (water) in a 10% (w/w) aqueous NaOH solution, calculated by determining the moles of solute and solvent.
130 Hard
130 of 436
Analyzing +4 | -1
0.1 M solution of KI reacts with excess of H2SO4\mathrm{H}_2 \mathrm{SO}_4 and KIO3\mathrm{KIO}_3 solution. According to equation

5I−+IO3−+6H+→3I2+3H2O5 \mathrm{I}^{-}+\mathrm{IO}_3^{-}+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}

Identify the correct statements:

(A) 200 mL of KI solution reacts with 0.004 mol of KIO3\mathrm{KIO}_3

(B) 200 mL of KI solution reacts with 0.006 mol of H2SO4\mathrm{H}_2 \mathrm{SO}_4

(3) 0.5 L of KI solution produced 0.005 mol of I2\mathrm{I}_2

(4) Equivalent weight of KIO3\mathrm{KIO}_3 is equal to (Molecular weight5)\left(\frac{\text {Molecular weight}}{5}\right)

Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
(A) and (D) only
The correct answer is A ((A) and (D) only) because the stoichiometry of the reaction confirms that 200 mL of 0.1 M KI reacts with 0.004 mol of KIO3, and the equivalent weight of KIO3 is Molecular weight/5.
131 Medium
131 of 436
Application +4 | -1
Chlorophyll contains 2.68%2.68 \% of magnesium by mass. Calculate the number of magnesium atoms in 3 g3 \mathrm{~g} of chlorophyll.
View answer and explanation
Correct option: A
2.01×10212.01 \times 10^{21} atoms
Option A (6) is correct because when the given redox reaction is balanced with the smallest integer coefficients, the sum of the coefficients is 6.
132 Easy
132 of 436
Application +4 | -1
Which has the maximum number of molecules among the following
View answer and explanation
Correct option: C
8 g  H28\,g\,\,{H_2}
Option C (8g  H28g\,\,{H_2}) is correct because it has the highest number of moles, and therefore the highest number of molecules, among the given options.
133 Easy
133 of 436
Understanding +4 | -1
In 4 moles of ethane, the number of molecules of ethane are
View answer and explanation
Correct option: A
4 NA4 {~N}_{{A}}
Option A (4 NA4 {~N}_{{A}}) is correct because the number of molecules in a given number of moles is calculated by multiplying the number of moles by Avogadro's number (NAN_A).
134 Medium
134 of 436
Understanding +4 | -1
Assertion (A): Hydrogen burns with a pop sound and oxygen is a supporter of combustion but water is used as a fire extinguisher.
Reason (R): The properties of a compound are different from those of its constituent elements.
View answer and explanation
Correct option: A
Both Assertion and Reason are correct; Reason is the correct explanation of Assertion
Option A is correct because both the assertion and reason are true, and the reason correctly explains why water, a compound of hydrogen and oxygen, has different properties than its constituent elements.
135 Easy
135 of 436
Application +4 | -1
24 g of metal react with an acid and liberated 22.4 litres of hydrogen at STP. The equivalent wt of the metal is
View answer and explanation
Correct option: B
12
Option B (12) is correct because the equivalent weight is calculated by dividing the weight of the metal by the number of moles of hydrogen liberated.
136 Medium
136 of 436
Analyzing +4 | -1
Statement I: Equimolar solutions of H3PO3\mathrm{H}_3 \mathrm{PO}_3 and H2SO4\mathrm{H}_2 \mathrm{SO}_4 are not equinormal.

Statement II: Both are dibasic acid and Molarity ×2=\times 2= normality.
View answer and explanation
Correct option: D
Statement I is false; Statement II is true
Option D (Statement I is false; Statement II is true) is correct because both H3PO3 and H2SO4 are dibasic, meaning their equimolar solutions are equinormal, making Statement I incorrect, while Statement II correctly states they are dibasic.
137 Hard
137 of 436
Application +4 | -1
How many millilitres of a 0.05 M KMnO40.05 \mathrm{~M} \mathrm{~KMnO}_4 solution are required to oxide 2.0 g2.0 \mathrm{~g} of FeSO4\mathrm{FeSO}_4 in a dilute acid solution? (Fe=56,S=32,O=16)(\mathrm{Fe}=56, \mathrm{S}=32, \mathrm{O}=16).
View answer and explanation
Correct option: A
53
Option A (53) is correct because the volume of KMnO4 required is calculated using stoichiometry and molarity, resulting in approximately 53 mL.
138 Hard
138 of 436
Application +4 | -1
The density of 3M3 \mathrm{M} solution of NaCl\mathrm{NaCl} is 1.25 g mL−11.25 \mathrm{~g} \mathrm{~mL}^{-1}. Calculate the molality of the solution.
View answer and explanation
Correct option: A
2.79 m
Option A (2.79 m) is correct because molality is calculated by dividing the moles of solute by the mass of the solvent in kg, considering the density of the solution.
139 Medium
139 of 436
Application +4 | -1
A sample of AlF3\mathrm{Al}{\mathrm{F}_3} contains 3.0×10243.0 \times {10^{24}} F ions. The number of formula units of this sample are
View answer and explanation
Correct option: D
1.0×10241.0 \times {10^{24}}
Option D (1.0×10241.0 \times {10^{24}}) is correct because each formula unit of AlF3\mathrm{Al}{\mathrm{F}_3} contains 3 fluoride ions, so the number of formula units is one-third the number of fluoride ions.
140 Medium
140 of 436
Analyzing +4 | -1
Assertion (A) : 4 moles of KClO3{KClO}_3 (50 % pure) gave 3 moles of O2\mathrm{O}_2 on heating strongly and thus yield is 100%100 \%

