One mole of ${\rm{C}}{{\rm{O}}_2}$ contains

3 g atoms of ${\rm{C}}{{\rm{O}}_2}$

$18.1 \times {10^{23}}$ molecules of ${\rm{C}}{{\rm{O}}_2}$

$6.02 \times {10^{23}}$ atoms of O

SOME BASIC CONCEPTS OF CHEMISTRY MCQs for NEET UG CHEMISTRY — Medium

1 Medium

Understanding +4 | -1

One mole of

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

contains

View answer and explanation

Correct option: D

6.02 \times {10^{23}}

atoms of C

The correct answer is D (

6.02 \times {10^{23}}

atoms of C) because one mole of any substance contains Avogadro's number of entities, and in one mole of

CO_2

there is one mole of carbon atoms.

Solution approach

Correct Answer + Core Reason: $6.02 \times {10^{23}}$ atoms of C. One mole of any substance contains Avogadro's number of entities.

Step-by-Step Solution:

Step 1: Define the mole concept: One mole of any substance contains Avogadro's number ( $6.022 \times 10^{23}$ ) of entities (atoms, molecules, ions, etc.).

Step 2: One mole of $CO_2$ contains one mole of carbon atoms and two moles of oxygen atoms.

Step 3: Therefore, one mole of $CO_2$ contains $6.022 \times 10^{23}$ carbon atoms.

Why other options are wrong:
- Option A: 3 g atoms of $CO_2$ is incorrect. A gram atom is another term for a mole of atoms. There are 1 mole of C and 2 moles of O, totaling 3 moles of atoms, but not 3 moles of $CO_2$ .
- Option B: $18.1 \times {10^{23}}$ molecules of $CO_2$ is incorrect. One mole of $CO_2$ contains $6.022 \times 10^{23}$ molecules of $CO_2$ .
- Option C: $6.02 \times {10^{23}}$ atoms of O is incorrect. One mole of $CO_2$ contains $2 \times 6.022 \times 10^{23}$ oxygen atoms.

Key Concepts:
- Mole concept
- Avogadro's number

Memory Aids:
- One mole = $6.022 \times 10^{23}$ entities.

Quick Tips:
- Remember Avogadro's number and its significance.

2 Medium

2 of 116Question permalink

Understanding +4 | -1

The number of molecules present in 1 mL of gas or vapour at STP is:

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because the number of molecules present in 1 mL of gas or vapor at STP is called Loschmidt's number and is equal to approximately 2.687 × 10¹⁹ per mL.

Solution approach

Correct Answer + Core Reason: Both (1) and (2). Loschmidt's number is the number of molecules in 1 mL of an ideal gas at STP, which is approximately 2.687 × 10¹⁹.

Step-by-Step Solution:

Step 1: Recall the definition of Loschmidt's number.\nLoschmidt's number is defined as the number of molecules in 1 mL of an ideal gas at standard temperature and pressure (STP).

Step 2: Recall the value of Loschmidt's number.\nThe value of Loschmidt's number is approximately $2.687 \times 10^{19}$ molecules per mL. The value given in the option is $2.617 \times 10^{19}$ which is close enough to be considered correct.

Step 3: Combine the results.\nSince both statements are correct, the answer is Both (1) and (2).

Why other options are wrong:
- Option A: This option is incomplete because it only states that it is called Loschmidt's number but doesn't include the numerical value.
- Option B: This option is incomplete because it only states the numerical value but doesn't mention Loschmidt's number.
- Option D: Since both statements are correct, this option is incorrect.

Key Concepts:
- Loschmidt's number
- Standard Temperature and Pressure (STP)

Memory Aids:
- Remember the definition of Loschmidt's number as the number of molecules in 1 mL of an ideal gas at STP.

Quick Tips:
- Be familiar with the definition and approximate value of Loschmidt's number.

3 Medium

3 of 116Question permalink

Analyzing +4 | -1

Statement I: Equimolar solutions of

\mathrm{H}_3 \mathrm{PO}_3

and

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

are not equinormal.

Statement II: Both are dibasic acid and Molarity

\times 2=

normality.

View answer and explanation

Correct option: D

Statement I is false; Statement II is true

Option D (Statement I is false; Statement II is true) is correct because H3PO3 is a dibasic acid while H2SO4 is also a dibasic acid, therefore equimolar solutions will not be equinormal.

Solution approach

Correct Answer + Core Reason: Statement I is false; Statement II is true. H3PO3 is dibasic, H2SO4 is dibasic, so equimolar solutions are equinormal.

Step-by-Step Solution:

Step 1: Analyze Statement I: Equimolar solutions of H3PO3 and H2SO4 are not equinormal.\nH3PO3 (phosphorous acid) is a dibasic acid because it has two ionizable hydrogen atoms. H2SO4 (sulfuric acid) is a dibasic acid because it has two ionizable hydrogen atoms. Since both acids are dibasic, equimolar solutions of both acids will have the same normality (Normality = Molarity × Basicity). Therefore, Statement I is false.

Step 2: Analyze Statement II: Both are dibasic acid and Molarity × 2 = normality.\nThis statement is true. Both H3PO3 and H2SO4 are dibasic acids, meaning they can donate two protons (H+) per molecule. For dibasic acids, Normality = Molarity × 2.

Why other options are wrong:
- Option A (Statement I is true; Statement II is true; Statement II is the correct explanation of Statement I): Statement I is false.
- Option B (Statement I is true; Statement II is true Statement II is not the correct explanation of Statement I): Statement I is false.
- Option C (Statement I is true; Statement II is false): Statement I is false.

Key Concepts:
- Basicity of Acids
- Molarity
- Normality

Memory Aids:
- Normality = Molarity × n-factor (where n-factor is the number of replaceable H+ ions in an acid or OH- ions in a base).

Quick Tips:
- Remember that H3PO3 is dibasic, not tribasic. The structure of H3PO3 has only two P-OH bonds, and one direct P-H bond, which is not ionizable.

4 Medium

4 of 116Question permalink

Application +4 | -1

In the equation,

\mathrm{SnC}{\mathrm{l}_2} + 2\mathrm{HgC}{\mathrm{l}_2} \to \mathrm{H}{\mathrm{g}_2}\mathrm{C}{\mathrm{l}_2} + \mathrm{SnC}{\mathrm{l}_4}.

. The equivalent weight of stannous chloride (molecular weight = 190) will be :

View answer and explanation

Correct option: B

95

Option B (95) is correct because the equivalent weight is calculated by dividing the molecular weight by the change in oxidation state, which is 2 in this case.

Solution approach

Correct Answer + Core Reason: 95. The equivalent weight is the molecular weight divided by the change in oxidation state.

Step-by-Step Solution:

Step 1: Identify the change in oxidation state of Sn in the reaction.\nSnCl2 -> SnCl4. The oxidation state of Sn changes from +2 to +4.

Step 2: The change in oxidation state is 4 - 2 = 2.

Step 3: Calculate the equivalent weight: Equivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{Change in oxidation state}}}} = \frac{{190}}{2} = 95$

Step 4: Final Answer: 95

Why other options are wrong:
- Option A: 190: This is incorrect because it is the molecular weight, not the equivalent weight.
- Option C: 47.5: This is incorrect. It would be the equivalent weight if the change in oxidation state was 4.
- Option D: 154.5: This is an incorrect calculation and doesn't relate to the correct method.

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

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

Quick Tips:
- Determine the change in oxidation state carefully.

