A 0.50 $M$ solution of $KI$ reacts with excess of ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$ and ${\rm{KI}}{{\rm{O}}_3}$ solutions according to the equation, $6{{\rm{H}}^ + } + 5{{\rm{I}}^ - } + {\rm{IO}}_3^ - \to 3{{\rm{I}}_2} + 3{{\rm{H}}_2}{\rm{O}}$. Which of the following statements is true?

0.5 litre of the $KI$ solution produces 0.15 mole of ${{\rm{I}}_2}$

200 mL of the $KI$ solution reacts with 0.10 mole ${\rm{KI}}{{\rm{O}}_3}$.

100 mL of the $KI$solution reacts with 0.060 $M$ of ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$.

In the reaction, $2{\rm{Al}}\left( s \right) + 6{\rm{HCl}}\left( {aq} \right) \to 2{\rm{A}}{{\rm{l}}^{3 + }}\left( {aq} \right) + 6{\rm{C}}{{\rm{l}}^ - }\left( {aq} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right)$

$11.2{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ at STP is produced for every mole HCl(aq) consumed

6 L HCl (aq) is consumed for every $3{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ produced

$33.6{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ is produced regardless of temperature and pressure for every mole Al that reacts

$67.2{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ at STP is produced for every mole Al that reacts

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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Hard

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1 Hard

1 of 77

Application +4 | -1

8 g of NaOH is dissolved in 18 g of

{H_2}O.

Mole fraction of NaOH in solution and molality (in

mol\,\,k{g^{ - 1}})

of the solution respectively are

View answer and explanation

Correct option: D

0.167, 11.11

Option D (0.167, 11.11) is correct because it accurately calculates the mole fraction of NaOH and the molality of the solution.

2 Hard

2 of 77

Application +4 | -1

A sample of sodium has a mass of

46 \mathrm{~g}

. What is the mass of the same number of calcium atoms as sodium atoms present in given sample?

View answer and explanation

Correct option: D

80 \mathrm{~g}

Option D (

80 \mathrm{~g}

) is correct because the mass of the same number of calcium atoms is twice the mass of sodium atoms since the atomic mass of calcium is approximately twice that of sodium.

3 Hard

3 of 77

Application +4 | -1

The most abundant elements by mass in the body of a healthy human adult are Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all

{}^1H

atoms are replaced by

{}^2H

atoms is

View answer and explanation

Correct option: C

7.5 kg

Option C (7.5 kg) is correct because it is the mass gain calculated by replacing all

{}^1H

atoms with

{}^2H

atoms.

4 Hard

4 of 77

Application +4 | -1

50 \mathrm{~mL}

0.01 \mathrm{~M} \mathrm{~FeSO}_4

will react with what volume of

0.01 \mathrm{~M} \mathrm{~KMnO}_4

solution in acid medium?

\left(1 \mathrm{~mol} \mathrm{~KMnO}_4\right.

requires

5 \mathrm{~mol}

\mathrm{FeSO}_4

for complete reaction)

View answer and explanation

Correct option: D

10 \mathrm{~mL}

Option D (

10 \mathrm{~mL}

) is correct because the volume of KMnO4 required is calculated using the stoichiometry of the reaction.

5 Hard

5 of 77

Application +4 | -1

6.25

is to be rounded off to

A

and if

6.35

is to be rounded off to

B

. Here,

A

and

B

refer to

View answer and explanation

Correct option: A

A \rightarrow 6.2, B \rightarrow 6.4

Option A (

A \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.

6 Hard

6 of 77

Application +4 | -1

Density of dry air containing only

\mathrm{N}_2

and

\mathrm{O}_2

1.15 \mathrm{~g} / \mathrm{L}

740 \mathrm{~mm}

\mathrm{Hg}

and

300 \mathrm{~K}

. What is

\%

composition of

\mathrm{N}_2

by weight in the air?

View answer and explanation

Correct option: C

70.02 \%

Option C (

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.

7 Hard

7 of 77

Application +4 | -1

2.4 kg of carbon is made to react with 1.35 kg of aluminium to form

Al_4C_3.

The maximum amount (in kg) of aluminium carbide formed is

View answer and explanation

Correct option: D

1.8

The correct answer is 1.8 kg because Aluminium is the limiting reagent. The maximum amount of aluminium carbide formed is determined by the limiting reactant.

8 Hard

8 of 77

Application +4 | -1

The amount of

{BaSO}_4

formed upon mixing

100 {~mL}

20.8 \% \,{BaCl}_2

solution with

50 {~mL}

9.8 \% {H}_2 {SO}_4

solution with

50 {~mL}

9.8 \%\,

{H}_2 {SO}_4

solution will be:

({Ba}=137, {Cl}=35.5, {~S}=32, {H}=1 \text { and } {O}=16 \text { ) }

View answer and explanation

Correct option: A

23.3 {~g}

Option A (

23.3 {~g}

) is correct because it is the mass of

{BaSO}_4

formed based on the stoichiometry of the reaction between

{BaCl}_2

and

{H}_2 {SO}_4

9 Hard

9 of 77

Analyzing +4 | -1

Assertion:-

44 \mathrm{~g}

\mathrm{CO}_2, 28 \mathrm{~g}

\mathrm{CO}

have same volume at STP.