2KClO3⟶2KCl+3O22{KClO}_3 \longrightarrow 2 {KCl}+3 {O}_2

Reason (R) : 3 moles of O2{O}_2 are obtained from 2 moles of KClO3{KClO}_3.
View answer and explanation
Correct option: B
Both A\mathbf{A} and R\mathbf{R} are true but R\mathbf{R} is not the correct explanation of A\mathbf{A}.
Option B (Both A\mathbf{A} and R\mathbf{R} are true but R\mathbf{R} is not the correct explanation of A\mathbf{A}) is correct because both the assertion and the reason are true statements, but the reason does not explain why the yield is 100%.
141 Medium
141 of 436
Understanding +4 | -1
The number of significant figures in 60.0001 is
View answer and explanation
Correct option: B
6
Option B (6) is correct because all digits in 60.0001 are significant, including the zeros between the non-zero digits and after the decimal point.
142 Easy
142 of 436
Application +4 | -1
4 g-atom of Ag contains:
View answer and explanation
Correct option: C
432 g
Option C (432 g) is correct because it is the mass of 4 gram-atoms of Ag, calculated using the molar mass of Ag.
143 Hard
143 of 436
Application +4 | -1
5 moles of AB2A{B_2} weight 125×10−3kg125 \times {10^{ - 3}}kg and 10 moles of A2B2{A_2}{B_2} weight 300×10−3kg300 \times {10^{ - 3}}kg. The molar mass of A(MA)A\left( {{M_A}} \right) and molar mass of B(MB)B\left( {{M_B}} \right) in kg mol−1\mathrm{mo}{\mathrm{l}^{ - 1}} are
View answer and explanation
Correct option: D
MA=5×10−3{M_A} = 5 \times {10^{ - 3}} and MB=10×10−3{M_B} = 10 \times {10^{ - 3}}
Option D (MA=5×10−3{M_A} = 5 \times {10^{ - 3}} and MB=10×10−3{M_B} = 10 \times {10^{ - 3}}) is correct because it is the solution to the system of equations derived from the given masses and moles of the compounds AB2AB_2 and A2B2A_2B_2.
144 Hard
144 of 436
Analyzing +4 | -1
Assertion: Sulphate is estimated as BaSO4\mathrm{BaSO}_4 and not as MgSO4\mathrm{MgSO}_4.
Reason: Ionic radius of Mg2+\mathrm{Mg}^{2+} is smaller than Ba2+\mathrm{Ba}^{2+}.
View answer and explanation
Correct option: D
Both Assertion and Reason are correct but Reason is not correct explanation of Assertion
Option D (Both Assertion and Reason are correct but Reason is not correct explanation of Assertion) is correct because while both statements are true, the smaller ionic radius of Mg2+ doesn't directly explain why sulfate is estimated as BaSO4.
145 Hard
145 of 436
Application +4 | -1
Mole fraction of A in water is 0.2. The molality of A in water is:
View answer and explanation
Correct option: A
13.8
Option A (13.8) is correct because it is calculated by converting mole fraction to molality using the molar mass of water.
146 Medium
146 of 436
Application +4 | -1
A quantity of 50 g50 \mathrm{~g} of water is saturated with HCl{HCl} gas to get 75 ml75 \mathrm{~ml} of solution containing 40% HCl40 \% \, {HCl} by mass. The density of solution formed is
View answer and explanation
Correct option: A
1.11 g/ml1.11 \mathrm{~g} / \mathrm{ml}
Option A (1.11 g/ml) is correct because it is calculated using the mass percentage and volume of the solution.
147 Medium
147 of 436
Application +4 | -1
An ethanolic solution of naphthalene contains 0.25 mole fraction of solute. Molality of solution is
View answer and explanation
Correct option: C
7.25 m
Option C (7.25 m) is correct because it is the molality of the solution calculated using the mole fraction of the solute and the molar mass of the solvent.
148 Medium
148 of 436
Application +4 | -1
Element X\mathrm{X} reacts with oxygen to produce a pure sample of X2O3.\mathrm{X}_2 \mathrm{O}_3. In an experiment it is found that 1 g1 \mathrm{~g} of X\mathrm{X} produces 1.16 g1.16 \mathrm{~g} of X2O3.\mathrm{X}_2 \mathrm{O}_3. Calculate the atomic weight of X.\mathrm{X}. (Given atomic weight of oxygen, 16.0 g mol−116.0 \mathrm{~g} \mathrm{~mol}^{-1})
View answer and explanation
Correct option: D
150
Option D (150) is correct because it is calculated using the principle of atom conservation (POAC) applied to element X.
149 Medium
149 of 436
Understanding +4 | -1
Potassium combines with two isotopes of chlorine (35Cl and 37Cl)\left({ }^{35} {Cl} \text{ and } { }^{37} {Cl}\right) respectively to form two samples of KCl.{KCl}. Their formation follows the law of
View answer and explanation
Correct option: A
Constant proportions
Option A (Constant proportions) is correct because the formation of KCl from potassium and either isotope of chlorine always follows the law of constant proportions.
150 Easy
150 of 436
Remembering +4 | -1
A molal solution is one that contains one mole of a solute in:
View answer and explanation
Correct option: A
1000 g of the solvent
Option A is correct because a molal solution is defined as one containing one mole of solute in 1000 g of the solvent.
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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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