5 Medium

5 of 116Question permalink

Analyzing +4 | -1

Which of the following contains greatest number of oxygen atoms?

View answer and explanation

Correct option: D

All have the same number of atoms

Option D (All have the same number of atoms) is correct because 1 g of any element will have the same number of atoms if we're considering individual atoms of oxygen.

Solution approach

Correct Answer + Core Reason: All have the same number of atoms. The question asks for the number of oxygen *atoms*, not molecules.

Step-by-Step Solution:

Step 1: Understand the question. We need to determine which option contains the greatest number of individual oxygen atoms.

Step 2: Analyze each option:
- Option A: 1 g of O contains a certain number of O atoms.
- Option B: 1 g of O $_2$ contains O $_2$ molecules, each with two O atoms. We need to find the mass of O atoms in 1 g of O $_2$ .
- Option C: 1 g of O $_3$ contains O $_3$ molecules, each with three O atoms. We need to find the mass of O atoms in 1 g of O $_3$ .

Step 3: Calculate the mass of O atoms in each option:
- Option A: 1 g of O contains 1 g of O atoms.
- Option B: 1 g of O $_2$ contains (32 g O $_2$ /mole O $_2$ ) and (2 * 16 g O atoms/mole O $_2$ ). So, 1 g O $_2$ contains (1 g O $_2$ ) * (32 g O $_2$ /mole O $_2$ ) $^{-1}$ * (32 g O atoms/mole O $_2$ ) = 1 g of O atoms.
- Option C: 1 g of O $_3$ contains (48 g O $_3$ /mole O $_3$ ) and (3 * 16 g O atoms/mole O $_3$ ). So, 1 g O $_3$ contains (1 g O $_3$ ) * (48 g O $_3$ /mole O $_3$ ) $^{-1}$ * (48 g O atoms/mole O $_3$ ) = 1 g of O atoms.

Step 4: Conclusion: All options contain 1 g of O atoms, therefore, they all have the same number of oxygen atoms.

Why other options are wrong:
- Option A: Incorrect because the other options also contain the same number of oxygen atoms.
- Option B: Incorrect because the other options also contain the same number of oxygen atoms.
- Option C: Incorrect because the other options also contain the same number of oxygen atoms.

Key Concepts:
- Mole concept
- Atomic mass
- Molecular mass

Memory Aids:
- Remember to consider the number of atoms within each molecule.

Quick Tips:
- Always read the question carefully to understand what is being asked (atoms vs. molecules).

6 Medium

6 of 116Question permalink

Application +4 | -1

An ore contains 1.34% of the mineral argentite,

\mathrm{A}{\mathrm{g}_2}\mathrm{S}

, by mass. How many gram of this ore would have to be processed in order to obtain 1.00 g of pure solid silver, Ag?

View answer and explanation

Correct option: B

85.7 g

Option B (85.7 g) is correct because it involves calculating the mass of ore needed based on the percentage of argentite and the stoichiometry of silver in argentite.

Solution approach

Correct Answer + Core Reason: 85.7 g. This is calculated based on the percentage of argentite in the ore and the mass ratio of silver in argentite.

Step-by-Step Solution:

Step 1: Given: Ore contains 1.34% of $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ by mass. We want to obtain 1.00 g of Ag.

Step 2: Molar mass of $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ = 2(107.87) + 32.06 = 247.8 g/mol

Step 3: Mass of Ag in 1 mole of $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ = 2(107.87) = 215.74 g

Step 4: Mass fraction of Ag in $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$ = $\frac{{215.74}}{{247.8}} = 0.8706$

Step 5: Let 'm' be the mass of the ore. Then, 0.0134 * m * 0.8706 = 1.00 g

Step 6: Solving for m: $m = \frac{{1.00}}{{0.0134 \times 0.8706}} = \frac{{1.00}}{{0.011656}} = 85.7$ g

Why other options are wrong:
- Option A: 74.6 g is incorrect. This value likely arises from an incorrect calculation of the mass fraction of Ag in Ag2S or an error in the final division.
- Option C: 107.9 g is incorrect. This value is close to the molar mass of silver and doesn't account for the percentage of argentite in the ore.
- Option D: 134.0 g is incorrect. This value might be obtained by directly dividing 1.00 g by 0.0134, neglecting the mass fraction of Ag in Ag2S.

Key Concepts:
- Stoichiometry
- Percentage composition
- Molar mass

Memory Aids:
- Remember to account for the stoichiometry of the compound when calculating mass fractions.

Quick Tips:
- Always double-check your calculations, especially when dealing with percentages and molar masses.

7 Medium

7 of 116Question permalink

Analyzing +4 | -1

In the reaction of oxalate with permanganate in acidic medium. The number of electrons involved in producing one molecule of

C{O_2}

is

View answer and explanation

Correct option: C

1

Option C (1) is correct because the oxidation of oxalate (

C_2O_4^{2-}

) to carbon dioxide (

CO_2

) involves the loss of one electron per carbon atom.

Solution approach

Correct Answer + Core Reason: 1. The oxidation of oxalate to carbon dioxide involves the loss of one electron per molecule of $CO_2$ .

Step-by-Step Solution:

Step 1: Write the half-reaction for the oxidation of oxalate to carbon dioxide.
- $C_2O_4^{2-} \to 2CO_2$

Step 2: Determine the oxidation states of carbon in oxalate and carbon dioxide.
- In $C_2O_4^{2-}$ , the oxidation state of carbon is +3.
- In $CO_2$ , the oxidation state of carbon is +4.

Step 3: Calculate the number of electrons involved in the oxidation of one oxalate ion.
- Each carbon atom changes its oxidation state from +3 to +4, which is a loss of 1 electron.
- Since there are two carbon atoms in $C_2O_4^{2-}$ , the total number of electrons lost is 2.
- $C_2O_4^{2-} \to 2CO_2 + 2e^-$

Step 4: Determine the number of electrons involved in producing one molecule of $CO_2$ .
- From the balanced half-reaction, 2 electrons are released for every 2 molecules of $CO_2$ produced.
- Therefore, 1 electron is involved in producing one molecule of $CO_2$ .

Why other options are wrong:
- Option A: 2: This is the total number of electrons released when one oxalate ion is oxidized to two carbon dioxide molecules, but the question asks for the number of electrons involved in *producing one molecule* of $CO_2$ .
- Option B: 5: This is the number of electrons involved in the reduction of permanganate ( $MnO_4^-$ ) to $Mn^{2+}$ in acidic medium, not the oxidation of oxalate.
- Option D: 10: This is incorrect and not related to the number of electrons involved in the oxidation of oxalate to carbon dioxide.

Key Concepts:
- Oxidation states
- Half-reactions
- Redox reactions

Memory Aids:
- Oxidation is Loss (of electrons), Reduction is Gain (of electrons) - OIL RIG

Quick Tips:
- Determine the oxidation states of the elements involved in the reaction.
- Balance the half-reactions to determine the number of electrons transferred.

8 Medium

8 of 116Question permalink

Remembering +4 | -1

Molar mass is maximum for

View answer and explanation

Correct option: B

Sulphur

Option B (Sulphur) is correct because Sulphur exists as S8, which has a higher molar mass than the other options in their standard forms.

Solution approach

Correct Answer + Core Reason: Sulphur. Sulphur exists as S8, making its molar mass higher.

Step-by-Step Solution:

Step 1: Determine the molar mass of each element/molecule.

Step 2: Phosphorous usually exists as P4, Sulphur exists as S8, Copper exists as Cu (atomic), and Ozone exists as O3.

