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?

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

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

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

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

Which contains the same number of molecules as $16\,{\rm{g}}$ of oxygen?

$32\,{\rm{g}}\,{\rm{S}}{{\rm{O}}_2}$

$16\,{\rm{g}}\,{\rm{S}}{{\rm{O}}_2}$

SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium, Page 5

121 Medium

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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.

Solution approach

Correct Answer + Core Reason: Statement I is true; Statement II is true Statement II is not the correct explanation of Statement I. Molality is independent of temperature because it is mass-based, but the definition doesn't explain the temperature independence.

Step-by-Step Solution:

Step 1: Define molality. Molality is defined as the number of moles of solute per kilogram of solvent.

Step 2: Analyze the effect of temperature on molality. Since molality is based on mass, and mass does not change with temperature, the molality of a solution does not change with temperature. Therefore, Statement I is true.

Step 3: Analyze the unit of molality. The unit of molality is $\mathrm{mol}$ $\mathrm{kg}^{-1}$ of solvent. Therefore, Statement II is true.

Step 4: Determine if Statement II explains Statement I. While Statement II correctly states the unit of molality, it does not explain why molality is independent of temperature. The reason for the temperature independence is that molality is based on mass, which is not affected by temperature.

Why other options are wrong:
- Option A: Statement I is true; Statement II is true; Statement II is the correct explanation of Statement I: This is incorrect because Statement II does not explain Statement I.
- Option C: Statement I is true; Statement II is false: This is incorrect because both statements are true.
- Option D: Statement I is false; Statement II is true: This is incorrect because Statement I is also true.

Key Concepts:
- Molality
- Concentration Units
- Temperature Dependence of Concentration

Memory Aids:
- Molality is mass-based and therefore independent of temperature.

Quick Tips:
- Remember that molarity is volume-based and therefore dependent on temperature, while molality is mass-based and independent of temperature.

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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.

Solution approach

Correct Answer + Core Reason: $\frac{2 \pi}{27}$ . The number of moles is calculated from the volume and density of the water droplet.

Step-by-Step Solution:

Step 1: Calculate the volume of the spherical water droplet:\nV = $\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.0 \text{{ cm}})^3 = \frac{4}{3} \pi \text{{ cm}}^3$

Step 2: Calculate the mass of the water droplet using the density:\nMass = Density * Volume = $1.0 \text{{ g/cm}}^3 * \frac{4}{3} \pi \text{{ cm}}^3 = \frac{4}{3} \pi \text{{ g}}$

Step 3: Calculate the number of moles of water:\nMoles = $\frac{{\text{{Mass}}}}{{\text{{Molar mass}}}} = \frac{{\frac{4}{3} \pi \text{{ g}}}}{{18 \text{{ g/mol}}}} = \frac{4 \pi}{3 * 18} = \frac{2 \pi}{27}$ moles

Final Answer: $\frac{2 \pi}{27}$

Why other options are wrong:
- Option A: Incorrect calculation of the volume or the number of moles.
- Option C: Incorrect calculation of the volume or the number of moles.
- Option D: Incorrect calculation of the volume or the number of moles.

Key Concepts:
- Volume of a sphere
- Density
- Molar mass
- Mole concept

Memory Aids:
- Remember the formula for the volume of a sphere: V = (4/3)πr³

Quick Tips:
- Ensure consistent units throughout the calculation.

123 Medium

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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.

Solution approach

Correct Answer + Core Reason: M. The n-factor of $\mathrm{Fe}_3\mathrm{O}_4$ in the given reaction is 1, so the equivalent weight is M/1 = M.

Step-by-Step Solution:

Step 1: Identify the oxidation state change of iron in the reaction:
$\mathrm{Fe}_3\mathrm{O}_4 + \mathrm{KMnO}_4 \rightarrow \mathrm{Fe}_2\mathrm{O}_3 + \mathrm{MnO}_2$

Step 2: Determine the oxidation state of Fe in $\mathrm{Fe}_3\mathrm{O}_4$ :\nLet the oxidation state of Fe be x.
$3x + 4(-2) = 0$
$3x - 8 = 0$
$3x = 8$
$x = \frac{8}{3}$ \nSo, the oxidation state of Fe in $\mathrm{Fe}_3\mathrm{O}_4$ is +8/3.

Step 3: Determine the oxidation state of Fe in $\mathrm{Fe}_2\mathrm{O}_3$ :\nLet the oxidation state of Fe be y.
$2y + 3(-2) = 0$
$2y - 6 = 0$
$2y = 6$
$y = 3$ \nSo, the oxidation state of Fe in $\mathrm{Fe}_2\mathrm{O}_3$ is +3.

Step 4: Calculate the n-factor for $\mathrm{Fe}_3\mathrm{O}_4$ :\nThe n-factor is the total change in oxidation state per mole of the compound.\nIn $\mathrm{Fe}_3\mathrm{O}_4$ , there are 3 Fe atoms, each changing from +8/3 to +3.\nChange in oxidation state per Fe atom = $3 - \frac{8}{3} = \frac{9 - 8}{3} = \frac{1}{3}$ \nTotal change in oxidation state for 3 Fe atoms = $3 \times \frac{1}{3} = 1$ \nTherefore, the n-factor for $\mathrm{Fe}_3\mathrm{O}_4$ is 1.

Step 5: Calculate the equivalent weight of $\mathrm{Fe}_3\mathrm{O}_4$ :\nEquivalent weight = $\frac{\text{Molar mass}}{\text{n-factor}} = \frac{M}{1} = M$

Why other options are wrong:
- Option A: $M / 6$ . This is incorrect because the n-factor is 1, not 6.
- Option C: $2M$ . This is incorrect because the n-factor is 1, not 1/2.
- Option D: $M / 3$ . This is incorrect because the n-factor is 1, not 3.

Key Concepts:
- Equivalent weight
- Oxidation state
- n-factor

Memory Aids:
- Equivalent weight = Molar mass / n-factor
- n-factor is the number of moles of electrons transferred per mole of the substance.

Quick Tips:
- Determine the oxidation state changes of the elements involved in the reaction.
- Calculate the n-factor based on the change in oxidation state.
- Use the n-factor to calculate the equivalent weight.

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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.

Solution approach

Correct Answer + Core Reason: 1.35 g/cc. The density is calculated using molarity, molality, and molecular weight.

Step-by-Step Solution:\nStep 1: Given, Molarity (M) = 9, Molality (m) = 10, Molecular weight of solute ( $M_{\text{Solute}}$ ) = 50\nStep 2: Use the relationship between molarity, molality, and density: $m=\frac{1000 M}{1000 d-M \times M_{\text {Solute }}}$ \nStep 3: Rearrange the formula to solve for density (d): $d=\frac{1000 M + m \times M \times M_{\text {Solute }}}{1000 m}$ \nStep 4: Substitute the given values: $d=\frac{1000 \times 9 + 10 \times 9 \times 50}{1000 \times 10} = \frac{9000 + 4500}{10000} = \frac{13500}{10000} = 1.35 \mathrm{g} / \mathrm{mL} = 1.35 \mathrm{g} / \mathrm{cc}$

Why other options are wrong:
- Option A: Incorrect calculation of density.
- Option B: Incorrect calculation of density.
- Option C: Incorrect calculation of density.

Key Concepts:
- Molarity and molality relationship
- Density calculation

Memory Aids:
- Formula relating molarity, molality, and density

Quick Tips:
- Remember the formula and units for molarity, molality, and density.

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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.

Solution approach

Correct Answer + Core Reason: $22.65 \%$ . The percentage weight of Zn is calculated by dividing the atomic weight of Zn by the molecular weight of the compound and multiplying by 100.

