SOME BASIC CONCEPTS OF CHEMISTRY MCQs for JEE CHEMISTRY — Medium, Page 9

241 Medium

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Analyzing +4 | -1

A gas mixture contains

\mathrm{CH}_4

and

\mathrm{C}_3 \mathrm{H}_6.

When this mixture undergo cracking into

\mathrm{C}(\mathrm{s})

and

\mathrm{H}_2(\mathrm{g}),

the total number of moles of

\mathrm{H}_2(\mathrm{g})

obtained is 42. If the total volume of the initial gas mixture at

1.5 \mathrm{~atm}

and

27^{\circ} \mathrm{C}

is

246.0 \mathrm{~L},

then what is the mole per cent of

\mathrm{CH}_4

gas in the initial mixture?

View answer and explanation

Correct option: A

20 \%

Option A (20%) is correct because by setting up and solving a system of equations based on the cracking reactions and the ideal gas law, the mole percent of CH4 is found to be 20%.

Solution approach

Correct Answer + Core Reason: 20 %. By setting up equations based on the cracking reactions and using the ideal gas law, the mole percent of CH4 is found to be 20%.

Step-by-Step Solution:

Step 1: Write the cracking reactions.
$\mathrm{CH}_4 \rightarrow \mathrm{C} + 2\mathrm{H}_2$
$\mathrm{C}_3\mathrm{H}_6 \rightarrow 3\mathrm{C} + 3\mathrm{H}_2$

Step 2: Define variables.\nLet $x$ be the number of moles of $\mathrm{CH}_4$ and $y$ be the number of moles of $\mathrm{C}_3\mathrm{H}_6$ .

Step 3: Set up equations based on the given information.\nFrom the cracking reactions, the total number of moles of $\mathrm{H}_2$ is $2x + 3y = 42$ .\nThe total number of moles of the initial gas mixture can be found using the ideal gas law: $PV = nRT$ , where $P = 1.5 \text{ atm}$ , $V = 246.0 \text{ L}$ , $R = 0.0821 \text{ L atm / (mol K)}$ , and $T = 27 + 273 = 300 \text{ K}$ .
$n = \frac{{PV}}{{RT}} = \frac{{1.5 \times 246}}{{0.0821 \times 300}} = \frac{{369}}{{24.63}} = 15 \text{ moles}$ \nSo, $x + y = 15$ .

Step 4: Solve the system of equations.\nWe have the following system of equations:
$2x + 3y = 42$
$x + y = 15$ \nMultiply the second equation by 2: $2x + 2y = 30$ .\nSubtract this from the first equation: $(2x + 3y) - (2x + 2y) = 42 - 30$ , which gives $y = 12$ .\nSubstitute $y = 12$ into $x + y = 15$ : $x + 12 = 15$ , so $x = 3$ .

Step 5: Calculate the mole percent of $\mathrm{CH}_4$ .\nMole percent of $\mathrm{CH}_4 = \frac{{x}}{{x + y}} \times 100 = \frac{{3}}{{15}} \times 100 = 20\%$ .

Why other options are wrong:
- Option B: Incorrect calculation of mole percent.
- Option C: Incorrect calculation of mole percent.
- Option D: Incorrect calculation of mole percent.

Key Concepts:
- Stoichiometry
- Ideal Gas Law

Memory Aids:
- PV = nRT

Quick Tips:
- Always write balanced chemical equations.
- Use the ideal gas law to find the number of moles.

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Understanding +4 | -1

Statement I : 1 mole of

{H_2}S{O_4}

is neutralised by 2 mol of NaOH but 1 equivalent of

{H_2}S{O_4}

is neutralised by 1 equivalent of NaOH
Statement II : Equivalent amount of

{H_2}S{O_4}

is the quantity that is equal to 1 g H and hence, each mol of

{H_2}S{O_4}

is equal to 2 equivalents of

{H_2}S{O_4}

Which statement(s) is/are correct ?

View answer and explanation

Correct option: C

Both I and II

Option C (Both I and II) is correct because 1 mole of H2SO4 requires 2 moles of NaOH for neutralization, and the equivalent amount of H2SO4 is indeed the quantity that is equal to 1 g H, meaning each mole of H2SO4 is equal to 2 equivalents.

Solution approach

Correct Answer + Core Reason: Both I and II. 1 mole of ${H_2}S{O_4}$ neutralizes 2 moles of NaOH, and 1 mole of ${H_2}S{O_4}$ is equal to 2 equivalents.