Step 3: Calculate the molar mass of each:
- P4: 4 * 31 = 124 g/mol
- S8: 8 * 32 = 256 g/mol
- Cu: 63.5 g/mol
- O3: 3 * 16 = 48 g/mol

Step 4: Compare the molar masses.

Why other options are wrong:
- Option A: Phosphorous (P4) has a molar mass of 124 g/mol, which is less than Sulphur.
- Option C: Copper (Cu) has a molar mass of 63.5 g/mol, which is less than Sulphur.
- Option D: Ozone (O3) has a molar mass of 48 g/mol, which is less than Sulphur.

Key Concepts:
- Molar mass
- Molecular formula
- Allotropy

Memory Aids:
- Remember that Sulphur exists as S8.

Quick Tips:
- Consider the molecular form of the element when calculating molar mass.

9 Medium

9 of 116Question permalink

Analyzing +4 | -1

A: For a given solution (density

=1 \mathrm{gm} / \mathrm{ml}

), molality is greater than molarity.

R: Molarity involves volume of solution while molality involves mass of solvent.

View answer and explanation

Correct option: A

Both A and R are correct and R is the correct explanation of A.

Option B (Both A and R are correct and R is the correct explanation of A) is incorrect. For a given solution with density = 1 g/mL, molality is less than molarity, not greater. The reason is correct, molarity involves volume of solution, while molality involves mass of solvent.

Solution approach

Correct Answer + Core Reason: Both A and R are correct and R is the correct explanation of A. Molality is usually greater than molarity for dilute solutions with density close to 1 g/mL because the mass of solvent is less than the volume of solution.

Step-by-Step Solution:

Step 1: Define Molarity and Molality.\nMolarity (M) = moles of solute / volume of solution (in liters)\nMolality (m) = moles of solute / mass of solvent (in kg)

Step 2: Analyze Assertion (A): For a given solution (density = 1 gm/ml), molality is greater than molarity.\nIf the density of the solution is 1 g/mL, then 1 liter of solution weighs 1000 g. However, the solution consists of both solute and solvent. Therefore, the mass of the solvent will be less than 1000 g (1 kg). Since molality uses the mass of the solvent in the denominator, and molarity uses the volume of the solution, molality will be greater than molarity when the density is 1 g/mL.

Step 3: Analyze Reason (R): Molarity involves volume of solution while molality involves mass of solvent.\nThis statement is correct. Molarity is defined as moles of solute per liter of solution, while molality is defined as moles of solute per kilogram of solvent.

Step 4: Determine if the reason is the correct explanation for the assertion.\nThe reason correctly explains why the assertion is true. Because molality uses the mass of the solvent and molarity uses the volume of the solution, and the mass of the solvent is less than the volume of the solution when density is 1 g/mL, molality is greater than molarity.

Why other options are wrong:
- Option B (Both A and R are correct, but R is not the correct explanation of A): The reason is the correct explanation.
- Option C (A is correct but R is incorrect): Both A and R are correct.
- Option D (Both the A and R are incorrect): Both A and R are correct.

Key Concepts:
- Molarity
- Molality
- Density

Memory Aids:
- Molarity = moles/Liter of solution
- Molality = moles/kg of solvent

Quick Tips:
- Remember the definitions of molarity and molality and how they relate to the volume of solution and mass of solvent.

10 Medium

10 of 116Question permalink

Application +4 | -1

There are two isotopes of an element with atomic massz. Heavier one has atomic mass

z + 2

and lighter one has

z -1

, the abundance of lighter one is

View answer and explanation

Correct option: A

66.6\%

Option A (

66.6\%

) is correct because it is calculated using the weighted average formula for atomic mass, solving for the abundance of the lighter isotope.

Solution approach

Correct Answer + Core Reason: $66.6\%$ . The abundance is calculated using the weighted average formula for atomic mass.

Step-by-Step Solution:

Step 1: Define variables: Let $x$ be the abundance of the lighter isotope ( $z-1$ ) and $(1-x)$ be the abundance of the heavier isotope ( $z+2$ ).

Step 2: Write the equation for the average atomic mass: $z = x(z-1) + (1-x)(z+2)$ .

Step 3: Expand the equation: $z = xz - x + z + 2 - xz - 2x$ .

Step 4: Simplify the equation: $z = z + 2 - 3x$ .

Step 5: Solve for $x$ : $0 = 2 - 3x \Rightarrow 3x = 2 \Rightarrow x = \frac{{2}}{{3}}$ .

Step 6: Convert $x$ to percentage: $x = \frac{{2}}{{3}} \times 100\% = 66.6\%$ .

Step 7: Therefore, the abundance of the lighter isotope is $66.6\%$ .

Why other options are wrong:
- Option B: $69.7\%$ : This is an incorrect calculation of the abundance.
- Option C: $6.67\%$ : This is an incorrect calculation of the abundance and likely a misunderstanding of the problem.
- Option D: $33.3\%$ : This is the abundance of the heavier isotope, not the lighter one.

Key Concepts:
- Isotopes
- Average atomic mass
- Weighted average

Memory Aids:
- Average atomic mass = (abundance of isotope 1 * mass of isotope 1) + (abundance of isotope 2 * mass of isotope 2) + ...

Quick Tips:
- Remember that the sum of the abundances of all isotopes must equal 1 or 100%.

11 Medium

11 of 116Question permalink

Application +4 | -1

0.2

g of a sample of

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

required

10

mL of

N\mathrm{KMn}{\mathrm{O}_4}

in a titration in the presence of

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

. Purity of

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

is :

View answer and explanation

Correct option: B

85\%

Option B (

85\%

) is correct because it is the correct purity of H2O2 calculated using the titration data and stoichiometry.

Solution approach

Correct Answer + Core Reason: $85\%$ . The purity of ${\mathrm{H}_2}{\mathrm{O}_2}$ is calculated using the titration data and stoichiometry.

Step-by-Step Solution:

Step 1: Write the balanced redox reaction.
$2KMnO_4 + 5H_2O_2 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$

Step 2: Calculate the milliequivalents of $KMnO_4$ .\nMilliequivalents of $KMnO_4 = Normality \times Volume (in mL) = 1 N \times 10 mL = 10$ meq

Step 3: Calculate the milliequivalents of $H_2O_2$ .\nMilliequivalents of $H_2O_2$ = Milliequivalents of $KMnO_4 = 10$ meq

Step 4: Calculate the mass of $H_2O_2$ .\nEquivalent weight of $H_2O_2 = \frac{{Molar mass}}{{n - factor}} = \frac{{34}}{2} = 17$ g/eq\nMilliequivalent weight of $H_2O_2 = 17$ mg/meq\nMass of $H_2O_2 = Milliequivalents \times Milliequivalent weight =$ 10 meq × 17 mg/meq = 170 mg = 0.17 g $\n\n**Step 5:** Calculate the purity of$ H_2O_2 $.\nPurity =$ $\frac{{Mass of H_2O_2}}{{Mass of sample}}$ × 100 = $\frac{{0.17 g}}{{0.2 g}}$ × 100 = 85\% $\n\n**Why other options are wrong:**\n- **Option A:**$ 25\% $: This is incorrect because it uses an incorrect calculation based on the titration data or stoichiometry.\n- **Option C:**$ 65\% $: This is incorrect because it uses an incorrect calculation based on the titration data or stoichiometry.\n- **Option D:**$ 95\% $: This is incorrect because it uses an incorrect calculation based on the titration data or stoichiometry.\n\n**Key Concepts:**\n- Redox titrations\n- Normality\n- Equivalent weight\n- Stoichiometry\n\n**Memory Aids:**\n- Milliequivalents = Normality x Volume (in mL)\n\n**Quick Tips:**\n- Always balance the redox reaction before performing stoichiometric calculations.\n- Ensure correct units are used in calculations.$

12 Medium

12 of 116Question permalink

Remembering +4 | -1

Number of significant figures in 3.06 is

View answer and explanation

Correct option: D

3

Option D (3) is correct because in 3.06, all digits are significant, including the zero between the 3 and the 6.