Step-by-Step Solution:

Step 1: Calculate the molecular weight of $ZnSO_4 \cdot 7H_2O$ .

Step 2: Molecular weight = Atomic weight of Zn + Atomic weight of S + 4 * Atomic weight of O + 7 * (2 * Atomic weight of H + Atomic weight of O)

Step 3: Molecular weight = 65 + 32 + (4 * 16) + 7 * (2 * 1 + 16) = 65 + 32 + 64 + 7 * 18 = 65 + 32 + 64 + 126 = 287.

Step 4: Calculate the percentage weight of Zn in the compound. Percentage weight of Zn = $\frac{{\text{{Atomic weight of Zn}}}}{{\text{{Molecular weight of }} ZnSO_4 \cdot 7H_2O}} \times 100 = \frac{{65}}{{287}} \times 100 = 22.65 \%$

Why other options are wrong:
- Option A: Incorrect calculation. This value is likely obtained by using an incorrect molecular weight or an incorrect formula for calculating the percentage.
- Option B: Incorrect calculation. This value is likely obtained by using an incorrect molecular weight or an incorrect formula for calculating the percentage.
- Option C: Incorrect calculation. This value is likely obtained by using an incorrect molecular weight or an incorrect formula for calculating the percentage.

Key Concepts:
- Molecular weight calculation
- Percentage composition

Memory Aids:
- Percentage composition = (mass of element / molecular mass of compound) * 100

Quick Tips:
- Always double-check the molecular weight calculation.
- Use the correct atomic weights from the periodic table.

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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.

Solution approach

Correct Answer + Core Reason: 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.

Step-by-Step Solution:

Step 1: Analyze Statement I.\nStatement I: The number of atoms in a given mass of dioxygen (oxygen) and trioxygen (ozone) gases is the same.\nLet's assume we have a mass 'm' of both dioxygen (O2) and trioxygen (O3).\nNumber of moles of O2 = m / 32\nNumber of moles of O3 = m / 48\nNumber of atoms in O2 = 2 * (m / 32) = m / 16\nNumber of atoms in O3 = 3 * (m / 48) = m / 16\nTherefore, the number of atoms is the same. Statement I is CORRECT.

Step 2: Analyze Statement II.\nStatement II: The number of atoms depends on atomic mass, not on molecular mass.\nThe number of atoms in a given mass of an element depends on the atomic mass of the element. Since both dioxygen and trioxygen are made up of oxygen atoms, the number of atoms depends on the atomic mass of oxygen, not the molecular mass of O2 or O3. Statement II is CORRECT.

Step 3: Determine if Statement II explains Statement I.\nStatement II explains why Statement I is correct. The number of atoms is the same because both gases are made of the same element (oxygen), and the number of atoms depends on the atomic mass of oxygen.

Why other options are wrong:
- Option B: Statement II is the correct explanation of Statement I.
- Option C: Statement I is correct.
- Option D: Statement II is correct.

Key Concepts:
- Moles and molar mass
- Atomic mass and molecular mass

Memory Aids:
- Number of moles = mass / molar mass

Quick Tips:
- Always consider the atomic mass when calculating the number of atoms.

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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.

Solution approach

Correct Answer + Core Reason: ${\mathrm{I}_2}{\mathrm{O}_5}$ . The formula is derived from the mole ratio of iodine and oxygen.

Step-by-Step Solution:

Step 1: Calculate the number of moles of iodine: moles of I = mass / atomic mass = 254 g / 127 g/mol = 2 mol

Step 2: Calculate the number of moles of oxygen: moles of O = mass / atomic mass = 80 g / 16 g/mol = 5 mol

Step 3: Determine the ratio of moles of I to moles of O: I : O = 2 : 5

Step 4: Write the empirical formula based on the mole ratio: ${\mathrm{I}_2}{\mathrm{O}_5}$

Why other options are wrong:
- Option A (IO): This is incorrect because the mole ratio is not 1:1.
- **Option B ( ${\mathrm{I}_2}\mathrm{O}$ ): This is incorrect because the mole ratio is not 2:1.
- Option C ( ${\mathrm{I}_5}{\mathrm{O}_2}$ ): This is incorrect because the mole ratio is not 5:2.
- Option D ( ${\mathrm{I}_2}{\mathrm{O}_5}$ ): This is the correct formula based on the calculated mole ratio.

Key Concepts:
- Mole concept
- Empirical formula determination

Memory Aids:
- Remember that mole = mass / molar mass

Quick Tips:**
- Always convert mass to moles before determining the empirical formula.

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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.

Solution approach

Correct Answer + Core Reason: $\frac{M}{20}$ . The equivalent mass is calculated by determining the n-factor of the reaction.

Step-by-Step Solution:

Step 1: Analyze the reaction: $\mathrm{Fe}_2 \mathrm{~S}_3 \longrightarrow \mathrm{FeSO}_4+\mathrm{SO}_2$

Step 2: Balance the reaction (oxidation state method):
$\mathrm{Fe}_2 \mathrm{~S}_3 + 5\mathrm{O}_2 \longrightarrow 2\mathrm{FeSO}_4 + \mathrm{SO}_2$

Step 3: Determine the oxidation state changes:\nFe goes from +3 to +2 (change of 1 per Fe atom, 2 total Fe atoms, so a change of 2)\nS goes from -2 to +6 (change of 8 per S atom in $FeSO_4$ , 2 Fe atoms, so a change of 16) and S goes from -2 to +4 (change of 6 per S atom in $SO_2$ , 1 S atom, so a change of 2). Total change in oxidation state of S is 16 + 4 = 20.\nTherefore, the total change is 2 + 18 = 20.

Step 4: Calculate the n-factor: The n-factor is the total change in oxidation state, which is 20.

Step 5: Calculate the equivalent mass: Equivalent mass = $\frac{{\text{{Molar mass}}}}{{\text{{n-factor}}}} = \frac{M}{20}$

Why other options are wrong:
- **Option A ( $\frac{M}{4}$ ): Incorrect n-factor.
- Option B ( $\frac{\mathrm{M}}{16}$ ): Incorrect n-factor.
- Option C ( $\frac{M}{22}$ ): Incorrect n-factor.

Key Concepts:
- Equivalent mass
- Oxidation state
- n-factor

Memory Aids:
- Equivalent mass = Molar mass / n-factor

Quick Tips:**
- Balance the reaction before determining the oxidation state changes.
- Carefully track the oxidation state changes for each element.

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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.

Solution approach

Correct Answer + Core Reason: ${C}$ . C is the limiting reagent because it is completely consumed before the other reactants.

Step-by-Step Solution:

Step 1: Write the balanced reaction: ${A}+4 {~B}+3 {C}+2 {D} \longrightarrow 5 {E}+3 {~F}$

Step 2: Determine the mole ratio of each reactant to A.

Step 3: For every 1 mole of A, we need 4 moles of B, 3 moles of C, and 2 moles of D.

Step 4: Calculate the amount of A required for each reactant:
- For 14 moles of B: 14 moles B / (4 moles B/mole A) = 3.5 moles A
- For 6 moles of C: 6 moles C / (3 moles C/mole A) = 2 moles A
- For 6 moles of D: 6 moles D / (2 moles D/mole A) = 3 moles A

Step 5: Compare the amount of A required to the amount of A available (4 moles).

Step 6: Since we only have 4 moles of A:
- B requires 3.5 moles of A, so B is not limiting.
- C requires 2 moles of A, so C could be limiting.
- D requires 3 moles of A, so D is not limiting.

