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

If the density of water is 1 g ${\rm{c}}{{\rm{m}}^3}$) then the volume occupied by one molecule of water is approximately

$3.0 \times {10^{ - 23}}\;{\rm{c}}{{\rm{m}}^3}$

$6.02 \times {10^{ - 23}}\;{\rm{c}}{{\rm{m}}^3}$

0.1 mole of a carbohydrate with empirical formula ${\rm{C}}{{\rm{H}}_2}{\rm{O}}$ contains 1 g of hydrogen. What is its molecular formula?

${{\rm{C}}_5}{{\rm{H}}_{10}}{{\rm{O}}_5}$

${{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}$

${{\rm{C}}_4}{{\rm{H}}_8}{{\rm{O}}_4}$

${{\rm{C}}_3}{{\rm{H}}_6}{{\rm{O}}_3}$

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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Mixed difficulty

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181 Medium

181 of 436

Application +4 | -1

If a piece of copper weighs

0.635 \mathrm{~g}

, how many atoms does it contain?

View answer and explanation

Correct option: A

6.023 \times 10^{21}

Option A (

6.023 \times 10^{21}

) is correct because it is calculated by dividing the given mass by the molar mass and multiplying by Avogadro's number.

182 Easy

182 of 436

Application +4 | -1

Four faradays of electricity is passed through a solution of

{CuSO}_4

. The mass of copper deposited at the cathode is: (At. mass of

{Cu}=63.5 \mathrm{amu}

View answer and explanation

Correct option: C

127 \mathrm{~g}

Option C (

127 \mathrm{~g}

) is correct because it represents the mass of copper deposited when four faradays of electricity are passed through a

CuSO_4

solution.

183 Medium

183 of 436

Application +4 | -1

An element

(\mathrm{X})

reacts with hydrogen leading to formation of a class of compounds that is analogous to hydrocarbons.

5 \mathrm{~g}

\mathrm{X}

forms

5.628 \mathrm{~g}

of a mixture of two compounds of

\mathrm{X}\left(\mathrm{XH}_4\right.

and

\left.\mathrm{X}_2 \mathrm{H}_6\right)

in the molar ratio of

2: 1.

Determine the molar mass of

\mathrm{X}.

View answer and explanation

Correct option: A

Option A (28) is correct because applying the Principle of Atomic Conservation (POAC) on element X allows us to determine its molar mass.

184 Medium

184 of 436

Application +4 | -1

Rest mass of 1 mole neutrons

\left(m_{\mathrm{n}}=1.675 \times 10^{-27} \mathrm{~kg}\right)

is:

View answer and explanation

Correct option: D

1.008 \times 10^{-3} \mathrm{~kg}

Option D (

1.008 \times 10^{-3} \mathrm{~kg}

) is correct because it is the product of the mass of one neutron and Avogadro's number.

185 Medium

185 of 436

Application +4 | -1

Number of moles of hydrogen atoms present in

60 \mathrm{~amu}

of acetic acid:

View answer and explanation

Correct option: A

4 / \mathrm{N}_{\mathrm{A}}

Option A (

4 / N_A

) is correct because one molecule of acetic acid contains 4 hydrogen atoms, and we are dealing with one molecule (60 amu).

186 Easy

186 of 436

Application +4 | -1

Vapour density of a volatile substance is

(\mathrm{C}{\mathrm{H}_4} = \mathrm{ }1\mathrm{ })

. Its molecular weight would be:

View answer and explanation

Correct option: C

Option C (64) is correct because the molecular weight is twice the vapor density, and the vapor density is given relative to methane (CH4 = 1).

187 Medium

187 of 436

Application +4 | -1

1.00 \times 10^{-3}

moles of

{Ag}^{+}

and

1.00 \times 10^{-3}

moles of

{CrO}_4^{2-}

react together to form solid

{Ag}_2 {CrO}_4

. Calculate the amount of

{Ag}_2 {CrO}_4

formed

\left({Ag}_2 {CrO}_4=331.73 \mathrm{~g} \mathrm{~mol}^{-1}\right)

View answer and explanation

Correct option: B

0.166 \mathrm{~g}

Option B (

0.166 \mathrm{~g}

) is correct because it is the mass of Ag₂CrO₄ formed, calculated using the limiting reagent concept.

188 Easy

188 of 436

Remembering +4 | -1

Molarity is expressed as:

View answer and explanation

Correct option: B

\mathrm{Mol\; litr}{\mathrm{e}^{ - 1}}

Option B (

\mathrm{Mol\; litr}{\mathrm{e}^{ - 1}}

) is correct because molarity is defined as moles of solute per liter of solution.