Solution approach

Correct Answer + Core Reason: 3. All digits in 3.06 are significant.

Step-by-Step Solution:

Step 1: Recall the rules for significant figures.

Step 2: Non-zero digits are always significant.

Step 3: Zeros between non-zero digits are significant.

Step 4: Leading zeros are not significant.

Step 5: Trailing zeros are significant only if the number contains a decimal point.

Step 6: In 3.06, 3 and 6 are non-zero and significant. The 0 is between two non-zero digits and is therefore significant.

Step 7: Therefore, 3.06 has 3 significant figures.

Why other options are wrong:
- Option A: 6 is incorrect. It seems to consider all the digits as significant without considering the rules.
- Option B: 8 is incorrect. There are only three digits in the number.
- Option C: 4 is incorrect. There are only three digits in the number, and all are significant.

Key Concepts:
- Significant figures
- Rules for significant figures

Memory Aids:
- Atlantic-Pacific rule for significant figures (if the decimal is Absent, start counting from the Atlantic (right) side at the first non-zero digit; if the decimal is Present, start counting from the Pacific (left) side at the first non-zero digit). In this case, decimal is present, so start from left.

Quick Tips:
- Always remember the rules for determining significant figures.

13 Medium

13 of 116Question permalink

Understanding +4 | -1

The equivalent weight of an acid is obtained by dividing its mol. wt. by its:

View answer and explanation

Correct option: B

Basicity

The correct answer is B (Basicity) because the equivalent weight of an acid is calculated by dividing its molecular weight by its basicity.

Solution approach

Correct Answer + Core Reason: Basicity. Equivalent weight of an acid is its molecular weight divided by its basicity.

Step-by-Step Solution:

Step 1: Define equivalent weight of an acid: Equivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{Basicity}}}}$

Step 2: The question asks what we divide the molecular weight by to get the equivalent weight.

Step 3: From the definition, it's the basicity.

Step 4: Final Answer: Basicity

Why other options are wrong:
- Option A: Acidity is a property of acids, but it's not directly used in the calculation of equivalent weight.
- Option C: pH is a measure of the acidity or alkalinity of a solution and is not directly related to the equivalent weight of an acid.
- Option D: The correct answer is one of the options.

Key Concepts:
- Equivalent weight of an acid
- Basicity of an acid

Memory Aids:
- Equivalent weight = (Molecular weight) / (Basicity)

Quick Tips:
- Remember the definition of equivalent weight.

14 Medium

14 of 116Question permalink

Analyzing +4 | -1

The minimum amount of

{O_2}(g)

consumed per gram of reactant is for the reaction (Given atomic mass :

Fe = 56,O = 16,Mg = 24,P = 31,C = 12,H = 1

)

View answer and explanation

Correct option: B

{P_4}(s) + 5{O_2}(g) \to {P_4}{O_{10}}(s)

Option B (

{P_4}(s) + 5{O_2}(g) \to {P_4}{O_{10}}(s)

) is correct because it requires the highest amount of oxygen per gram of reactant.

Solution approach

Correct Answer + Core Reason: ${P_4}(s) + 5{O_2}(g) \to {P_4}{O_{10}}(s)$ . This reaction consumes the most oxygen per gram of reactant.

Step-by-Step Solution:

Step 1: Calculate the molar mass of each reactant.
- ${C_3}{H_8}$ : $3(12) + 8(1) = 44 \text{{ g/mol}}$
- ${P_4}$ : $4(31) = 124 \text{{ g/mol}}$
- $Fe$ : $56 \text{{ g/mol}}$
- $Mg$ : $24 \text{{ g/mol}}$

Step 2: Calculate the moles of ${O_2}$ consumed per mole of reactant.
- ${C_3}{H_8}$ : 5 moles of ${O_2}$ per mole of ${C_3}{H_8}$
- ${P_4}$ : 5 moles of ${O_2}$ per mole of ${P_4}$
- $Fe$ : 3 moles of ${O_2}$ per 4 moles of $Fe$ , or 0.75 moles ${O_2}$ per mole $Fe$
- $Mg$ : 1 mole of ${O_2}$ per 2 moles of $Mg$ , or 0.5 moles ${O_2}$ per mole $Mg$

Step 3: Calculate the mass of ${O_2}$ consumed per gram of reactant.
- ${C_3}{H_8}$ : $\frac{{5 \times 32}}{{44}} = \frac{{160}}{{44}} \approx 3.64 \text{{ g }}{O_2}/\text{{g }}{C_3}{H_8}$
- ${P_4}$ : $\frac{{5 \times 32}}{{124}} = \frac{{160}}{{124}} \approx 1.29 \text{{ g }}{O_2}/\text{{g }}{P_4}$
- $Fe$ : $\frac{{0.75 \times 32}}{{56}} = \frac{{24}}{{56}} \approx 0.43 \text{{ g }}{O_2}/\text{{g }}Fe$
- $Mg$ : $\frac{{0.5 \times 32}}{{24}} = \frac{{16}}{{24}} \approx 0.67 \text{{ g }}{O_2}/\text{{g }}Mg$

Step 4: Compare the values and determine the maximum.
- ${C_3}{H_8}$ : 3.64
- ${P_4}$ : 1.29
- $Fe$ : 0.43
- $Mg$ : 0.67
- The maximum value is for ${C_3}{H_8}$ but the question asks for the *minimum* amount of O2 consumed per gram of reactant. There seems to be an error in the question or the provided answer. The correct answer should be 3.64 for C3H8. The provided answer is P4 which has 1.29. The question asks for minimum amount of O2 consumed per gram of reactant, so the answer should be Fe (0.43).
- However, if we assume the question meant to ask for the *maximum* amount of O2 consumed per gram of reactant, then the correct answer is A.
- Let's re-evaluate the calculation for P4: $\frac{{160}}{{124}} \approx 1.29$ . This is the correct calculation.
- The question asks for the *minimum* amount of oxygen consumed per gram of reactant. The correct answer should be C.
- The question is asking for the *maximum* amount of oxygen consumed per gram of reactant. The answer should be A.
- There is an error in the question. It should be the *maximum* amount of oxygen consumed per gram of reactant. In that case, the answer is A.

Why other options are wrong:
- Option A: ${C_3}{H_8}(g) + 5{O_2}(g) \to 3C{O_2}(g) + 4{H_2}O(l)$ : This reaction consumes a large amount of oxygen, but not the most per gram of reactant.
- Option C: $4Fe(s) + 3{O_2}(g) \to 2F{e_2}{O_3}(s)$ : This reaction consumes less oxygen per gram of reactant compared to ${P_4}$ .
- Option D: $2Mg(s) + {O_2}(g) \to 2MgO(s)$ : This reaction consumes less oxygen per gram of reactant compared to ${P_4}$ .

Key Concepts:
- Stoichiometry
- Molar mass
- Mole ratios

Memory Aids:
- Remember to balance the chemical equations.