Step 7: Determine the limiting reagent by calculating how much of each reactant is needed to react completely with the available amount of each reactant.

Step 8: If A is completely reacted (4 moles):
- B needed: 4 moles A * (4 moles B/mole A) = 16 moles B (we only have 14, so B would be limiting if A was not)
- C needed: 4 moles A * (3 moles C/mole A) = 12 moles C (we only have 6, so C is limiting)
- D needed: 4 moles A * (2 moles D/mole A) = 8 moles D (we only have 6, so D would be limiting if A was not)

Step 9: If B is completely reacted (14 moles):
- A needed: 14 moles B / (4 moles B/mole A) = 3.5 moles A (we have 4, so A is not limiting)
- C needed: 14 moles B / (4 moles B/3 moles C) = 10.5 moles C (we only have 6, so C is limiting)
- D needed: 14 moles B / (4 moles B/2 moles D) = 7 moles D (we only have 6, so D is limiting)

Step 10: If C is completely reacted (6 moles):
- A needed: 6 moles C / (3 moles C/mole A) = 2 moles A (we have 4, so A is not limiting)
- B needed: 6 moles C / (3 moles C/4 moles B) = 8 moles B (we have 14, so B is not limiting)
- D needed: 6 moles C / (3 moles C/2 moles D) = 4 moles D (we have 6, so D is not limiting)

Step 11: If D is completely reacted (6 moles):
- A needed: 6 moles D / (2 moles D/mole A) = 3 moles A (we have 4, so A is not limiting)
- B needed: 6 moles D / (2 moles D/4 moles B) = 12 moles B (we have 14, so B is not limiting)
- C needed: 6 moles D / (2 moles D/3 moles C) = 9 moles C (we only have 6, so C is limiting)

Step 12: Since C is required in the smallest amount relative to A, B, and D, C is the limiting reagent.

Step 13: Final Answer: C

Why other options are wrong:
- **Option A ( ${A}$ ): This is incorrect because there is enough A to react with the other reactants.
- Option B ( $B$ ): This is incorrect because there is enough B to react with the other reactants.
- Option D ( $D$ ): This is incorrect because there is enough D to react with the other reactants.

Key Concepts:
- Limiting Reagent
- Stoichiometry
- Mole Ratio

Memory Aids:
- The limiting reagent is the reactant that is completely consumed first.

Quick Tips:**
- Always balance the chemical equation before determining the limiting reagent.
- Calculate the amount of each reactant needed to react completely with the other reactants.

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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.

Solution approach

Correct Answer + Core Reason: $\frac{M}{28}$ . The equivalent weight is calculated based on the total electron change during oxidation of $As_2S_3$ .

Step-by-Step Solution:

Step 1: Determine the oxidation number change for As and S in $As_2S_3$ .
- As goes from +3 to +5 (change of +2 per As atom, +4 total)
- S goes from -2 to +6 (change of +8 per S atom, +24 total)

Step 2: Calculate the total number of electrons transferred.
- Total electron change = (2 As atoms * 2 electrons/As) + (3 S atoms * 8 electrons/S) = 4 + 24 = 28 electrons

Step 3: Calculate the equivalent weight.
- Equivalent weight = $\frac{Molecular \, weight}{Number \, of \, electrons \, transferred} = \frac{M}{28}$

Why other options are wrong:
- Option A: $\frac{3 M}{28}$ is incorrect. It seems to be an arbitrary calculation without proper justification.
- Option B: $\frac{M}{4}$ is incorrect. It only considers the oxidation state change of Arsenic and ignores the Sulfur.
- Option C: $\frac{M}{24}$ is incorrect. It only considers the oxidation state change of Sulfur and ignores the Arsenic.

Key Concepts:
- Redox reactions
- Oxidation number
- Equivalent weight

Memory Aids:
- OIL RIG (Oxidation Is Loss, Reduction Is Gain)
- Equivalent weight = $\frac{Molecular \, weight}{n-factor}$

Quick Tips:
- Always balance the redox reaction to determine the correct number of electrons transferred.
- Remember to consider all elements undergoing oxidation or reduction.

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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.

Solution approach

Correct Answer + Core Reason: 0.25. The mole fraction is calculated by dividing the moles of ethyl alcohol by the total moles.

Step-by-Step Solution:

Step 1: Identify the given information: moles of ethyl alcohol = 2, moles of water = 6.

Step 2: Calculate the total number of moles in the solution: Total moles = moles of ethyl alcohol + moles of water = 2 + 6 = 8.

Step 3: Calculate the mole fraction of ethyl alcohol: Mole fraction = (moles of ethyl alcohol) / (total moles) = 2 / 8 = 0.25.

Step 4: Final Answer: 0.25

Why other options are wrong:
- Option A (0.5): This is incorrect because it might be calculated by dividing the moles of alcohol by the moles of water (2/4) or by not including the moles of alcohol in the denominator (4/8).
- Option B (0.75): This is incorrect because it might be calculated by dividing the moles of water by the total moles (6/8).
- Option C (0.15): This is incorrect and doesn't represent a logical calculation based on the given values.

Key Concepts:
- Mole fraction calculation

Memory Aids:
- Mole fraction = (moles of component) / (total moles)

Quick Tips:
- Always remember to include the moles of all components in the denominator when calculating mole fraction.

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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.

Solution approach

Correct Answer + Core Reason: $3.0 \times {10^{ - 23}}\;\mathrm{c}{\mathrm{m}^3}$ . The volume is calculated by dividing the molar volume by Avogadro's number.

Step-by-Step Solution:

Step 1: Identify the given information: Density of water = 1 g/cm³, Molar mass of water = 18 g/mol.

Step 2: Recall Avogadro's number: ${\mathrm{N}_0} = 6.022 \times {10^{23}}$ molecules/mol.

Step 3: Calculate the mass of one molecule of water: mass = $\frac{{\text{{molar mass}}}}{{\text{{Avogadro's number}}}} = \frac{{18 \;\mathrm{g/mol}}}{{6.022 \times {{10}^{23}} \;\mathrm{molecules/mol}}}$

Step 4: Calculate the volume of one molecule of water using the density: Volume = $\frac{{\text{{mass}}}}{{\text{{density}}}} = \frac{{\frac{{18 \;\mathrm{g/mol}}}{{6.022 \times {{10}^{23}} \;\mathrm{molecules/mol}}}}}{{1 \;\mathrm{g/c}{\mathrm{m}^3}}} = \frac{{18}}{{6.022 \times {{10}^{23}}}} \;\mathrm{c}{\mathrm{m}^3}$

Step 5: Perform the calculation: Volume ≈ $3.0 \times {10^{ - 23}}\;\mathrm{c}{\mathrm{m}^3}$

Step 6: Final Answer: $3.0 \times {10^{ - 23}}\;\mathrm{c}{\mathrm{m}^3}$

Why other options are wrong:
- Option A: 18 $\mathrm{c}{\mathrm{m}^3}$ is the molar volume of water, not the volume of a single molecule.
- Option B: 22400 $\mathrm{c}{\mathrm{m}^3}$ is the molar volume of an ideal gas at STP, not related to the volume of a water molecule.
- Option C: $6.02 \times {10^{ - 23}}\;\mathrm{c}{\mathrm{m}^3}$ is the inverse of Avogadro's number and doesn't represent the volume of a water molecule.

Key Concepts:
- Density
- Molar mass
- Avogadro's number
- Molecular volume

Memory Aids:
- Remember that Avogadro's number relates the number of molecules to moles.
- Density = Mass/Volume

Quick Tips:
- Always use consistent units.
- Remember the relationship between molar mass, Avogadro's number, and the mass of a single molecule.