Step-by-Step Solution:

Step 1: Analyze Statement I.\nStatement I: 1 mole of ${H_2}S{O_4}$ is neutralised by 2 mol of NaOH but 1 equivalent of ${H_2}S{O_4}$ is neutralised by 1 equivalent of NaOH
${H_2}S{O_4} + 2NaOH \rightarrow N{a_2}S{O_4} + 2{H_2}O$ . This statement is correct.

Step 2: Analyze Statement II.\nStatement II : Equivalent amount of ${H_2}S{O_4}$ is the quantity that is equal to 1 g H and hence, each mol of ${H_2}S{O_4}$ is equal to 2 equivalents of ${H_2}S{O_4}$ \nEquivalent weight of ${H_2}S{O_4} = \frac{{Molecular Weight}}{{Basicity}} = \frac{{98}}{2} = 49$ . This statement is also correct.

Step 3: Determine the correctness of both statements.\nBoth statements are correct.

Why other options are wrong:
- Option A: Statement II is also correct.
- Option B: Statement I is also correct.
- Option D: Both statements are correct.

Key Concepts:
- Neutralization Reactions
- Equivalent Weight
- Molarity and Normality

Memory Aids:
- Basicity of an acid is the number of replaceable hydrogen ions.

Quick Tips:
- Remember the relationship between molarity and normality.

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Application +4 | -1

The total number of valence electrons in

33.6 \mathrm{~g} \mathrm{~N}_3^{-}

ion is

View answer and explanation

Correct option: C

12.8 \mathrm{~N}_{\mathrm{A}}

The correct answer is 12.8 NA because the total number of valence electrons in 33.6 g of N3- ion is 12.8 times Avogadro's number, calculated based on the molar mass and valence electrons of N3-.

Solution approach

Correct Answer + Core Reason: $12.8 \mathrm{{N}}_{\mathrm{{A}}}$ . The total number of valence electrons in 33.6 g of N3- ion is 12.8 times Avogadro's number, calculated based on the molar mass and valence electrons of N3-.

Step-by-Step Solution:\nStep 1: Calculate the molar mass of N3-: $3 \times 14 = 42 \text{{ g/mol}}$ \nStep 2: Calculate the number of moles of N3-: $\frac{{33.6 \text{{ g}}}}{{42 \text{{ g/mol}}}} = 0.8 \text{{ mol}}$ \nStep 3: Determine the number of valence electrons in N3-: Each nitrogen atom has 5 valence electrons, so $3 \times 5 = 15$ . The ion has a -1 charge, so add 1 electron: $15 + 1 = 16$ valence electrons.\nStep 4: Calculate the total number of valence electrons: $0.8 \text{{ mol}} \times 16 \frac{{\text{{electrons}}}}{{\text{{molecule}}}} \times N_A = 12.8 N_A$

Why other options are wrong:
- Option A: Incorrect, as it doesn't match the calculated number of valence electrons based on the given mass and the number of valence electrons in N3-.
- Option B: Incorrect, as it doesn't match the calculated number of valence electrons based on the given mass and the number of valence electrons in N3-.
- Option C: Incorrect, as it doesn't match the calculated number of valence electrons based on the given mass and the number of valence electrons in N3-.
- Option D: Incorrect, as it doesn't match the calculated number of valence electrons based on the given mass and the number of valence electrons in N3-.

Key Concepts:
- Molar mass
- Number of moles
- Valence electrons
- Avogadro's number

Memory Aids:
- Remember that Avogadro's number is the number of entities in one mole of a substance.

Quick Tips:
- Always consider the charge of an ion when determining the number of valence electrons.

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Application +4 | -1

The strengths of 5.6 volume hydrogen peroxide (of density 1 g/mL ) in terms of mass percentage and molarity

(M),

respectively are (Take molar mass of hydrogen peroxide as 34 g/mol

View answer and explanation

Correct option: C

1.7 and 0.5

Option C (1.7 and 0.5) is correct because the mass percentage and molarity are correctly calculated based on the volume strength of hydrogen peroxide.

Solution approach

Correct Answer + Core Reason: 1.7 and 0.5. The mass percentage and molarity are calculated using the volume strength of hydrogen peroxide.

Step-by-Step Solution:

Step 1: Calculate the molarity (M) using the formula: Volume strength = 11.2 × M.\nGiven volume strength = 5.6, so $M = \frac{{5.6}}{{11.2}} = 0.5 \,M$ .

Step 2: Assume 1 L of solution. The mass of the solution is 1000 g (since density = 1 g/mL).

Step 3: Calculate the mass of ${H_2}{O_2}$ in 1 L of solution. Moles of ${H_2}{O_2} = 0.5 \,mol$ . Molar mass of ${H_2}{O_2} = 34 \,g/mol$ .\nMass of ${H_2}{O_2} = 0.5 \times 34 = 17 \,g$ .