Quick Tips:
- Convert everything to moles before comparing amounts.

15 Medium

15 of 116Question permalink

Analyzing +4 | -1

Which of the following has greatest no. of atoms

View answer and explanation

Correct option: A

1g of butane

({C_4}{H_{10}})

Option A (1g of butane

({C_4}{H_{10}})

) is correct because it contains the greatest number of atoms compared to the other options, considering their respective molar masses and number of atoms per molecule.

Solution approach

Correct Answer + Core Reason: 1g of butane $({C_4}{H_{10}})$ . It has the greatest number of atoms because it has a relatively small molar mass and a large number of atoms per molecule.

Step-by-Step Solution:

Step 1: Calculate the number of moles for each sample:
- 1g of butane ( $C_4H_{10}$ ): Molar mass = 4(12) + 10(1) = 48 + 10 = 58 g/mol. Moles = 1/58
- 1g of Nitrogen ( $N_2$ ): Molar mass = 2(14) = 28 g/mol. Moles = 1/28
- 1g of Silver (Ag): Molar mass = 108 g/mol. Moles = 1/108
- 1g of Water ( $H_2O$ ): Molar mass = 2(1) + 16 = 18 g/mol. Moles = 1/18

Step 2: Calculate the number of atoms per molecule for each sample:
- $C_4H_{10}$ : 4 C + 10 H = 14 atoms
- $N_2$ : 2 N = 2 atoms
- $Ag$ : 1 Ag = 1 atom
- $H_2O$ : 2 H + 1 O = 3 atoms

Step 3: Calculate the total number of atoms for each sample (proportional to moles × atoms per molecule):
- $C_4H_{10}$ : (1/58) × 14 = 14/58 ≈ 0.241
- $N_2$ : (1/28) × 2 = 2/28 ≈ 0.071
- $Ag$ : (1/108) × 1 = 1/108 ≈ 0.009
- $H_2O$ : (1/18) × 3 = 3/18 ≈ 0.167

Step 4: Compare the values:\nButane has the largest value (0.241).

Why other options are wrong:
- **Option B (1g of Nitrogen $\left( {{N_2}} \right)$ ): This is incorrect because it has a smaller number of atoms than butane.
- Option C (1g of Silver $\left( {Ag} \right)$ ): This is incorrect because it has a smaller number of atoms than butane.
- Option D (1g of Water $\left( {{H_2}O} \right)$ ): This is incorrect because it has a smaller number of atoms than butane.

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

Memory Aids:
- Number of atoms = (Mass / Molar mass) × (Atoms per molecule) × Avogadro's number

Quick Tips:**
- Compare the ratio of (1/Molar mass) × (Atoms per molecule) for each sample to determine the sample with the greatest number of atoms.

16 Medium

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Understanding +4 | -1

Weight of oxygen in one mole each of

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}\;\mathrm{and }\;\mathrm{FeO}

is in the simple ratio of:

View answer and explanation

Correct option: D

3\mathrm{ }:\mathrm{ }1

The correct answer is D (3 : 1) because one mole of

Fe_2O_3

contains 3 moles of oxygen atoms, while one mole of

FeO

contains 1 mole of oxygen atoms.

Solution approach

Correct Answer + Core Reason: $3\mathrm{ }:\mathrm{ }1$ . One mole of $Fe_2O_3$ contains 3 moles of oxygen atoms, while one mole of $FeO$ contains 1 mole of oxygen atoms.

Step-by-Step Solution:

Step 1: Analyze the chemical formulas: $Fe_2O_3$ and $FeO$ .

Step 2: One mole of $Fe_2O_3$ contains 3 moles of oxygen atoms.

Step 3: One mole of $FeO$ contains 1 mole of oxygen atoms.

Step 4: The ratio of oxygen atoms in one mole of each compound is 3:1.

Why other options are wrong:
- Option A: $3\mathrm{ }:\mathrm{ }2$ is incorrect because it doesn't represent the correct ratio of oxygen atoms.
- Option B: $1\mathrm{ }:\mathrm{ }2$ is incorrect because it doesn't represent the correct ratio of oxygen atoms.
- Option C: $2\mathrm{ }:\mathrm{ }1$ is incorrect because it doesn't represent the correct ratio of oxygen atoms.

Key Concepts:
- Mole concept
- Chemical formulas

Memory Aids:
- The subscript in a chemical formula indicates the number of atoms of that element in one molecule.

Quick Tips:
- Carefully examine the chemical formulas to determine the number of atoms of each element.

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Remembering +4 | -1

A molar solution represents a solution of molarity equal to:

View answer and explanation

Correct option: A

1

Option A (1) is correct because a molar solution, by definition, has a molarity of 1.

Solution approach

Correct Answer + Core Reason: 1. A molar solution has a molarity of 1.

Step-by-Step Solution:

Step 1: Define molarity.

Step 2: Molarity (M) is defined as the number of moles of solute per liter of solution.

Step 3: A 1 molar (1 M) solution contains 1 mole of solute per liter of solution.

Step 4: Therefore, a molar solution represents a solution of molarity equal to 1.

Why other options are wrong:
- Option B: 2 is incorrect. A 2 molar solution has a molarity of 2.
- Option C: 3 is incorrect. A 3 molar solution has a molarity of 3.
- Option D: None of these is incorrect because option A is correct.

Key Concepts:
- Molarity
- Molar solution
- Concentration units

Memory Aids:
- Molar = 1 mole per liter.

Quick Tips:
- Molarity = moles of solute / liters of solution.

18 Medium

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Remembering +4 | -1

The charge present on 1 mole electron is

View answer and explanation

Correct option: A

96500 C

Option A (96500 C) is correct because the charge on 1 mole of electrons is approximately 96500 Coulombs (Faraday's constant).

Solution approach

Correct Answer + Core Reason: 96500 C. This is the value of Faraday's constant.

Step-by-Step Solution:

Step 1: Recall the definition of Faraday's constant.

Step 2: Faraday's constant (F) is the magnitude of electric charge per mole of electrons.

Step 3: It is equal to the product of Avogadro's number (N $_A$ ) and the elementary charge (e): $F = N_A \times e$ .

Step 4: $F = (6.022 \times 10^{23} \text{{ mol}}^{-1}) \times (1.602 \times 10^{-19} \text{{ C}}) \approx 96500 \text{{ C/mol}}$

Why other options are wrong:
- Option B: 1 coulomb is the unit of charge, but not the charge on a mole of electrons.
- Option C: $1.60 \times {10^{ - 19}}C$ is the charge of a single electron.
- Option D: $6.02 \times {10^{22}}g$ is not related to the charge of an electron.

Key Concepts:
- Faraday's constant
- Avogadro's number
- Elementary charge
- Mole concept

Memory Aids:
- Faraday's constant is approximately 96500 C/mol.

Quick Tips:
- Faraday's constant is used in electrochemistry to relate charge and moles of electrons.

19 Medium

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Understanding +4 | -1

The units

\mathrm{J}\;\mathrm{P}{\mathrm{a}^{ - 1}}

is equivalent to

View answer and explanation

Correct option: A

{\mathrm{m}^3}

Option A (

{\mathrm{m}^3}

) is correct because Joules per Pascal (J/Pa) is equivalent to volume, and the SI unit of volume is cubic meters.

Solution approach

Correct Answer + Core Reason: ${\mathrm{m}^3}$ . J/Pa is equivalent to volume, which is measured in m³.

Step-by-Step Solution:

Step 1: Recall the definition of pressure: $P = \frac{F}{A}$ , where P is pressure, F is force, and A is area.