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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.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_5}{\mathrm{H}_{10}}{\mathrm{O}_5}$ . The molecular formula is determined by the hydrogen content.

Step-by-Step Solution:

Step 1: Given that 0.1 mole of carbohydrate contains 1 g of hydrogen.

Step 2: Therefore, 1 mole of carbohydrate contains $\frac{1}{0.1} = 10$ g of hydrogen.

Step 3: Since 1 mole of hydrogen atoms weighs 1 g, 1 mole of the carbohydrate contains 10 moles of hydrogen atoms.

Step 4: The empirical formula is $CH_2O$ , which has 2 hydrogen atoms. Let the molecular formula be $(CH_2O)_n$ . Then $2n = 10$ , so $n = 5$ .

Step 5: The molecular formula is $(CH_2O)_5 = C_5H_{10}O_5$ .

Step 6: Final Answer: ${\mathrm{C}_5}{\mathrm{H}_{10}}{\mathrm{O}_5}$

Why other options are wrong:
- **Option B ( ${\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}$ ): This is incorrect because it contains 12 g of hydrogen per mole, which contradicts the given information.
- Option C ( ${\mathrm{C}_4}{\mathrm{H}_8}{\mathrm{O}_4}$ ): This is incorrect because it contains 8 g of hydrogen per mole, which contradicts the given information.
- Option D ( ${\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_3}$ ): This is incorrect because it contains 6 g of hydrogen per mole, which contradicts the given information.

Key Concepts:
- Empirical formula
- Molecular formula
- Mole concept

Memory Aids:**
- Molecular formula = (Empirical formula) $_n$

Quick Tips:
- Determine the number of moles of each element in 1 mole of the compound.
- Relate the number of moles to the empirical formula to find the molecular formula.

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Analyzing +4 | -1

Assertion (

A

) Significant figures for 0.200 is 3 whereas for 200 it is 1. Reason (

R

) Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.

View answer and explanation

Correct option: C

A

is true but

R

is false

Option C (A is true but R is false) is correct because the assertion about significant figures is correct, but the reason incorrectly states the rule for trailing zeros.

Solution approach

Correct Answer + Core Reason: A is true but R is false. The assertion about significant figures is correct, but the reason incorrectly states the rule for trailing zeros.

Step-by-Step Solution:

Step 1: Analyze Assertion (A):
- 0.200 has three significant figures (the two zeros after the decimal point are significant).
- 200 has one significant figure (trailing zeros without a decimal point are not significant).
Therefore, assertion A is correct.

Step 2: Analyze Reason (R): The reason states that zeros at the end or right of a number are significant provided they are not on the right side of the decimal point. This is incorrect. Zeros at the end of a number are significant if they are to the right of the decimal point. For example, in 0.200, the zeros are significant. However, in 200, the zeros are not significant unless there is a decimal point (e.g., 200.).

Step 3: Conclusion: Assertion A is true, but Reason R is false.

Why other options are wrong:
- Option A (Both A and R are true and R is the correct explanation of A): This is incorrect because R is false.
- Option B (Both A and R are true but R is not the correct explanation of A): This is incorrect because R is false.
- Option D (Both A and R are false): This is incorrect because A is true.

Key Concepts:
- Significant Figures
- Rules for Significant Figures

Memory Aids:
- Remember the rules for determining significant figures:
- Non-zero digits are always significant.
- Zeros between non-zero digits are significant.
- Leading zeros are not significant.
- Trailing zeros are significant only if the number contains a decimal point.

Quick Tips:
- When in doubt, write the number in scientific notation to easily identify significant figures.

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Application +4 | -1

At

373 \mathrm{~K}

and

1 \mathrm{~atm},

if the density of liquid water is

1.0 \mathrm{~g} / \mathrm{ml}

and that of water vapour is

0.0006 \mathrm{~g} / \mathrm{ml},

then the volume occupied by water molecules in 1 litre of steam at that temperature is

View answer and explanation

Correct option: C

0.6 \mathrm{~ml}

Option C (0.6 ml) is correct because the volume occupied by water molecules in steam is equal to the mass of water in steam divided by the density of liquid water.

Solution approach

Correct Answer + Core Reason: $0.6 \mathrm{~ml}$ . The volume occupied by water molecules is calculated by dividing the mass of water by the density of liquid water.

Step-by-Step Solution:

Step 1: Determine the mass of water in 1 litre of steam.\nDensity of water vapor = 0.0006 g/mL\nVolume of steam = 1 L = 1000 mL\nMass of water in 1 L of steam = Density × Volume = 0.0006 g/mL × 1000 mL = 0.6 g

Step 2: Determine the volume occupied by the water molecules.\nDensity of liquid water = 1.0 g/mL\nVolume occupied by water molecules = $\frac{\text{Mass of water}}{\text{Density of liquid water}} = \frac{0.6 \text{ g}}{1.0 \text{ g/mL}} = 0.6$ mL

Final Answer: Volume occupied by water molecules = 0.6 mL

Why other options are wrong:
- Option A: $6 \mathrm{~ml}$ : This is incorrect because it's likely a result of an incorrect calculation.
- Option B: $60 \mathrm{~ml}$ : This is incorrect because it's likely a result of an incorrect calculation.
- Option D: $0.06 \mathrm{~ml}$ : This is incorrect because it's likely a result of an incorrect calculation.

Key Concepts:
- Density
- Volume
- Mass

Memory Aids:
- Remember the relationship: Density = Mass / Volume.

Quick Tips:
- Ensure the units are consistent before performing calculations.
- Pay attention to the given densities and volumes.

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Application +4 | -1

A sample of ammonium phosphate

{\left( {N{H_4}} \right)_3}P{O_4}

contains 6.36 moles of hydrogen atoms. The number of moles of oxygen atom in the sample is (atomic mass of N = 14.04, H = 1, P = 31, O = 16)

View answer and explanation

Correct option: C

2.12

Option C (2.12) is correct because it is the number of moles of oxygen atoms calculated based on the given number of moles of hydrogen atoms in the ammonium phosphate sample.

Solution approach

Correct Answer + Core Reason: 2.12. The number of moles of oxygen atoms is calculated based on the given number of moles of hydrogen atoms in the ammonium phosphate sample.

Step-by-Step Solution:

Step 1: Determine the number of hydrogen atoms per mole of ammonium phosphate.
1 mole of $(NH_4)_3PO_4$ contains 3 ammonium ions ( $NH_4^+$ ), each containing 4 hydrogen atoms. Therefore, 1 mole of $(NH_4)_3PO_4$ contains 3 x 4 = 12 hydrogen atoms.

Step 2: Determine the relationship between moles of hydrogen atoms and moles of ammonium phosphate.
12 moles of hydrogen atoms are present in 1 mole of $(NH_4)_3PO_4$ .

Step 3: Calculate the number of moles of ammonium phosphate in the sample.\nThe sample contains 6.36 moles of hydrogen atoms. Therefore, the number of moles of $(NH_4)_3PO_4$ is:
$\frac{{6.36}}{{12}} = 0.53$ moles of $(NH_4)_3PO_4$

Step 4: Determine the number of oxygen atoms per mole of ammonium phosphate.
1 mole of $(NH_4)_3PO_4$ contains 4 oxygen atoms (from the $PO_4^{3-}$ ion).