189 Medium

189 of 436

Application +4 | -1

Number of molecules in 1 litre of oxygen at NTP is:

View answer and explanation

Correct option: B

\frac{6.02 \times 10^{23}}{22.4}

Option B is correct because at NTP, 1 mole of any gas occupies 22.4 L, so the number of molecules in 1 L is Avogadro's number divided by 22.4.

190 Medium

190 of 436

Analyzing +4 | -1

100 \mathrm{~mL}

10 \% \mathrm{~NaOH}(w / \mathrm{V})

is added to

100 \mathrm{~mL}

10 \% \mathrm{HCl}(w / \mathrm{V})

. The resultant solution becomes:

View answer and explanation

Correct option: C

acidic

Option C (acidic) is correct. The number of moles of HCl is greater than the number of moles of NaOH, resulting in an acidic solution.

191 Medium

191 of 436

Understanding +4 | -1

Statement I: The molality of the solution does not change with temperature.

Statement II: The molality is expressed in the unit of

\mathrm{mol}

\mathrm{kg}^{-1}

of solvent.

View answer and explanation

Correct option: B

Statement I is true; Statement II is true Statement II is not the correct explanation of Statement I

Option B (Statement I is true; Statement II is true Statement II is not the correct explanation of Statement I) is correct because molality is defined as moles of solute per kg of solvent, which is mass-based and thus independent of temperature, but this definition doesn't explain the temperature independence.

192 Hard

192 of 436

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.

193 Medium

193 of 436

Application +4 | -1

The number of moles of water present in a spherical water droplet of radius

1.0 \mathrm{~cm}

is

[Given: density of water in the droplet

=1.0 \mathrm{~g}

\left.\mathrm{cm}^{-3}\right]

View answer and explanation

Correct option: B

\frac{2 \pi}{27}

Option B (2π/27) is correct because the number of moles of water is calculated from the volume of the sphere, density of water, and molar mass of water.

194 Medium

194 of 436

Analyzing +4 | -1

The equivalent weight of

\mathrm{Fe}_3 \mathrm{O}_4

in the reaction,

\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{KMnO}_4 \rightarrow \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{MnO}_2

would be

View answer and explanation

Correct option: B

M

Option B (M) is correct because the n-factor for Fe3O4 in the given reaction is 1, so the equivalent weight is M/1 = M.

195 Medium

195 of 436

Application +4 | -1

Molarity and molality of a solution of an liquid (mol. wt.

=50

) in aqueous solution is 9 and 10 respectively. What is the density of solution?

View answer and explanation

Correct option: D

1.35 \mathrm{~g} / \mathrm{cc}

Option D (1.35 g/cc) is correct because the density is calculated using the molarity, molality, and molecular weight of the solute.

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196 Easy

196 of 436

Application +4 | -1

The number of oxygen atoms in

24.9 \mathrm{~g}

\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}

is (molar mass of

\mathrm{Cu}=63 \mathrm{~g} \mathrm{~mol}^{-1}

)

View answer and explanation

Correct option: C

5.42 \times 10^{23}

The correct answer is

5.42 \times 10^{23}

because the number of oxygen atoms is calculated from the moles of the compound and the number of oxygen atoms per molecule.

197 Easy

197 of 436

Application +4 | -1

Active mass of

0.64 \mathrm{~g}\, {SO}_2

in 10 lit vessel is

View answer and explanation

Correct option: B

10^{-3}\, \mathrm{M}

Option B (

10^{-3}

M) is correct because the active mass is calculated by dividing the number of moles by the volume in liters.

198 Hard

198 of 436

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

199 Medium

199 of 436

Application +4 | -1

The percentage weight of

Zn

in white vitriol

\left[ZnSO_4 \cdot 7H_2O\right]

is approximately equal to

\left(Zn=65, \, S=32, \, O=16 \, \text{and} \, H=1\right)

View answer and explanation

Correct option: D

22.65 \%

Option D (

22.65 \%

) is correct because it is calculated by dividing the mass of Zn by the molecular weight of the compound and multiplying by 100.

200 Medium

200 of 436

Analyzing +4 | -1

Statement I: The number of atoms in a given mass of dioxygen (oxygen) and trioxygen (ozone) gases is same.

Statement II: The number of atoms depends on atomic mass, not on molecular mass.

View answer and explanation

Correct option: A

If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I

If both statements are CORRECT, and Statement II is the CORRECT explanation of Statement I. Same mass, same element and hence, same number of atoms.

201 Medium

201 of 436

Application +4 | -1

On analysis a certain compound was found to contain iodine and oxygen in the ration 254 g of iodine and 80 g of oxygen. The atomic mass of iodine is 127 and that of oxygen is 16. Which is the formula of the compound?