Reason:- Both

\mathrm{CO}_2

and

\mathrm{CO}

are formed by

\mathrm{C}

and oxygen.

View answer and explanation

Correct option: B

If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion

Option B is correct. The assertion is true because equal moles of gases occupy the same volume at STP. The reason is also true but doesn't explain the assertion.

10 Hard

10 of 77

Application +4 | -1

Calculate the molality of 1 litre solution of

93 \% \, \mathrm{H}_2 \mathrm{SO}_4

(weight/volume). The density of the solution is

1.84 \mathrm{~g} / \mathrm{mL}

View answer and explanation

Correct option: C

10.42 m

Option C (10.42 m) is correct because the molality is calculated using the mass of solute and solvent in the solution.

11 Hard

11 of 77

Application +4 | -1

Complete combustion of

0.858 \mathrm{~g}

of compound

X

gives

2.64 \mathrm{~g}

\mathrm{CO}_2

and

1.26 \mathrm{~g}_{\text {of }} \mathrm{H}_2 \mathrm{O}

. The lowest molecular mass

X

can have :

View answer and explanation

Correct option: A

43 \mathrm{~g}

Option B (21) is correct because it is the equivalent weight calculated based on the change in oxidation state of iron.

12 Hard

12 of 77

Application +4 | -1

4 \mathrm{~g}

mixture of

\mathrm{Na}_2 \mathrm{O}

and

\mathrm{NaCl}

was dissolved in

400 \mathrm{~mL}

0.3 \mathrm{~M} \mathrm{~HCl}

. After complete reaction, the resulting solution required

200 \mathrm{~mL}

0.2 \mathrm{~M} \mathrm{~NaOH}

for complete neutralization of the excess

\mathrm{HCl}

. What mass

\%

\mathrm{Na}_2 \mathrm{O}

is present in the mixture?

View answer and explanation

Correct option: C

62 \%

Option C (62%) is correct because it is calculated by finding the moles of Na2O that reacted with HCl and then calculating the mass percent.

13 Hard

13 of 77

Application +4 | -1

A spherical ball of radius

7 \mathrm{~cm}

contains

56 \%

iron. If density is

1.4 \mathrm{~g} \mathrm{~cm}^{-3},

number of moles of

\text{Fe}

present approximately is

View answer and explanation

Correct option: C

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.

14 Hard

14 of 77

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.

15 Hard

15 of 77

Application +4 | -1

A compound used in making nylon, is

43.8 \%

oxygen. There are four oxygen atoms per molecule. What is the molecular weight of compound?

View answer and explanation

Correct option: D

146

Option D (

146

) 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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16 Hard

16 of 77

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.

17 Hard

17 of 77

Analyzing +4 | -1

0.1 M solution of KI reacts with excess of

\mathrm{H}_2 \mathrm{SO}_4

and

\mathrm{KIO}_3

solution. According to equation

5 \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

\mathrm{KIO}_3

(B) 200 mL of KI solution reacts with 0.006 mol of

\mathrm{H}_2 \mathrm{SO}_4

(3) 0.5 L of KI solution produced 0.005 mol of

\mathrm{I}_2

(4) Equivalent weight of

\mathrm{KIO}_3

is equal to

\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.

18 Hard

18 of 77

Application +4 | -1

How many millilitres of a

0.05 \mathrm{~M} \mathrm{~KMnO}_4

solution are required to oxide

2.0 \mathrm{~g}

\mathrm{FeSO}_4

in a dilute acid solution?

(\mathrm{Fe}=56, \mathrm{S}=32, \mathrm{O}=16)

View answer and explanation

Correct option: A

Option A (53) is correct because the volume of KMnO4 required is calculated using stoichiometry and molarity, resulting in approximately 53 mL.

19 Hard

19 of 77

Application +4 | -1

The density of

3 \mathrm{M}

solution of

\mathrm{NaCl}

1.25 \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.

20 Hard

20 of 77

Application +4 | -1

5 moles of

A{B_2}

weight

125 \times {10^{ - 3}}kg

and 10 moles of

{A_2}{B_2}

weight

300 \times {10^{ - 3}}kg

. The molar mass of

A\left( {{M_A}} \right)

and molar mass of

B\left( {{M_B}} \right)

in kg

\mathrm{mo}{\mathrm{l}^{ - 1}}

are

View answer and explanation

Correct option: D

{M_A} = 5 \times {10^{ - 3}}

and

{M_B} = 10 \times {10^{ - 3}}

Option D (

{M_A} = 5 \times {10^{ - 3}}

and

{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

AB_2

and

A_2B_2

21 Hard

21 of 77

Analyzing +4 | -1

Assertion: Sulphate is estimated as

\mathrm{BaSO}_4

and not as

\mathrm{MgSO}_4

.
Reason: Ionic radius of

\mathrm{Mg}^{2+}

is smaller than

\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.