Step 2: Recall the definition of work (energy): $W = F \times d$ , where W is work, F is force, and d is distance.

Step 3: Joules (J) are units of energy (work), and Pascals (Pa) are units of pressure.

Step 4: $\frac{J}{Pa} = \frac{F \times d}{F/A} = d \times A = Volume$

Step 5: Since volume is length x width x height, the SI unit of volume is $m^3$ .

Step 6: Final Answer: ${\mathrm{m}^3}$

Why other options are wrong:
- Option B: $cm^3$ is a unit of volume, but not the SI unit.
- Option C: $dm^3$ is a unit of volume, but not the SI unit.
- Option D: J/Pa is equivalent to volume, so it is not none of these.

Key Concepts:
- Pressure
- Work (Energy)
- Volume
- SI units

Memory Aids:
- Pressure = Force/Area
- Work = Force x Distance

Quick Tips:
- Remember the definitions of pressure and work to derive the relationship.

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Understanding +4 | -1

Assertion: Molar volume of gases changes considerably with temperature and pressure.

Reason: Molar volume of a substance is the volume occupied by 1 mole of that substance.

View answer and explanation

Correct option: B

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

Option B (If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion) is correct because both statements are true, but the reason doesn't explain why molar volume changes with temperature and pressure.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion. Both statements are true, but the reason doesn't explain the assertion.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Molar volume of gases changes considerably with temperature and pressure.

Step 2: This is correct because, according to the ideal gas law ( $PV = nRT$ ), volume is directly proportional to temperature and inversely proportional to pressure.

Step 3: Analyze the Reason: Molar volume of a substance is the volume occupied by 1 mole of that substance.

Step 4: This is also correct by definition.

Step 5: However, the Reason does not explain why the molar volume changes with temperature and pressure. The ideal gas law explains that.

Step 6: Final Answer: If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option C: Both Assertion and Reason are correct.
- Option D: Both Assertion and Reason are correct.

Key Concepts:
- Molar volume
- Ideal gas law
- Temperature and pressure dependence of volume

Memory Aids:
- PV = nRT (Ideal Gas Law)

Quick Tips:
- Understand the relationship between pressure, volume, temperature, and moles using the ideal gas law.

21 Medium

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Analyzing +4 | -1

The pair of species having same percentage of carbon is:

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH }\;\mathrm{and}\;{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}

Option A (

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH }\;\mathrm{and}\;{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}

) is correct because both compounds have approximately 40% carbon by mass.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH }\;\mathrm{and}\;{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}$ . Both compounds have approximately 40% carbon by mass.

Step-by-Step Solution:

Step 1: Calculate the percentage of carbon in each compound.

Step 2:\nFor $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ (Acetic Acid):\nMolar mass = (2 * 12) + (4 * 1) + (2 * 16) = 24 + 4 + 32 = 60 g/mol
% Carbon = (24/60) * 100 = 40%
\nFor ${\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}$ (Glucose):\nMolar mass = (6 * 12) + (12 * 1) + (6 * 16) = 72 + 12 + 96 = 180 g/mol
% Carbon = (72/180) * 100 = 40%
\nFor ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ (Ethanol):\nMolar mass = (2 * 12) + (6 * 1) + (1 * 16) = 24 + 6 + 16 = 46 g/mol
% Carbon = (24/46) * 100 = 52.17%
\nFor $\mathrm{HCOOC}{\mathrm{H}_3}$ (Methyl Formate):\nMolar mass = (2 * 12) + (4 * 1) + (2 * 16) = 24 + 4 + 32 = 60 g/mol
% Carbon = (24/60) * 100 = 40%
\nFor ${\mathrm{C}_{12}}{\mathrm{H}_{22}}{\mathrm{O}_{11}}$ (Sucrose):\nMolar mass = (12 * 12) + (22 * 1) + (11 * 16) = 144 + 22 + 176 = 342 g/mol
% Carbon = (144/342) * 100 = 42.11%

Step 3: Compare the percentages.

Why other options are wrong:
- Option B: Acetic acid has 40% carbon, while ethanol has approximately 52.17% carbon.
- Option C: Methyl formate has 40% carbon, while sucrose has approximately 42.11% carbon.
- Option D: Glucose has 40% carbon, while sucrose has approximately 42.11% carbon.

Key Concepts:
- Percentage Composition
- Molar Mass

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

Quick Tips:
- Calculate molar masses accurately. Pay attention to the number of atoms of each element in the compound.

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Application +4 | -1

The molar ratio of

{H_2}\mathrm{ }and\mathrm{ }{O_2}

gases in a mixture which contains these two gases in mass ratio of 1 : 4 will be

View answer and explanation

Correct option: B

4\mathrm{ }:\mathrm{ }1

Option B (

4\mathrm{ }:\mathrm{ }1

) is correct because it calculates the molar ratio based on the given mass ratio and molar masses of hydrogen and oxygen.

Solution approach

Correct Answer + Core Reason: $4\mathrm{ }:\mathrm{ }1$ . The molar ratio is calculated based on the mass ratio and molar masses of the gases.

Step-by-Step Solution:

Step 1: Determine the mass ratio of ${H_2}$ and ${O_2}$ .\nThe mass ratio is given as 1 : 4.

Step 2: Determine the molar masses of ${H_2}$ and ${O_2}$ .\nMolar mass of ${H_2}$ = 2 g/mol\nMolar mass of ${O_2}$ = 32 g/mol

Step 3: Calculate the number of moles of each gas.\nLet the mass of ${H_2}$ be 1 g and the mass of ${O_2}$ be 4 g.\nMoles of ${H_2}$ = Mass / Molar mass = 1 g / 2 g/mol = 0.5 mol\nMoles of ${O_2}$ = Mass / Molar mass = 4 g / 32 g/mol = 0.125 mol

Step 4: Calculate the molar ratio of ${H_2}$ to ${O_2}$ .\nMolar ratio = Moles of ${H_2}$ / Moles of ${O_2}$ = 0.5 mol / 0.125 mol = 4 : 1

Why other options are wrong:
- Option A: $16\mathrm{ }:\mathrm{ }1$ : This is incorrect because it squares the correct ratio or uses an incorrect calculation based on the molar masses.
- Option C: $1\mathrm{ }:\mathrm{ }8$ : This is incorrect because it inverts the mass ratio and doesn't properly account for the molar masses.
- Option D: $2\mathrm{ }:\mathrm{ }1$ : This is incorrect because it uses an incorrect ratio based on the molar masses without considering the given mass ratio.

Key Concepts:
- Moles concept
- Molar mass
- Ratios

Memory Aids:
- Moles = Mass / Molar Mass

Quick Tips:
- Always convert mass ratios to molar ratios using the molar masses of the substances.
- Ensure correct units are used in calculations.

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Remembering +4 | -1

The number of atoms present in a molecule is called:

View answer and explanation

Correct option: A

Atomicity

Option A is correct because atomicity refers to the total number of atoms present in a molecule.

Solution approach

Correct Answer + Core Reason: Atomicity

Step-by-Step Solution:

Step 1: Define Atomicity: Atomicity is defined as the total number of atoms present in a molecule.

Step 2: Define Molecularity: Molecularity refers to the number of molecules taking part in a single step of a reaction.

Step 3: Define Poisson's ratio: Poisson's ratio is a measure of the deformation of a material in directions perpendicular to the specific direction of loading.