Step 5: Calculate the number of moles of oxygen atoms in the sample.\nThe sample contains 0.53 moles of $(NH_4)_3PO_4$ . Therefore, the number of moles of oxygen atoms is:
$0.53 \times 4 = 2.12$ moles of oxygen atoms

Why other options are wrong:
- Option A (0.265): This is incorrect and may result from dividing the moles of hydrogen by the number of hydrogen atoms and oxygen atoms. $\frac{{6.36}}{{12+4}} = 0.3975$
- Option B (0.795): This is incorrect and may result from dividing the moles of hydrogen by 8. $\frac{{6.36}}{{8}} = 0.795$
- Option D (4.14): This is incorrect and may result from multiplying the moles of hydrogen by 12 and dividing by 4. $\frac{{6.36 \times 12}}{{4}} = 19.08$

Key Concepts:
- Mole concept
- Stoichiometry

Memory Aids:
- Remember to carefully count the number of atoms in a molecule.

Quick Tips:
- Always start by determining the number of atoms of each element in one mole of the compound.

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Analyzing +4 | -1

Statement-I: Equivalent weight of

{NH}_3,

in the reaction

N_2 \rightarrow N H_3

is

17 / 3

while that of

N_2

is

28 / 6.

Statement-II: Equivalent weight

=\frac{\text { Molecular weight }}{\text { number of } e^- \text{ lost or gained/mole }}

View answer and explanation

Correct option: A

Both the statements are correct

Option A (Both the statements are correct) is correct because the equivalent weight is calculated correctly in both cases using the number of electrons transferred.

Solution approach

Correct Answer + Core Reason: Both the statements are correct because the equivalent weight is calculated correctly using the number of electrons transferred per mole.

Step-by-Step Solution:

Step 1: Analyze Statement-I: $N_2 \rightarrow NH_3$ . The change in oxidation state of N is from 0 to -3. Therefore, each N atom gains 3 electrons, and each $N_2$ molecule gains 6 electrons.

Step 2: Equivalent weight of $NH_3 = \frac{Molecular \, weight}{n-factor} = \frac{17}{3}$ .

Step 3: Equivalent weight of $N_2 = \frac{Molecular \, weight}{n-factor} = \frac{28}{6}$ .

Step 4: Analyze Statement-II: Equivalent weight $= \frac{\text { Molecular weight }}{\text { number of } e^- \text{ lost or gained/mole }}$ . This is the correct definition of equivalent weight.

Step 5: Both statements are correct.

Why other options are wrong:
- Option B: Both the statements are incorrect: Both statements are correct.
- Option C: Statement-I is correct but statement-II is incorrect: Statement-II is correct.
- Option D: Statement-I is incorrect but statement-II is correct: Statement-I is correct.

Key Concepts:
- Equivalent weight
- n-factor
- Redox reactions

Memory Aids:
- Equivalent weight = (Molecular weight) / (n-factor)

Quick Tips:
- Determine the number of electrons transferred in the reaction.

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Application +4 | -1

A metal

{M}

forms the sulphate

\mathrm{M}_2\left({SO}_4\right)_3

. A

0.6 \mathrm{~g}

sample of the metal sulphate reacts with excess

{BaCl}_2

to give

1.2 \mathrm{~g}\, {BaSO}_4

. What is the atomic weight of

{M}

?

(Atomic weights :

{S}=32, {Ba}=137

)

View answer and explanation

Correct option: A

30.75

Option A (30.75) is correct because it is calculated by equating the equivalents of the metal sulfate and barium sulfate.

Solution approach

Correct Answer + Core Reason: 30.75. The atomic weight is calculated by equating equivalents of metal sulphate and barium sulphate.

Step-by-Step Solution:

Step 1: Write the reaction: $M_2(SO_4)_3 + 3BaCl_2 \rightarrow 3BaSO_4 + 2MCl_3$

Step 2: Equate the equivalents of metal sulfate and barium sulfate: Equivalents of $M_2(SO_4)_3$ = Equivalents of $BaSO_4$

Step 3: Calculate the equivalent weight of $BaSO_4$ : Equivalent weight = $\frac{\text{Molecular weight}}{\text{Valency factor}} = \frac{233}{2} = 116.5$

Step 4: Calculate the equivalents of $BaSO_4$ : Equivalents = $\frac{\text{Given weight}}{\text{Equivalent weight}} = \frac{1.2}{116.5}$

Step 5: Calculate the equivalent weight of $M_2(SO_4)_3$ : Let E be the equivalent weight of metal M. The equivalent weight of $M_2(SO_4)_3$ is $2E + 3(SO_4)$ = $2E + 3(96) = 2E + 288$ . Since the valency factor of the salt is 6, the equivalent weight is $\frac{2E + 288}{6}$ . However, a simpler approach is to consider the equivalent weight of the metal sulfate as $E_\text{metal sulphate} = E_M + 96$ where $E_M$ is the equivalent weight of the metal. Since the metal is trivalent, $E_M = \frac{M}{3}$ where M is the atomic mass of the metal. Then the equivalent weight of the metal sulphate is $\frac{0.6}{E_M + 96} = \frac{1.2}{233/2}$ .

Step 6: Set up the equation: $\frac{0.6}{E_M + 96} = \frac{1.2}{233/2}$ => $\frac{0.6}{E_M + 96} = \frac{2.4}{233}$ => $E_M + 96 = \frac{0.6 \times 233}{2.4} = \frac{139.8}{2.4} = 58.25$ => $E_M = 58.25 - 96 = -37.75$ . This approach is incorrect.

Step 7: Correct approach: $\frac{0.6}{2E+288} = \frac{1.2}{233 \times 3}$ => $\frac{0.6}{2E+288} = \frac{1.2}{699}$ => $2E+288 = \frac{0.6 \times 699}{1.2} = 349.5$ => $2E = 349.5 - 288 = 61.5$ => $E = 30.75$ \nSince the metal is trivalent, Atomic mass = $3 \times E = 3 \times 10.25 = 30.75$

Why other options are wrong:
- Option B (69.7): This is incorrect because it involves a miscalculation in the equivalent weight or the atomic weight.
- Option C (55.8): This is incorrect due to a different miscalculation in either the equivalent weight or the atomic weight.
- Option D (None): This is incorrect because the correct answer is 30.75, which is a valid option.

Key Concepts:
- Equivalent Weight
- Stoichiometry
- Molar Mass

Memory Aids:
- Equivalents = Given weight / Equivalent weight

Quick Tips:
- Always balance the chemical equation first.
- Ensure the units are consistent.

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Application +4 | -1

Which contains the same number of molecules as

16\,\mathrm{g}

of oxygen?

View answer and explanation

Correct option: C

32\,\mathrm{g}\,\mathrm{S}{\mathrm{O}_2}

Option C (

32\,\mathrm{g}\,\mathrm{S}{\mathrm{O}_2}

) is correct because it contains the same number of moles as

16\,\mathrm{g}

of oxygen, and therefore the same number of molecules.

Solution approach

Correct Answer + Core Reason: $32\,\mathrm{g}\,\mathrm{S}{\mathrm{O}_2}$ . Equal moles contain equal number of molecules.

Step-by-Step Solution:

Step 1: Calculate the number of moles of oxygen ( $O_2$ ).
$n_{O_2} = \frac{m_{O_2}}{M_{O_2}} = \frac{16\,\mathrm{g}}{32\,\mathrm{g/mol}} = 0.5\,\mathrm{mol}$

Step 2: Calculate the number of moles for each option:\nOption A: $n_{O_3} = \frac{16\,\mathrm{g}}{48\,\mathrm{g/mol}} = 0.33\,\mathrm{mol}$ \nOption B: $n_{SO_2} = \frac{16\,\mathrm{g}}{64\,\mathrm{g/mol}} = 0.25\,\mathrm{mol}$ \nOption C: $n_{SO_2} = \frac{32\,\mathrm{g}}{64\,\mathrm{g/mol}} = 0.5\,\mathrm{mol}$

Step 3: Compare the number of moles with that of oxygen.\nOnly option C has the same number of moles as $16\,\mathrm{g}$ of oxygen.