22 Hard

22 of 77

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.

23 Hard

23 of 77

Application +4 | -1

In the balanced chemical reaction,

\mathrm{IO}_3^{-}+\mathrm{al}^{-}+\mathrm{bH}^{+} \longrightarrow \mathrm{CH}_2 \mathrm{O}+\mathrm{dl}_2

. a, b, c and d respectively correspond to

View answer and explanation

Correct option: A

5,6,3,3

Option A (5,6,3,3) is correct because it represents the stoichiometric coefficients that balance the redox reaction.

24 Hard

24 of 77

Application +4 | -1

Suppose that

A

and

B

form the compounds

B_2 A_3

and

B_2 A

. If 0.05 mole of

B_2 A_3

weighs

9 \mathrm{~g}

and 0.1 mole of

B_2 A

weighs

10 \mathrm{~g}

, the atomic weight of

A

and

B

respectively are

View answer and explanation

Correct option: B

40 and 30

Option B (40 and 30) is correct because solving the system of equations derived from the molar masses of the compounds yields these atomic weights.

25 Hard

25 of 77

Application +4 | -1

Amount of

{FeSO}_4

solution required containing

20 \%(w / w) {FeSO}_4

for complete titration of

100 \,{ml}

0.1 \,\,{M}\,\, {KMnO}_4

solution in acidic medium is

View answer and explanation

Correct option: B

38 \,{~g}

Option B (110.2) is correct because the change in standard entropy is calculated using the formula

\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}

26 Hard

26 of 77

Application +4 | -1

A quantity of

27.6 \mathrm{~g}

\mathrm{K}_2 \mathrm{CO}_3

was treated by a series of reagent so as to convert all of its carbon to

\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2.

The mass of the product formed is

(\mathrm{K}=39, \mathrm{Zn}=65.4, \mathrm{Fe}=56)

View answer and explanation

Correct option: B

11.6 \mathrm{~g}

Option B (11.6 g) is correct because it is calculated based on the stoichiometry of the reaction and the molar masses of the compounds involved.

27 Hard

27 of 77

Application +4 | -1

21.75g of

Mn{O_2}

on reaction with HCl forms 2.8 L of

C{l_2}\left( g \right)

at STP, the percentage purity of

Mn{O_2}

is (atomic mass of Mn = 55u)

Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O

View answer and explanation

Correct option: D

0.5

Option D (0.5) is correct because the percentage purity is calculated by comparing the actual yield of

Cl_2

to the theoretical yield based on the amount of

MnO_2

used.

28 Hard

28 of 77

Application +4 | -1

For the reaction:

2 \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}, 8

mole of

\mathrm{A}

and excess of

\mathrm{B}

will produce

View answer and explanation

Correct option: B

4 mole of C

Option B (4 mole of C) is correct because according to the stoichiometry of the reaction, 2 moles of A produce 1 mole of C, so 8 moles of A will produce 4 moles of C.

29 Hard

29 of 77

Application +4 | -1

A 0.50

M

solution of

KI

reacts with excess of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

and

\mathrm{KI}{\mathrm{O}_3}

solutions according to the equation,

6{\mathrm{H}^ + } + 5{\mathrm{I}^ - } + \mathrm{IO}_3^ - \to 3{\mathrm{I}_2} + 3{\mathrm{H}_2}\mathrm{O}

. Which of the following statements is true?

View answer and explanation

Correct option: C

0.5 litre of the

KI

solution produces 0.15 mole of

{\mathrm{I}_2}

Option C (0.5 litre of the

KI

solution produces 0.15 mole of

{\mathrm{I}_2}

) is correct because 5 moles of

I^-

produces 3 moles of

I_2

30 Hard

30 of 77

Analyzing +4 | -1

In the reaction,

2\mathrm{Al}\left( s \right) + 6\mathrm{HCl}\left( {aq} \right) \to 2\mathrm{A}{\mathrm{l}^{3 + }}\left( {aq} \right) + 6\mathrm{C}{\mathrm{l}^ - }\left( {aq} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right)

View answer and explanation

Correct option: D

11.2\mathrm{L}{\mathrm{H}_2}\left( \mathrm{g} \right)

at STP is produced for every mole HCl(aq) consumed

The correct answer is D because the stoichiometry of the balanced equation shows that 6 moles of HCl produce 3 moles of H2, meaning 1 mole of HCl produces 0.5 moles of H2, which is 11.2 L at STP.

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