Step 4: Calculate the mass percentage of ${H_2}{O_2}$ :\nMass percentage = $\frac{{Mass\,of\,{H_2}{O_2}}}{{Mass\,of\,solution}} \times 100 = \frac{{17}}{{1000}} \times 100 = 1.7\%$ .

Why other options are wrong:
- Option A (0.85 and 0.25): This is incorrect. The molarity and mass percentage are both calculated incorrectly.
- Option B (0.85 and 0.5): This is incorrect. The mass percentage is calculated incorrectly.
- Option D (1.7 and 0.25): This is incorrect. The molarity is calculated incorrectly.

Key Concepts:
- Volume strength of hydrogen peroxide
- Molarity calculation
- Mass percentage calculation

Memory Aids:
- Volume strength = 11.2 × M

Quick Tips:
- Remember the relationship between volume strength and molarity.
- Assume a convenient volume (e.g., 1 L) to simplify calculations.

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Application +4 | -1

For the reaction,

2\mathrm{F}{\mathrm{e}^{3 + }} + \mathrm{S}{\mathrm{n}^{2 + }} \to 2\mathrm{F}{\mathrm{e}^{2 + }} + \mathrm{S}{\mathrm{n}^{4 + }}

The normality of

\mathrm{SnC}{\mathrm{l}_2}

(mol.wt. = 189.7) solution prepared by dissolving 47.5 g in acid solution and diluting with

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

to a total of 2.25 litre is :

View answer and explanation

Correct option: A

0.222

N

Option A (0.222

N

) is correct because the normality is calculated by dividing the given mass by the equivalent weight and the volume of the solution.

Solution approach

Correct Answer + Core Reason: 0.222 $N$ . The normality is calculated using the formula $N = \frac{{\text{{mass}}}}{{\text{{equivalent weight}} \times \text{{volume}}}}$ .

Step-by-Step Solution:

Step 1: Identify the given reaction: $2\mathrm{F}{\mathrm{e}^{3 + }} + \mathrm{S}{\mathrm{n}^{2 + }} \to 2\mathrm{F}{\mathrm{e}^{2 + }} + \mathrm{S}{\mathrm{n}^{4 + }}$ .

Step 2: Determine the n-factor for $SnCl_2$ . In this reaction, $Sn^{2+}$ is oxidized to $Sn^{4+}$ , so the change in oxidation state is 2. Therefore, the n-factor is 2.

Step 3: Calculate the equivalent weight of $SnCl_2$ :\nEquivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{n-factor}}}} = \frac{{189.7}}{2}$ .

Step 4: Calculate the normality of the $SnCl_2$ solution:\nNormality (N) = $\frac{{\text{{mass of solute}}}}{{\text{{Equivalent weight}} \times \text{{Volume of solution in liters}}}}$
$N = \frac{{47.5}}{{\frac{{189.7}}{2} \times 2.25}} = \frac{{47.5 \times 2}}{{189.7 \times 2.25}} = \frac{{95}}{{426.825}} = 0.222 N$ .

Why other options are wrong:
- **Option B (0.111 $N$ ): This is incorrect because it does not account for the n-factor of 2 in the calculation of normality.
- Option C (0.333 $N$ ): This is incorrect and likely arises from a calculation error or using the wrong formula.
- Option D (0.444 $N$ ): This is incorrect and likely arises from a calculation error or using the wrong formula.

Key Concepts:
- Normality
- Equivalent weight
- Redox reactions

Memory Aids:**
- Normality = $\frac{{\text{{grams of solute}}}}{{\text{{Equivalent weight}} \times \text{{Volume in liters}}}}$

Quick Tips:
- Determine the n-factor correctly based on the change in oxidation state.

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Analyzing +4 | -1

Statement I : Values of R can be used to convert one energy parameter into another
Statement II : R has the dimension of work

{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\,\,{{\text{K}}^{{\text{ - 1}}}}

.
Which statement(s) is/are correct?

View answer and explanation

Correct option: B

Only II

Option B (Only II) is correct because Statement II accurately describes the dimensions of the ideal gas constant R, while Statement I is too general and not always applicable.

Solution approach

Correct Answer + Core Reason: Only II. Statement II correctly describes the dimensions of R, while Statement I is not always true.

Step-by-Step Solution:

Step 1: Analyze Statement I: Values of R can be used to convert one energy parameter into another.\nWhile R is related to energy, it doesn't directly convert all energy parameters. It's primarily used in equations involving gases and thermodynamics. This statement is too general and not always applicable.