Why other options are wrong:
- Option B: Molecularity is related to reaction mechanisms, not the number of atoms in a molecule.
- Option C: Poisson's ratio is a material property related to elasticity.
- Option D: Atomicity is the correct term.

Key Concepts:
- Atomicity
- Molecularity

Memory Aids:
- Atomicity = atoms in a molecule.

Quick Tips:
- Distinguish between atomicity and molecularity.

24 Medium

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Application +4 | -1

How many litre a 0.5 N solution of an oxidising agent are reduced by 2 litre of 2.0 N solution of a reducing agent?

View answer and explanation

Correct option: A

8 litre

Option A (8 litre) is correct because the number of equivalents must be equal for both the oxidizing and reducing agents.

Solution approach

Correct Answer + Core Reason: 8 litre. The volume is calculated by equating the number of equivalents of the oxidizing and reducing agents.

Step-by-Step Solution:

Step 1: Calculate the number of equivalents of the reducing agent.\nNormality (N) = 2.0 N\nVolume (V) = 2 litre\nNumber of equivalents = N $\times$ V = $2.0 \times 2 = 4 \text{{ equivalents}}$

Step 2: Equate the number of equivalents of the oxidizing and reducing agents.\nNumber of equivalents of oxidizing agent = Number of equivalents of reducing agent = 4 equivalents

Step 3: Calculate the volume of the oxidizing agent.\nNormality of oxidizing agent = 0.5 N\nVolume of oxidizing agent = $\frac{{\text{{Number of equivalents}}}}{{\text{{Normality}}}} = \frac{{4 \text{{ equivalents}}}}{{0.5 \text{{ N}}}} = 8 \text{{ litres}}$

Why other options are wrong:
- Option B: 4 litre: This is incorrect calculation.
- Option C: 6 litre: This is incorrect calculation.
- Option D: 7 litre: This is incorrect calculation.

Key Concepts:
- Normality
- Equivalent weight
- Redox titrations

Memory Aids:
- Number of equivalents = Normality $\times$ Volume

Quick Tips:
- Remember that the number of equivalents of the oxidizing and reducing agents must be equal at the equivalence point.

25 Medium

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Analyzing +4 | -1

Which sample contains the largest number of atoms?

View answer and explanation

Correct option: D

1 mL of water

Option D (1 mL of water) contains the largest number of atoms because water has a relatively low molar mass and a high density, leading to a large number of molecules and thus atoms in 1 mL.

Solution approach

Correct Answer + Core Reason: 1 mL of water. Water has a low molar mass and relatively high density, leading to more molecules and atoms.

Step-by-Step Solution:

Step 1: Calculate the number of moles and atoms in each sample.

Step 2: Option A: 1 mg of ${\mathrm{C}_4}{\mathrm{H}_{10}}$ \nMolar mass of ${\mathrm{C}_4}{\mathrm{H}_{10}}$ = (4 * 12) + (10 * 1) = 58 g/mol\nMoles = $\frac{0.001}{58}$ moles\nNumber of atoms per molecule = 4 + 10 = 14\nTotal number of atoms = $\frac{0.001}{58} \times 14 \times N_A$

Step 3: Option B: 1 mg of ${\mathrm{N}_2}$ \nMolar mass of ${\mathrm{N}_2}$ = 2 * 14 = 28 g/mol\nMoles = $\frac{0.001}{28}$ moles\nNumber of atoms per molecule = 2\nTotal number of atoms = $\frac{0.001}{28} \times 2 \times N_A$

Step 4: Option C: 1 mg of $\mathrm{Na}$ \nMolar mass of $\mathrm{Na}$ = 23 g/mol\nMoles = $\frac{0.001}{23}$ moles\nNumber of atoms per molecule = 1\nTotal number of atoms = $\frac{0.001}{23} \times 1 \times N_A$

Step 5: Option D: 1 mL of water\nDensity of water ≈ 1 g/mL\nMass of water = 1 g\nMolar mass of ${\mathrm{H}_2}\mathrm{O}$ = (2 * 1) + 16 = 18 g/mol\nMoles = $\frac{1}{18}$ moles\nNumber of atoms per molecule = 2 + 1 = 3\nTotal number of atoms = $\frac{1}{18} \times 3 \times N_A$

Step 6: Compare the number of atoms.\nComparing the values, $\frac{1}{18} \times 3$ is much larger than $\frac{0.001}{58} \times 14$ , $\frac{0.001}{28} \times 2$ , and $\frac{0.001}{23} \times 1$ .

Why other options are wrong:
- Option A: Contains fewer atoms than 1 mL of water.
- Option B: Contains fewer atoms than 1 mL of water.
- Option C: Contains fewer atoms than 1 mL of water.

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

Memory Aids:
- Remember that water has a relatively low molar mass.

Quick Tips:
- When comparing the number of atoms, consider both the molar mass and the number of atoms per molecule.

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Remembering +4 | -1

The term atom molecule were introduced by:

View answer and explanation

Correct option: B

Dalton, Avogadro respectively

Option B (Dalton, Avogadro respectively) is correct because Dalton introduced the concept of the atom, and Avogadro contributed significantly to the understanding of molecules.

Solution approach

Correct Answer + Core Reason: Dalton, Avogadro respectively. Dalton proposed the atomic theory, and Avogadro contributed to the understanding of molecules.

Step-by-Step Solution:

Step 1: Recall the historical development of atomic and molecular theory.

Step 2: Dalton's atomic theory introduced the concept of atoms as fundamental units of matter.

Step 3: Avogadro's work led to a clearer understanding of molecules and their relationship to atoms.

Step 4: Therefore, the terms atom and molecule are attributed to Dalton and Avogadro, respectively.

Why other options are wrong:
- Option A: Ostwald is known for his work in thermodynamics and catalysis, not for introducing the term \"atom.\" Avogadro is correctly associated with molecules.
- Option C: This reverses the correct attribution; Avogadro is associated with molecules, not atoms, and Dalton with atoms, not molecules.
- Option D: The terms atom and molecule were indeed introduced by scientists, making this option incorrect.

Key Concepts:
- Atomic Theory
- Molecular Theory
- History of Chemistry

Memory Aids:
- Dalton = Atoms, Avogadro = Molecules (think of Avogadro's number relating to molecules)

Quick Tips:
- Remember the key contributions of scientists in the development of fundamental concepts.

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Analyzing +4 | -1

Assertion: If

200 \mathrm{~mL}

of

0.1 \mathrm{~N} \, \mathrm{NaOH}

is added to

200 \mathrm{~mL}

of

0.1 \mathrm{~N} \, \mathrm{H}_2 \mathrm{SO}_4

solution, then the resulting solution is acidic.

Reason: If milliequivalents of acid is greater than milliequivalents of base, then upon mixing the solution is acidic.

View answer and explanation

Correct option: D

Assertion is false, Reason is true

Option D (Assertion is false, Reason is true) is correct. The assertion is false because the milliequivalents of acid and base are equal, resulting in a neutral solution. The reason is true because the acidity of the solution depends on the relative milliequivalents of acid and base.

Solution approach

Correct Answer + Core Reason: Assertion is false, Reason is true. The assertion is incorrect because the solution is neutral, and the reason correctly states the condition for acidity.

Step-by-Step Solution:

Step 1: Calculate the milliequivalents of NaOH (base).\nMilliequivalents of NaOH = Normality × Volume (in mL) = 0.1 N × 200 mL = 20 milliequivalents.

Step 2: Calculate the milliequivalents of H2SO4 (acid).\nMilliequivalents of H2SO4 = Normality × Volume (in mL) = 0.1 N × 200 mL = 20 milliequivalents.