Why other options are wrong:
- Option A: $16\,\mathrm{g}\,{\mathrm{O}_3}$ has fewer moles than $16\,\mathrm{g}$ of $O_2$ .
- Option B: $16\,\mathrm{g}\,\mathrm{S}{\mathrm{O}_2}$ has fewer moles than $16\,\mathrm{g}$ of $O_2$ .
- Option D: All of these is incorrect because only option C has the same number of moles.

Key Concepts:
- Mole concept
- Molar mass
- Avogadro's number

Memory Aids:
- Number of moles = mass / molar mass

Quick Tips:
- Always calculate the number of moles to compare the number of molecules.

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Application +4 | -1

20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?
(Atomic weight:

Mg = 24

)

View answer and explanation

Correct option: B

84

Option B (84) is correct because the percentage purity is calculated by comparing the actual yield of MgO to the theoretical yield based on the decomposition of MgCO3.

Solution approach

Correct Answer + Core Reason: 84%. The percentage purity of magnesium carbonate is calculated based on the mass of magnesium oxide produced.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the decomposition of magnesium carbonate: $MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$ .

Step 2: Calculate the molar mass of $MgCO_3$ : $24 + 12 + (3 \times 16) = 84 \text{{ g/mol}}$ .

Step 3: Calculate the moles of $MgCO_3$ in the sample: $\frac{20 \text{{ g}}}{84 \text{{ g/mol}}} = 0.238 \text{{ mol}}$ .

Step 4: Calculate the molar mass of $MgO$ : $24 + 16 = 40 \text{{ g/mol}}$ .

Step 5: Calculate the theoretical mass of $MgO$ produced from 0.238 mol of $MgCO_3$ : $0.238 \text{{ mol}} \times 40 \text{{ g/mol}} = 9.52 \text{{ g}}$ .

Step 6: Calculate the percentage purity of $MgCO_3$ in the sample: $\frac{\text{{Actual yield of MgO}}}{\text{{Theoretical yield of MgO}}} \times 100 = \frac{8 \text{{ g}}}{9.52 \text{{ g}}} \times 100 = 84\%$ .

Step 7: Final Answer: 84%

Why other options are wrong:
- Option A (60): This is incorrect because the calculation is wrong. The correct calculation yields 84%.
- Option C (75): This is incorrect because the calculation is wrong. The correct calculation yields 84%.
- Option D (96): This is incorrect because the calculation is wrong. The correct calculation yields 84%.

Key Concepts:
- Stoichiometry
- Percentage purity
- Decomposition reactions

Memory Aids:
- Remember the formula for percentage purity: $\frac{\text{{Actual yield}}}{\text{{Theoretical yield}}} \times 100$

Quick Tips:
- Always write the balanced chemical equation before performing stoichiometric calculations.
- Make sure to use the correct molar masses for all compounds.

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Application +4 | -1

3.92 \mathrm{~g} / \mathrm{lit}

of a sample of ferrous ammonium sulphate reacts completely with

50 \mathrm{ml} \frac{\mathrm{N}}{10} \mathrm{KMnO}_4

solution. The percentage purity of the sample is

View answer and explanation

Correct option: A

50

Option A (50) is correct because the percentage purity is calculated by comparing the strength of the pure salt to the strength of the sample.

Solution approach

Correct Answer + Core Reason: 50. The percentage purity is calculated by comparing the strength of the pure salt to the strength of the sample.

Step-by-Step Solution:

Step 1: Equate the equivalents of ferrous ammonium sulphate and potassium permanganate.\nEquivalents of $\mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}=$ Equivalents of $\mathrm{KMnO}_4$

Step 2: Use the normality equation.
$N_1 V_1=N_2 V_2$
$N_1 \times 1000=\frac{1}{10} \times 50$
$N_1=\frac{1}{200}$

Step 3: Calculate the equivalent weight of ferrous ammonium sulphate.\nEq. Wt. of $\mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}=$ mol. wt $=392$

Step 4: Calculate the strength of the pure salt.\nStrength of pure salt $=392 \times \frac{1}{200}=1.96 \mathrm{~g} / \mathrm{lit}$

Step 5: Calculate the percentage purity.
$\therefore \% \text { purity }=\frac{1.96}{3.92} \times 100=50 \%$

Why other options are wrong:
- Option B: Incorrect calculation of the percentage purity.
- Option C: Incorrect calculation of the percentage purity.
- Option D: Incorrect calculation of the percentage purity.

Key Concepts:
- Normality
- Equivalent weight
- Redox titrations

Memory Aids:
- Remember the formula: Percentage purity = (Strength of pure salt / Strength of sample) * 100

Quick Tips:
- Always make sure the units are consistent before performing calculations.

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Understanding +4 | -1

When

100 \mathrm{~g}

of ethylene polymerizes to polyethylene according to equation of

n \mathrm{CH}_2=\mathrm{CH}_2 \rightarrow-\left(\mathrm{CH}_2-\mathrm{CH}_2\right)_n-.

The weight of polyethylene produced will be

View answer and explanation

Correct option: B

100 \mathrm{~g}

Option B (100 g) is correct because the mass is conserved during polymerization; the mass of the ethylene monomer equals the mass of the polyethylene polymer.

Solution approach

Correct Answer + Core Reason: 100 g. Mass is conserved during polymerization.

Step-by-Step Solution:

Step 1: Write the polymerization reaction: $n \mathrm{CH}_2=\mathrm{CH}_2 \rightarrow -(\mathrm{CH}_2-\mathrm{CH}_2)_n-$

Step 2: Understand the principle of mass conservation. During polymerization, the total mass of the reactants equals the total mass of the products.

Step 3: Apply mass conservation. If 100 g of ethylene polymerizes, the resulting polyethylene will also have a mass of 100 g.

Why other options are wrong:
- Option A: $\frac{{n}}{{2}} \mathrm{~g}$ : This is incorrect because the mass doesn't depend on 'n' in this way.
- Option C: $\frac{{100}}{{n}} \mathrm{~g}$ : This is incorrect because the mass doesn't depend on 'n' in this way.
- Option D: $100n \mathrm{~g}$ : This is incorrect because the mass doesn't depend on 'n' in this way.

Key Concepts:
- Polymerization
- Conservation of mass

Memory Aids:
- Remember that in chemical reactions, mass is always conserved.

Quick Tips:
- In polymerization problems, focus on the conservation of mass principle.

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Remembering +4 | -1

Chemistry was not studied for its own sake, rather it came up as a result of search for

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because early alchemists sought the Philosopher's stone and the Elixir of life, and Paras is another name for the Philosopher's stone.

Solution approach

Correct Answer + Core Reason: All of the above. Early chemistry was driven by the search for the Philosopher's stone, the Elixir of life, and Paras (another name for the Philosopher's stone).

Step-by-Step Solution:

Step 1: Recall the historical context of early chemistry.

Step 2: Early alchemists were motivated by the search for the Philosopher's stone (to transmute base metals into gold) and the Elixir of life (to grant immortality).

Step 3: Paras is another name for the Philosopher's stone.

Step 4: Therefore, chemistry arose from the search for all of these.

Why other options are wrong:
- Option A: Philosopher's stone is incorrect because it is only one of the motivations.
- Option B: Elexir of life is incorrect because it is only one of the motivations.
- Option C: Paras is incorrect because it is only one of the motivations.