Step 2: Analyze Statement II: R has the dimension of work ${\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\,{{\text{K}}^{{\text{ - 1}}}}$ .\nFrom the ideal gas law, $PV = nRT$ , so $R = \frac{PV}{nT}$ .\nPressure (P) has dimensions of force per area ( $F/A$ ), and volume (V) has dimensions of area times length ( $AL$ ). Therefore, $PV$ has dimensions of force times length ( $FL$ ), which is work or energy.\nn (number of moles) has units of mol, and T (temperature) has units of K.\nThus, R has dimensions of $\frac{\text{work}}{\text{mol} \cdot \text{K}}$ , or work ${\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\,{{\text{K}}^{{\text{ - 1}}}}$ . This statement is correct.

Step 3: Determine the correct option: Since Statement I is not always correct and Statement II is correct, the correct option is Only II.

Why other options are wrong:
- Option A (Only I): Incorrect because Statement I is not always correct.
- Option C (Both I and II): Incorrect because Statement I is not always correct.
- Option D (None of these): Incorrect because Statement II is correct.

Key Concepts:
- Ideal gas law
- Dimensions of physical quantities
- Gas constant (R)

Memory Aids:
- Remember the ideal gas law: $PV = nRT$ .

Quick Tips:
- Analyze the dimensions of physical quantities to check the validity of equations.

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Application +4 | -1

\mathrm{NaClO}

solution reacts with

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3},as,\mathrm{NaClO} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3} \to \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

A solution of

\mathrm{NaClO}

used in the above reaction contained 15 g of

\mathrm{NaClO}

per litre. The normality of the solution would be :

View answer and explanation

Correct option: A

0.4

Option A (0.4) is correct because the normality is calculated by dividing the mass of NaClO per liter by its equivalent weight.

Solution approach

Correct Answer + Core Reason: 0.4. The normality is calculated by dividing the mass of $\mathrm{NaClO}$ per liter by its equivalent weight.

Step-by-Step Solution:

Step 1: Write the reaction: $\mathrm{NaClO} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3} \to \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$

Step 2: Determine the change in oxidation state of chlorine.\nIn $\mathrm{NaClO}$ , the oxidation state of Cl is +1.\nIn $\mathrm{NaCl}$ , the oxidation state of Cl is -1.\nTherefore, the change in oxidation state is 2.

Step 3: Calculate the equivalent weight of $\mathrm{NaClO}$ .\nEquivalent weight = (Molecular weight) / (Change in oxidation state)\nMolecular weight of $\mathrm{NaClO}$ = 23 + 35.5 + 16 = 74.5 g/mol\nEquivalent weight = 74.5 / 2 = 37.25 g/equivalent

Step 4: Calculate the normality of the solution.\nNormality = (Mass of $\mathrm{NaClO}$ per liter) / (Equivalent weight)\nNormality = 15 g/L / 37.25 g/equivalent = 0.4 N

Why other options are wrong:
- Option B (0.2): This is incorrect because it might be calculated using the molar mass directly instead of the equivalent mass.
- Option C (0.6): This is incorrect and may result from an error in calculating the equivalent weight or in the division.
- Option D (0.8): This is incorrect and may result from an error in calculating the equivalent weight or in the division.

Key Concepts:
- Normality
- Equivalent Weight
- Redox Reactions

Memory Aids:
- Normality = (Number of gram equivalents of solute) / (Volume of solution in liters)

Quick Tips:
- Always determine the change in oxidation state to calculate the equivalent weight correctly.
- Remember to divide the molecular weight by the change in oxidation state to get the equivalent weight.

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Remembering +4 | -1

The prefix

{10^{18}}

is

View answer and explanation

Correct option: C

exa

Option C (exa) is correct because the prefix exa represents

{10^{18}}

.

Solution approach

Correct Answer + Core Reason: exa. The prefix exa corresponds to ${10^{18}}$ .

Step-by-Step Solution:
Step 1: Recall the standard prefixes for powers of 10.
Step 2: Match the given power, ${10^{18}}$ , with its corresponding prefix.
Step 3: The prefix for ${10^{18}}$ is exa.

Why other options are wrong:
- Option A: giga represents ${10^9}$ .
- Option B: kilo represents ${10^3}$ .
- Option C: exa represents ${10^{18}}$ .
- Option D: nano represents ${10^{-9}}$ .

Key Concepts:
- SI prefixes
- Powers of 10

Memory Aids:
- Use a table of SI prefixes to memorize the values.

Quick Tips:
- Familiarize yourself with common SI prefixes and their corresponding powers of 10.