View answer and explanation

Correct option: D

{\mathrm{I}_2}{\mathrm{O}_5}

Option D (

{\mathrm{I}_2}{\mathrm{O}_5}

) is correct because the ratio of iodine to oxygen atoms is 2:5 after converting masses to moles.

202 Medium

202 of 436

Analyzing +4 | -1

For the reaction:

\mathrm{Fe}_2 \mathrm{~S}_3 \longrightarrow \mathrm{FeSO}_4+\mathrm{SO}_2

. The equivalent mass of

\mathrm{Fe}_2 \mathrm{~S}_3

is (M is the mol wt of

\mathrm{Fe}_2 \mathrm{~S}_3

)

View answer and explanation

Correct option: D

\frac{M}{20}

Option D (

\frac{M}{20}

) is correct because the n-factor for the given reaction is 20, and the equivalent mass is calculated as M/n.

203 Hard

203 of 436

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.

204 Medium

204 of 436

Application +4 | -1

Consider the following reaction

{A}+4 {~B}+3 {C}+2 {D} \longrightarrow 5 {E}+3 {~F}

If 4 mole of

A, 14

mole of

B, 6

mole of

C

reacts with 6 mole of

D

, then the limiting reagent is

View answer and explanation

Correct option: C

{C}

Option C (C) is correct because C is the limiting reagent as it is completely consumed before the other reactants.

205 Medium

205 of 436

Analyzing +4 | -1

In the following

\mathrm{NO}_3^{-}+\mathrm{As}_2 \mathrm{~S}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{AsO}_4^{3-}+\mathrm{NO}+\mathrm{SO}_4^{2-}+\mathrm{H}^{+}

Equivalent weight of

A s_2 S_3

(Molecular weight M) is

View answer and explanation

Correct option: D

\frac{M}{28}

Option D (

\frac{M}{28}

) is correct because the equivalent weight is the molecular weight divided by the total number of electrons transferred in the redox reaction.

206 Medium

206 of 436

Application +4 | -1

2 mole of ethyl alcohol are present with 6 mole of water. The mole fraction of alcohol is:

View answer and explanation

Correct option: D

0.25

Option D (0.25) is correct because the mole fraction is calculated by dividing the moles of alcohol by the total moles of the solution.

207 Medium

207 of 436

Application +4 | -1

If the density of water is 1 g

\mathrm{c}{\mathrm{m}^3}

) then the volume occupied by one molecule of water is approximately

View answer and explanation

Correct option: D

3.0 \times {10^{ - 23}}\;\mathrm{c}{\mathrm{m}^3}

Option D (

3.0 \times {10^{ - 23}}\;\mathrm{c}{\mathrm{m}^3}

) is correct because it is the approximate volume calculated by dividing the molar volume by Avogadro's number.

208 Medium

208 of 436

Application +4 | -1

0.1 mole of a carbohydrate with empirical formula

\mathrm{C}{\mathrm{H}_2}\mathrm{O}

contains 1 g of hydrogen. What is its molecular formula?

View answer and explanation

Correct option: A

{\mathrm{C}_5}{\mathrm{H}_{10}}{\mathrm{O}_5}

Option A (

{\mathrm{C}_5}{\mathrm{H}_{10}}{\mathrm{O}_5}

) is correct because the molecular formula is determined by finding the number of moles of hydrogen and relating it to the empirical formula.

209 Hard

209 of 436

Analyzing +4 | -1

One mole of mixture of

\mathrm{N}_2, \mathrm{NO}_2

and

\mathrm{N}_2 \mathrm{O}_4

has a mean molar mass of

55.4 \mathrm{~g}.

On heating to a temperature, at which all the

\mathrm{N}_2 \mathrm{O}_4

is dissociated into

\mathrm{NO}_2,

the mean molar mass tends to a lower value of

39.6 \mathrm{~g}.

What is the mole ratio of

\mathrm{N}_2, \mathrm{NO}_2

and

\mathrm{N}_2 \mathrm{O}_4

in the original mixture?

View answer and explanation

Correct option: A

5: 1: 4

Option A (5: 1: 4) is correct because solving the system of equations derived from the given molar mass data yields the mole ratio of N2, NO2, and N2O4 as 5:1:4.

210 Hard

210 of 436

Application +4 | -1

What will be the minimum molar mass of a compound containing

0.39 \%

of nickel?

View answer and explanation

Correct option: B

15 kg

Option B (15 kg) is correct because it is calculated by assuming the compound contains at least one nickel atom and then determining the molar mass based on the given percentage.

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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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