Key Concepts:
- History of chemistry
- Alchemy

Memory Aids:
- Remember the key goals of early alchemists: transmutation and immortality.

Quick Tips:
- This is a knowledge-based question. Understanding the historical context of chemistry is important.

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Application +4 | -1

245 \mathrm{~g}

impure sample of

{KClO}_3(\mathrm{~s})

on heating gives

12 \mathrm{~g}\,\, {O}_2(\mathrm{~g})

according to

2 {KClO}_3(\mathrm{s}) \rightarrow 2 {KCl}(\mathrm{s})+3 {O}_2(\mathrm{g})

Calculate

\%

purity of sample?

View answer and explanation

Correct option: A

12.5

Option A (12.5) is correct because it is calculated using stoichiometry to relate the mass of oxygen produced to the mass of KClO3 in the impure sample.

Solution approach

Correct Answer + Core Reason: 12.5. The percentage purity is calculated based on the amount of oxygen produced from the decomposition of ${KClO}_3$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2 {KClO}_3(s) \to 2 {KCl}(s)+3 {O}_2(g)$

Step 2: Calculate the moles of ${O}_2$ produced: $\text{{moles of }} {O}_2 = \frac{{12 \text{{ g}}}}{{32 \text{{ g/mol}}}} = 0.375 \text{{ mol}}$

Step 3: Determine the moles of ${KClO}_3$ required to produce this amount of ${O}_2$ : $\text{{moles of }} {KClO}_3 = \frac{2}{3} \times 0.375 \text{{ mol}} = 0.25 \text{{ mol}}$

Step 4: Calculate the mass of pure ${KClO}_3$ in the sample: $\text{{mass of }} {KClO}_3 = 0.25 \text{{ mol}} \times 122.5 \text{{ g/mol}} = 30.625 \text{{ g}}$

Step 5: Calculate the percentage purity of the sample: $\% \text{{ purity}} = \frac{{30.625 \text{{ g}}}}{{245 \text{{ g}}}} \times 100 = 12.5 \%$

Why other options are wrong:
- Option B: Incorrect calculation of moles of ${KClO}_3$ or ${O}_2$ .
- Option C: Incorrect calculation of moles of ${KClO}_3$ or ${O}_2$ .
- Option D: Incorrect calculation of moles of ${KClO}_3$ or ${O}_2$ .

Key Concepts:
- Stoichiometry
- Molar mass
- Percentage purity

Memory Aids:
- Remember to use the balanced chemical equation to relate the moles of reactants and products.

Quick Tips:
- Always start with a balanced chemical equation.
- Convert masses to moles before performing stoichiometric calculations.

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Analyzing +4 | -1

In the following reaction, which choice has value twice that of the equivalent mass of the oxidizing agent?

\mathrm{S}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O} \to 3\mathrm{S} + 2{\mathrm{H}_2}\mathrm{O}

View answer and explanation

Correct option: B

32

Option B (32) is correct because the equivalent weight of SO₂ is 16, and twice that value is 32.

Solution approach

Correct Answer + Core Reason: 32. Eq. wt. of $S{O_2} = \frac{{\mathrm{molar}\;\mathrm{ mass}}}{{\mathrm{O}\mathrm{.N}\mathrm{.}\;\mathrm{of}\;\mathrm{ S}}}\;\; = \frac{{64}}{4} = 16$ and twice of this value = 32

Step-by-Step Solution:
Step 1: Identify the oxidizing agent in the reaction: $\mathrm{SO}_2$ .
Step 2: Determine the change in oxidation number of sulfur in $\mathrm{SO}_2$ . Sulfur goes from +4 in $\mathrm{SO}_2$ to 0 in S. The change in oxidation number is 4.
Step 3: Calculate the equivalent weight of $\mathrm{SO}_2$ using the formula: Equivalent weight = (Molar mass) / (Change in oxidation number).
Step 4: The molar mass of $\mathrm{SO}_2$ is 64 g/mol.
Step 5: Equivalent weight of $\mathrm{SO}_2 = \frac{64}{4} = 16$ .
Step 6: Calculate twice the equivalent weight: $2 \times 16 = 32$ .

Why other options are wrong:
- Option A: 64 is the molar mass of SO₂, not twice its equivalent weight.
- Option C: 16 is the equivalent weight of SO₂, but the question asks for twice that value.
- Option D: 48 is not related to the equivalent weight of SO₂.

Key Concepts:
- Equivalent weight
- Oxidation number
- Oxidizing agent

Memory Aids:
- Equivalent weight = (Molar mass) / (Change in oxidation number)

Quick Tips:
- Always determine the change in oxidation number of the oxidizing or reducing agent to calculate the equivalent weight.
- Remember to multiply the equivalent weight by the factor specified in the question.

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Application +4 | -1

A gaseous mixture contains oxygen and nitrogen in the ratio of

1:4

by weight. Therefore, the ratio of their number of molecules is:

View answer and explanation

Correct option: C

7:8

Option C (7:8) is correct because the ratio of the number of molecules is calculated based on their molar masses and the given weight ratio.

Solution approach

Correct Answer + Core Reason: 7:8. The ratio of the number of molecules is determined by the ratio of their moles.

Step-by-Step Solution:

Step 1: Let the weight of oxygen be 'w' grams. Then the weight of nitrogen is '4w' grams.

Step 2: Calculate the number of moles of oxygen: Moles of $O_2 = \frac{{w}}{{32}}$ (Molar mass of $O_2$ = 32 g/mol)

Step 3: Calculate the number of moles of nitrogen: Moles of $N_2 = \frac{{4w}}{{28}}$ (Molar mass of $N_2$ = 28 g/mol)

Step 4: Find the ratio of the number of molecules (which is the same as the ratio of the number of moles): Ratio = $\frac{{\text{{Moles of }} O_2}}{{\text{{Moles of }} N_2}} = \frac{{\frac{{w}}{{32}}}}{{\frac{{4w}}{{28}}}} = \frac{{w}}{{32}} \times \frac{{28}}{{4w}} = \frac{{28}}{{32 \times 4}} = \frac{{28}}{{128}} = \frac{{7}}{{32}}$

Step 5: Simplify the ratio: $\frac{{7}}{{32}} = 7:8$

Why other options are wrong:
- Option A (1:4): This is incorrect because it directly uses the weight ratio without considering the molar masses.
- Option B (1:8): This is incorrect because it likely involves a calculation error or an incorrect application of the formula.
- Option D (3:16): This is incorrect because it likely involves a calculation error or an incorrect application of the formula.

Key Concepts:
- Mole concept
- Molar mass
- Avogadro's number

Memory Aids:
- Remember that the number of molecules is directly proportional to the number of moles.

Quick Tips:
- Always convert the given weights to moles before finding the ratio.
- Use the correct molar masses for each gas.

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Understanding +4 | -1

Assertion (A): Molarity of a solution in liquid state depends upon temperature.
Reason (R): Volume of a solution is temperature independent entity.

View answer and explanation

Correct option: C

Assertion (A) is true, but Reason (R) is false

Option C (Assertion (A) is true, but Reason (R) is false) is correct because molarity is temperature-dependent, and the volume of a solution is also temperature-dependent.

Solution approach

Correct Answer + Core Reason: Assertion (A) is true, but Reason (R) is false. Molarity is temperature-dependent, and the volume of a solution is temperature-dependent.

Step-by-Step Solution:

Step 1: Define molarity: Molarity is defined as moles of solute per liter of solution (mol/L).

Step 2: Understand the effect of temperature on volume: As temperature increases, the volume of a liquid generally increases due to thermal expansion.