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Analyzing +4 | -1

Which of the following is not a disproportion atom reaction

View answer and explanation

Correct option: D

{F_2} + O{H^ - } \to {F^ - } + O{F_2} + {H_2}O

Option D (

{F_2} + O{H^ - } \to {F^ - } + O{F_2} + {H_2}O

) is correct because fluorine cannot exhibit a positive oxidation state, so it cannot be both oxidized and reduced in the same reaction.

Solution approach

Correct Answer + Core Reason: ${F_2} + O{H^ - } \to {F^ - } + O{F_2} + {H_2}O$ . Fluorine does not show disproportionation as it cannot exhibit any positive oxidation state.

Step-by-Step Solution:

Step 1: Define disproportionation reaction: A reaction in which the same element is both oxidized and reduced.

Step 2: Analyze each reaction to determine if disproportionation occurs.

Step 3: Reaction A: $C{l_2} + O{H^ - } \to C{l^ - } + Cl{O^ - }_3 + {H_2}O$ . Chlorine goes from 0 to -1 (reduction) and 0 to +5 (oxidation). Disproportionation.

Step 4: Reaction B: ${P_4} + O{H^ - } + {H_2}O \to P{H_3} + {H_2}P{O^ - }_2$ . Phosphorus goes from 0 to -3 (reduction) and 0 to +1 (oxidation). Disproportionation.

Step 5: Reaction C: ${S_8} + O{H^ - } \to 4{S^2}^ - + 2\,{S_2}{O_3}^{2 - } + 6{H_2}O$ . Sulfur goes from 0 to -2 (reduction) and 0 to +2 (oxidation). Disproportionation.

Step 6: Reaction D: ${F_2} + O{H^ - } \to {F^ - } + O{F_2} + {H_2}O$ . Fluorine goes from 0 to -1 (reduction) and 0 to +2 (oxidation). However, fluorine is the most electronegative element and cannot have a positive oxidation state. Therefore, this is not a disproportionation reaction.

Why other options are wrong:
- **Option A ( $C{l_2} + O{H^ - } \to C{l^ - } + Cl{O^ - }_3 + {H_2}O$ ): Chlorine undergoes disproportionation.
- Option B ( ${P_4} + O{H^ - } + {H_2}O \to P{H_3} + {H_2}P{O^ - }_2$ ): Phosphorus undergoes disproportionation.
- Option C ( ${S_8} + O{H^ - } \to 4{S^2}^ - + 2\,{S_2}{O_3}^{2 - } + 6{H_2}O$ ): Sulfur undergoes disproportionation.

Key Concepts:
- Disproportionation reaction
- Oxidation states
- Electronegativity

Memory Aids:
- Fluorine is the most electronegative element.

Quick Tips:**
- Identify the oxidation states of the element before and after the reaction.
- Remember that fluorine cannot have a positive oxidation state.

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Understanding +4 | -1

Assertion (A): Laboratory reagents are usually made up to a specific molarity rather than a given molality.

Reason (R): The volume of a liquid is more easily measured than its mass.

View answer and explanation

Correct option: A

Both (A) & (R) are true and the (R) is the correct explanation of the (A)

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because molarity is preferred due to the ease of volume measurement, which is the reason why laboratory reagents are made to specific molarities.

Solution approach

Correct Answer + Core Reason: Both (A) & (R) are true and the (R) is the correct explanation of the (A). Molarity is preferred due to the ease of volume measurement, which is the reason why laboratory reagents are made to specific molarities.

Step-by-Step Solution:

Step 1: Understand the definitions of molarity and molality.\nMolarity (M) is defined as moles of solute per liter of solution.\nMolality (m) is defined as moles of solute per kilogram of solvent.

Step 2: Consider the ease of measurement. Measuring the volume of a liquid is generally easier and more precise than measuring its mass in a laboratory setting.

Step 3: Relate the ease of measurement to the choice of concentration unit. Because volume is easier to measure, molarity is more convenient for preparing solutions in the lab.

Step 4: Evaluate the assertion and reason. The assertion is true because laboratory reagents are usually made to specific molarities. The reason is also true because volume is easier to measure than mass. The reason correctly explains the assertion.

Why other options are wrong:
- Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)): The reason *is* the correct explanation.
- Option C ((A) is true but (R) is false): Both the assertion and reason are true.
- Option D (Both (A) and (R) are false): Both the assertion and reason are true.

Key Concepts:
- Molarity
- Molality
- Solution preparation

Memory Aids:
- Molarity: Moles per Liter; Molality: Moles per kg

Quick Tips:
- Remember the definitions of molarity and molality and their practical implications in the lab.