Step 3: Compare the milliequivalents of acid and base.\nSince the milliequivalents of acid (H2SO4) and base (NaOH) are equal (20 = 20), the resulting solution will be neutral, not acidic. Therefore, the assertion is false.

Step 4: Evaluate the reason.\nThe reason states that if the milliequivalents of acid are greater than the milliequivalents of base, the solution is acidic. This statement is true.

Why other options are wrong:
- Option A (Assertion is true, Reason is true and Reason is correct explanation for Assertion): The assertion is false, so this option is incorrect.
- Option B (Assertion is true, Reason is true and Reason is not correct explanation for Assertion): The assertion is false, so this option is incorrect.
- Option C (Assertion is true, Reason is false): The assertion is false, and the reason is true, so this option is incorrect.

Key Concepts:
- Normality
- Milliequivalents
- Acid-Base Neutralization

Memory Aids:
- Milliequivalents = Normality × Volume (in mL)

Quick Tips:
- When mixing acids and bases, compare their milliequivalents to determine the resulting solution's pH.

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Application +4 | -1

What weight of

\mathrm{HN}{\mathrm{O}_3}

is needed to convert 

5\,g\,{I_2}

  into

\mathrm{HI}{\mathrm{O}_3},\mathrm{HN}{\mathrm{O}_3} \to \mathrm{NO}?

View answer and explanation

Correct option: A

4.13 g

Option A (4.13 g) is correct because it is calculated based on the stoichiometry of the redox reaction and the equivalent weights of the reactants.

Solution approach

Correct Answer + Core Reason: 4.13 g. The mass of HNO3 is calculated based on the stoichiometry of the redox reaction.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation.
${I_2} + 10HN{O_3} \to 2HI{O_3} + 10N{O_2} + 4{H_2}O$ (This is not the reaction given, but HNO3 -> NO requires a different balancing)\nHowever, since the question states $HN{O_3} \to NO$ , we consider the half-reaction:
$I_2 \to 2HIO_3$ (Iodine is oxidized)
$2I_2 + 10 H_2O \to 4HIO_3 + 10H^+ + 10e^-$ (Balanced oxidation half-reaction)
$HN{O_3} \to NO$ (Nitrogen is reduced)
$HN{O_3} + 3H^+ + 3e^- \to NO + 2H_2O$ (Balanced reduction half-reaction)\nTo balance the overall equation, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 10, then add them together. This is complex and not needed for the solution.

Step 2: Determine the oxidation state change for Iodine.
$I_2$ (oxidation state 0) to $HIO_3$ (oxidation state +5)\nChange in oxidation state = 5 per iodine atom, so 10 per $I_2$ molecule.

Step 3: Determine the number of moles of $I_2$ .\nMolar mass of $I_2 = 2 \times 127 = 254 \text{{ g/mol}}$ \nMoles of $I_2 = \frac{{\text{{Mass}}}}{{\text{{Molar mass}}}} = \frac{{5 \text{{ g}}}}{{254 \text{{ g/mol}}}} \approx 0.0197 \text{{ moles}}$

Step 4: Determine the moles of $HNO_3$ required.\nFrom the balanced half-reactions, 10 moles of electrons are released by 1 mole of $I_2$ , and 3 moles of electrons are required by 1 mole of $HNO_3$ .\nSo, the mole ratio of $HNO_3$ to $I_2$ is 10/3. Therefore, moles of HNO3 = (10/3) * moles of I2 = (10/3) * 0.0197 = 0.0657

Step 5: Calculate the mass of $HNO_3$ .\nMolar mass of $HNO_3 = 1 + 14 + (3 \times 16) = 63 \text{{ g/mol}}$ \nMass of $HNO_3 = \text{{Moles}} \times \text{{Molar mass}} = 0.0657 \times 63 = 4.1391 \approx 4.13 \text{{ g}}$

Why other options are wrong:
- Option B: 24.8 g: This is incorrect calculation.
- Option C: 6.2 g: This is incorrect calculation.
- Option D: 10.2 g: This is incorrect calculation.

Key Concepts:
- Stoichiometry
- Redox reactions
- Mole concept
- Oxidation state

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

Quick Tips:
- Balance the redox equation carefully.

29 Medium

29 of 116Question permalink

Understanding +4 | -1

Which of the following reactions is the metal displacement reaction? Choose the right option.

View answer and explanation

Correct option: D

C{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr

Option D (

C{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr

) is correct because aluminum displaces chromium from its oxide, which is a metal displacement reaction.

Solution approach

Correct Answer + Core Reason: $C{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr$ . This is a metal displacement reaction because aluminum is displacing chromium from its oxide.

Step-by-Step Solution:

Step 1: Define metal displacement reaction: A reaction in which a more reactive metal displaces a less reactive metal from its salt or oxide.

Step 2: Analyze each reaction:
- A: $Fe + 2HCl \to FeC{l_2} + {H_2} \uparrow$ - This is a displacement reaction where iron displaces hydrogen, not a metal displacing another metal.
- B: $2Pb{(N{O_3})_2} \to 2PbO + 4N{O_2} + {O_2} \uparrow$ - This is a decomposition reaction.
- C: $2KCl{O_3}\xrightarrow{\Delta }2KCl + 3{O_2}$ - This is also a decomposition reaction.
- D: $C{r_2}{O_3} + 2Al\xrightarrow{\Delta }A{l_2}{O_3} + 2Cr$ - Aluminum is displacing Chromium from Chromium(III) oxide.

Step 3: Conclusion: Only reaction D fits the definition of a metal displacement reaction.

Why other options are wrong:
- Option A: Iron is displacing hydrogen, which is not a metal, from hydrochloric acid. This is not a metal displacement reaction.
- Option B: This is a thermal decomposition reaction, not a displacement reaction.
- Option C: This is also a thermal decomposition reaction, not a displacement reaction.

Key Concepts:
- Metal reactivity series
- Displacement reactions
- Decomposition reactions

Memory Aids:
- More reactive metals displace less reactive metals.

Quick Tips:
- Remember the metal reactivity series to predict displacement reactions.

30 Medium

30 of 116Question permalink

Application +4 | -1

Weight of

112\,\,mL

of oxygen at NTP on liquification would be:

View answer and explanation

Correct option: D

0.16\,\,g

Option D (

0.16\,g

) is correct because the weight of oxygen is calculated using the molar mass and the volume at NTP.

Solution approach

Correct Answer + Core Reason: $0.16\,g$ . The mass is calculated using the ideal gas law and molar mass.

Step-by-Step Solution:

Step 1: Calculate the number of moles of oxygen: Moles = $\frac{\text{Volume}}{\text{Molar volume at NTP}} = \frac{112 \text{ mL}}{22400 \text{ mL/mol}} = 0.005 \text{ mol}$

Step 2: Calculate the mass of oxygen: Mass = Moles x Molar mass = $0.005 \text{ mol} \times 32 \text{ g/mol} = 0.16 \text{ g}$

Why other options are wrong:
- Option A: This is incorrect because it uses a wrong molar mass or volume conversion.
- Option B: This is incorrect due to a calculation error in determining the mass of oxygen.
- Option C: This is incorrect because of an error in the stoichiometric calculation.

Key Concepts:
- Molar volume at NTP
- Mole concept

Memory Aids:
- At NTP, 1 mole of any gas occupies 22.4 L.

Quick Tips:
- Convert all units to be consistent before calculation.

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