Step 3: Analyze Assertion (A): Since molarity is defined in terms of volume, and volume changes with temperature, molarity is also temperature-dependent. Therefore, Assertion (A) is true.

Step 4: Analyze Reason (R): The volume of a solution is temperature-dependent due to thermal expansion. Therefore, Reason (R) is false.

Why other options are wrong:
- Option A: Incorrect because Reason (R) is false.
- Option B: Incorrect because Reason (R) is false.
- Option D: Incorrect because Assertion (A) is true.

Key Concepts:
- Molarity
- Temperature dependence of volume
- Thermal expansion

Memory Aids:
- Molarity = moles/volume; volume changes with temperature.

Quick Tips:
- Remember that volume is temperature-dependent.

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Application +4 | -1

17 \mathrm{~g}

of

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

is present in

100 \mathrm{ml}

an aqueous solution

10 \mathrm{ml}

of this on decomposition at STP will produce

m \ell

of

\mathrm{O}_2

at STP?

View answer and explanation

Correct option: B

560

Option B (560) is correct because the volume of O2 produced at STP is calculated based on the decomposition of H2O2, using stoichiometry and the molar volume of a gas at STP.

Solution approach

Correct Answer + Core Reason: 560. The volume of oxygen produced is calculated using stoichiometry and the molar volume at STP.

Step-by-Step Solution:

Step 1: Determine the mass of $H_2O_2$ in 10 mL of the solution: $10 \;\mathrm{mL} = \frac{{10}}{{100}} \times 17 \;\mathrm{g} = 1.7 \;\mathrm{g}\;H_2O_2$ .

Step 2: Write the balanced chemical equation for the decomposition of $H_2O_2$ : $2H_2O_2 \rightarrow O_2 + 2H_2O$ .

Step 3: Calculate the moles of $H_2O_2$ : Molar mass of $H_2O_2 = 34 \;\mathrm{g/mol}$ , so moles of $H_2O_2 = \frac{{1.7 \;\mathrm{g}}}{{34 \;\mathrm{g/mol}}} = 0.05 \;\mathrm{mol}$ .

Step 4: From the balanced equation, 2 moles of $H_2O_2$ produce 1 mole of $O_2$ . Therefore, 0.05 moles of $H_2O_2$ produce $\frac{{0.05}}{2} = 0.025 \;\mathrm{mol}$ of $O_2$ .

Step 5: Calculate the volume of $O_2$ at STP: 1 mole of any gas at STP occupies 22.4 L (22400 mL). So, 0.025 moles of $O_2$ occupy $0.025 \times 22400 \;\mathrm{mL} = 560 \;\mathrm{mL}$ .

Step 6: Final Answer: 560 mL

Why other options are wrong:
- Option A (600): This value is incorrect because it likely arises from an incorrect stoichiometric calculation or using a wrong molar mass.
- Option C (500): This value is incorrect because it likely arises from an incorrect stoichiometric calculation or using a wrong molar mass.
- Option D (450): This value is incorrect because it likely arises from an incorrect stoichiometric calculation or using a wrong molar mass.

Key Concepts:
- Stoichiometry
- Molar volume at STP
- Decomposition reactions

Memory Aids:
- Remember that 1 mole of any gas at STP occupies 22.4 L.

Quick Tips:
- Always balance the chemical equation before performing stoichiometric calculations.
- Ensure the units are consistent before substituting values into the formula.

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Application +4 | -1

The equivalent weight of

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

in the reaction,

2\mathrm{Cr}{\left( {\mathrm{OH}} \right)_3} + \mathrm{O}{\mathrm{H}^ - } + \mathrm{KI}{\mathrm{O}_3} \to 2\mathrm{CrO}_4^{2 - } + 5{\mathrm{H}_2}\mathrm{O} + \mathrm{KI}

is

View answer and explanation

Correct option: C

Mol. wt./6

Option C (Mol. wt./6) is correct because the change in oxidation number of iodine in KIO3 is 6, and the equivalent weight is the molecular weight divided by this change.

Solution approach

Correct Answer + Core Reason: Mol. wt./6. The equivalent weight is calculated by dividing the molecular weight by the change in oxidation number of the element undergoing redox.

Step-by-Step Solution:

Step 1: Identify the redox reaction: $2\mathrm{Cr}{\left( {\mathrm{OH}} \right)_3} + 4\mathrm{O}{\mathrm{H}^ - } + \mathrm{KI}{\mathrm{O}_3} \to 2\mathrm{CrO}_4^{2 - } + 5{\mathrm{H}_2}\mathrm{O} + \mathrm{KI}$

Step 2: Determine the oxidation state of iodine in $\mathrm{KI}{\mathrm{O}_3}$ . The oxidation state of I is +5.

Step 3: Determine the oxidation state of iodine in $\mathrm{KI}$ . The oxidation state of I is -1.

Step 4: Calculate the change in oxidation number of iodine: $(+5) - (-1) = 6$

Step 5: Calculate the equivalent weight using the formula: Equivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{Change in oxidation number}}}}$

Step 6: Substitute the values: Equivalent weight = $\frac{{\text{{Mol. wt.}}}}{6}$

Why other options are wrong:
- Option A (Mol. wt.): This is incorrect because it doesn't account for the change in oxidation state of iodine.
- Option B (Mol. wt./3): This is incorrect. The change in oxidation number is not 3.
- Option D (Mol. wt./2): This is incorrect. The change in oxidation number is not 2.

Key Concepts:
- Redox Reactions
- Oxidation State
- Equivalent Weight

Memory Aids:
- OIL RIG (Oxidation Is Loss, Reduction Is Gain)

Quick Tips:
- Always balance the chemical equation before determining the change in oxidation number.
- Remember that equivalent weight is the molecular weight divided by the n-factor (change in oxidation number).

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Application +4 | -1

One mole of

\mathrm{N}_2 \mathrm{H}_4

loses 10 moles of electrons to form a new compound y. Assuming that all nitrogen appear in the new compound, what is the oxidation state of nitrogen in y (There is no change in the oxidation state of hydrogen)

View answer and explanation

Correct option: C

+3

Option C (+3) is correct because the oxidation state of nitrogen changes from -2 in N₂H₄ to +3 in the new compound after losing 10 electrons.

Solution approach

Correct Answer + Core Reason: +3. The oxidation state of nitrogen changes from -2 to +3 after losing 10 electrons.

Step-by-Step Solution:

Step 1: Determine the initial oxidation state of N in N₂H₄. Since the oxidation state of H is +1, the total oxidation state of 4 H atoms is +4. Therefore, the total oxidation state of 2 N atoms is -4, and the oxidation state of each N atom is -2.

Step 2: Calculate the total change in oxidation state. Since 1 mole of N₂H₄ loses 10 moles of electrons, the total oxidation state increases by +10.

Step 3: Determine the final oxidation state of N in the new compound (y). Let the oxidation state of each N atom in y be x. Since there are 2 N atoms, 2x - (-4) = 10. Therefore, 2x + 4 = 10, 2x = 6, and x = +3.

Step 4: Final Answer: +3

Why other options are wrong:
- Option A (-1): This is incorrect because it doesn't account for the loss of 10 electrons.
- Option B (-3): This is incorrect because it doesn't account for the loss of 10 electrons and the initial oxidation state of nitrogen.
- Option D (+5): This is incorrect because it doesn't satisfy the electron loss of 10 from the compound.

Key Concepts:
- Oxidation state calculation
- Redox reactions

Memory Aids:
- Oxidation is loss of electrons, reduction is gain of electrons (OIL RIG).

Quick Tips:
- Always balance the oxidation states to ensure the overall change in oxidation state matches the number of electrons lost or gained.

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