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Application +4 | -1

If

N_A

is Avogadro's number then number of valence electrons in

4.2 \mathrm{~g}

of nitride ions

\left(\mathrm{N}^{3-}\right)

is

View answer and explanation

Correct option: B

2.4 \mathrm{~N}_{\mathrm{A}}

Option B (

2.4 N_A

) is correct because each nitride ion has 8 valence electrons, and the number of moles of nitride ions is 0.3.

Solution approach

Correct Answer + Core Reason: $2.4 N_A$ . The number of valence electrons is calculated by finding the moles of nitride ions and multiplying by the number of valence electrons per ion.

Step-by-Step Solution:

Step 1: Calculate the number of moles of nitride ions ( $N^{3-}$ ): Moles = $\frac{{mass}}{{molar \, mass}} = \frac{{4.2 \, g}}{{14 \, g/mol}} = 0.3 \, mol$

Step 2: Convert moles to number of ions: Number of ions = Moles $\times$ Avogadro's number = $0.3 \times N_A$

Step 3: Determine the number of valence electrons per nitride ion. Nitrogen has 5 valence electrons. With a 3- charge, it gains 3 more electrons, resulting in 5 + 3 = 8 valence electrons.

Step 4: Calculate the total number of valence electrons: Total valence electrons = (Number of ions) $\times$ (Valence electrons per ion) = $0.3 N_A \times 8 = 2.4 N_A$

Why other options are wrong:
- **Option A ( $4.2 N_A$ ): This is incorrect because it directly uses the mass of the nitride ions without considering the number of valence electrons per ion.
- Option C ( $1.6 N_A$ ): This is incorrect and doesn't account for the 8 valence electrons in the nitride ion. It might be a result of using an incorrect number of valence electrons or a calculation error.
- Option D ( $3.2 N_A$ ): This is incorrect and likely a calculation error or misunderstanding of how to calculate the total number of valence electrons.

Key Concepts:
- Moles
- Avogadro's number
- Valence electrons
- Ionic charge

Memory Aids:**
- Moles = Mass / Molar mass
- Number of particles = Moles $\times$ Avogadro's number

Quick Tips:
- Remember to account for the charge when determining the number of valence electrons in an ion.
- Always convert mass to moles before calculating the number of particles.

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Application +4 | -1

Number of significant figures in the answer of following will be:

\frac{(29.2-20.2)\left(1.79 \times 10^5\right)}{1.39}

View answer and explanation

Correct option: B

3

Option B (3) is correct because the result of the calculation has three significant figures, determined by the least precise number in the calculation.

Solution approach

Correct Answer + Core Reason: 3. The number of significant figures is determined by the least number of significant figures in the calculation.

Step-by-Step Solution:

Step 1: Calculate the subtraction in the numerator: 29.2 - 20.2 = 9.0. This has 2 significant figures.

Step 2: Identify the number of significant figures in 1.79 × 10⁵: This has 3 significant figures.

Step 3: Identify the number of significant figures in 1.39: This has 3 significant figures.

Step 4: Perform the calculation: (9.0 * 1.79 × 10⁵) / 1.39 = (16.11 × 10⁵) / 1.39 = 11.59 × 10⁵. Since 9.0 has the least number of significant figures (2), the result should be rounded to 2 significant figures after multiplication and division. However, the intermediate result (9.0 * 1.79 × 10⁵) has to be considered. The result is 16.11 × 10⁵. When dividing by 1.39, which has 3 significant figures, the final result should have 3 significant figures. So, the correct approach is to round the final answer to 3 significant figures.

Step 5: Round the result to the least number of significant figures (which is effectively 3 due to the intermediate calculation): 11.59 × 10⁵ becomes 1.16 × 10⁶. So, the number of significant figures is 3.

Step 6: Final Answer: 3

Why other options are wrong:
- Option A: 4: This is incorrect because it doesn't consider the least number of significant figures in the subtraction result.
- Option C: 2: This is incorrect because it only considers the subtraction result and doesn't account for the other numbers in the calculation.
- Option D: 1: This is incorrect because it is a gross underestimation of the significant figures.

Key Concepts:
- Significant figures
- Rules for significant figures in calculations

Memory Aids:
- In multiplication and division, the result should have the same number of significant figures as the number with the fewest significant figures.
- In addition and subtraction, the result should have the same number of decimal places as the number with the fewest decimal places.

Quick Tips:
- Always perform the calculations first and then round the final answer to the appropriate number of significant figures.
- Pay attention to the number with the least significant figures.

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Application +4 | -1

How much

\mathrm{Ca}\left(\mathrm{NO}_3\right)_2,

in

\mathrm{mg},

must be present in

50 \mathrm{~ml}

of a solution with

2.35 \mathrm{~ppm}

of

\mathrm{Ca}?

View answer and explanation

Correct option: D

0.48

Option D (0.48) is correct because it represents the calculated mass of

\mathrm{Ca(NO_3)_2}

in mg, present in the given solution based on the ppm of Ca.

Solution approach

Correct Answer + Core Reason: 0.48. The mass of $\mathrm{Ca(NO_3)_2}$ is calculated from the given ppm of Ca in the solution.

Step-by-Step Solution:

Step 1: Convert ppm of Ca to mass of Ca in the solution: $2.35 \text{{ ppm}} = \frac{{2.35 \text{{ g Ca}}}}{{10^6 \text{{ g solution}}}}$

Step 2: Calculate the mass of Ca in 50 ml of solution (assuming density of solution is 1 g/ml, so 50 ml = 50 g): $\frac{{2.35 \text{{ g Ca}}}}{{10^6 \text{{ g solution}}}} \times 50 \text{{ g solution}} = 1.175 \times 10^{-4} \text{{ g Ca}}$

Step 3: Calculate the molar mass of Ca: $40 \text{{ g/mol}}$

Step 4: Calculate the molar mass of $\mathrm{Ca(NO_3)_2}$ : $40 + 2(14 + 3(16)) = 40 + 2(62) = 40 + 124 = 164 \text{{ g/mol}}$

Step 5: Calculate the mass of $\mathrm{Ca(NO_3)_2}$ corresponding to the mass of Ca: $\frac{{1.175 \times 10^{-4} \text{{ g Ca}}}}{{40 \text{{ g/mol Ca}}}} \times 164 \text{{ g/mol Ca(NO_3)_2}} = 4.8175 \times 10^{-4} \text{{ g Ca(NO_3)_2}}$

Step 6: Convert the mass to mg: $4.8175 \times 10^{-4} \text{{ g}} \times 1000 \text{{ mg/g}} = 0.48175 \text{{ mg}}$ (approximately 0.48 mg)

Why other options are wrong:
- Option A: 0.1175 is incorrect. This might be the mass of Ca in grams, but it's not the mass of $\mathrm{Ca(NO_3)_2}$ in mg.
- Option B: 770.8 is incorrect. This value is significantly larger and doesn't follow from the correct calculation.
- Option C: 4.7 is incorrect. This value is off by an order of magnitude and doesn't follow the correct calculation.

Key Concepts:
- Parts per million (ppm)
- Stoichiometry
- Molar Mass

Memory Aids:
- Remember that ppm is a ratio of mass of solute to mass of solution.

Quick Tips:
- Pay attention to units and convert them appropriately.

254 Medium

254 of 272

Application +4 | -1

What volume of 0.40

M

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}

would be required to react with the

{\mathrm{I}_2}

liberated by adding excess of KI to 50 mL of 0.20

M

\mathrm{CuS}{\mathrm{O}_4}

solution?

View answer and explanation

Correct option: B

25 mL

Option B (25 mL) is correct because the milliequivalents of Na2S2O3 equals the milliequivalents of CuSO4.

Solution approach

Correct Answer + Core Reason: 25 mL. The volume is calculated by equating the milliequivalents of $Na_2S_2O_3$ and $CuSO_4$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equations:
$2CuSO_4 + 4KI \rightarrow 2CuI + I_2 + 2K_2SO_4$
$I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$

Step 2: Determine the milliequivalents of $CuSO_4$ : $Meq \, of \, CuSO_4 = Volume \times Molarity \times n-factor = 50 \, mL \times 0.20 \, M \times 1 = 10 \, meq$
(n-factor for CuSO4 is 1 because Cu2+ changes to Cu+)

Step 3: Determine the milliequivalents of $Na_2S_2O_3$ : $Meq \, of \, Na_2S_2O_3 = Volume \times Molarity \times n-factor = V \times 0.40 \, M \times 1$
(n-factor for Na2S2O3 is 1 because S changes from +2 to +2.5, change of 0.5 per S atom, and there are 2 S atoms)

Step 4: Equate the milliequivalents: $Meq \, of \, Na_2S_2O_3 = Meq \, of \, CuSO_4$
$V \times 0.40 \times 1 = 10$

Step 5: Solve for V: $V = \frac{10}{0.40} = 25 \, mL$

Why other options are wrong:
- Option A (12.5 mL): This is incorrect. It might be a miscalculation of the stoichiometry or n-factors.
- Option C (50 mL): This is incorrect. It might be a miscalculation of the stoichiometry or n-factors.
- Option D (2.5 mL): This is incorrect. It might be a miscalculation of the stoichiometry or n-factors.

Key Concepts:
- Redox titrations
- Milliequivalents
- Stoichiometry

Memory Aids:
- Remember to balance the chemical equations before performing calculations.

Quick Tips:
- Always determine the n-factors correctly for each species involved in the redox reaction.

255 Medium

255 of 272

Application +4 | -1

What is the percentage of 'free

\mathrm{SO}_3

' in a sample of oleum labelled as

104.5 \%?

View answer and explanation

Correct option: A

20 \%

Option A (20%) is correct because the percentage of free SO3 is calculated based on the excess percentage of oleum.

Solution approach

Correct Answer + Core Reason: $20 \%$ . The percentage of free $SO_3$ is calculated based on the excess percentage of oleum.

Step-by-Step Solution:

Step 1: Understand the concept of oleum labeling. Oleum is labeled based on the percentage of $H_2SO_4$ that can be formed when $SO_3$ is dissolved in water.

Step 2: The oleum is labeled as 104.5%, which means there is 4.5% excess $SO_3$ .

Step 3: The reaction is $SO_3 + H_2O \rightarrow H_2SO_4$ .

Step 4: Calculate the percentage of free $SO_3$ : $\text{{Percentage of free SO}}_3 = \frac{80}{18} \times 4.5 = 20 \%$ .

Why other options are wrong:
- Option B (40 %): Incorrect calculation of the percentage of free $SO_3$ .
- Option C (60 %): Incorrect calculation of the percentage of free $SO_3$ .
- Option D (80 %): Incorrect calculation of the percentage of free $SO_3$ .

Key Concepts:
- Oleum
- Chemical Composition
- Percentage Composition

Memory Aids:
- Oleum contains $SO_3$ dissolved in $H_2SO_4$ .

Quick Tips:
- Remember the formula for calculating the percentage of free $SO_3$ in oleum.

256 Medium

256 of 272

Analyzing +4 | -1

A closed vessel contains equal numbers of

{O_2}

and

{H_2}

molecules at same

T

. Which of the following is not true?

View answer and explanation

Correct option: D

The weight of

{H_2}

is the same as the weight of oxygen

Option D (The weight of

{H_2}

is the same as the weight of oxygen) is correct because if the number of molecules is equal, the weights are not the same due to different molar masses.

Solution approach

Correct Answer + Core Reason: The weight of ${H_2}$ is the same as the weight of oxygen. Equal number of molecules implies unequal weights due to different molar masses.

Step-by-Step Solution:

Step 1: Analyze the given information: Equal numbers of ${O_2}$ and ${H_2}$ molecules at the same temperature (T).

Step 2: Consider the average speed of gas molecules: $v_{avg} = \sqrt{{\frac{{8RT}}{{\pi M}}}}$ where M is the molar mass. Since the molar mass of ${H_2}$ is less than that of ${O_2}$ , the average speed of ${H_2}$ is greater.

Step 3: Consider the frequency of collisions with the walls: The lighter ${H_2}$ molecules will strike the walls more often due to their higher average speed.

Step 4: Consider the average kinetic energy: At the same temperature, the average kinetic energy of the two gases is the same ( $\frac{{3}}{{2}}kT$ ).

Step 5: Consider the weights: Since the number of molecules is the same, the number of moles is the same. Let n be the number of moles. Then, the weight of ${O_2}$ is $32n$ and the weight of ${H_2}$ is $2n$ . Therefore, the weights are not the same.

Step 6: Conclusion: The statement "The weight of ${H_2}$ is the same as the weight of oxygen" is not true.

Why other options are wrong:
- Option A: The average speed of hydrogen molecules is greater due to their lower molar mass.
- Option B: The hydrogen molecules strike the walls of the vessel more often due to their higher speed.
- Option C: The average kinetic energy of the two gases is the same because they are at the same temperature.

Key Concepts:
- Kinetic theory of gases
- Average speed of gas molecules
- Average kinetic energy of gas molecules
- Molar mass

Memory Aids:
- Kinetic energy depends only on temperature.
- Lighter molecules move faster at the same temperature.

Quick Tips:
- Remember the relationship between kinetic energy, temperature, and molar mass.

257 Medium

257 of 272

Application +4 | -1

When

\mathrm{BrO}_3^{-}

ion reacts with

\mathrm{Br}^{-}

in acid medium,

\mathrm{Br}_2

is liberated. The equivalent weight of

\mathrm{Br}_2

in this reaction is:

View answer and explanation

Correct option: C

\frac{3 M}{5}

Option C (

\frac{3 M}{5}

) is correct because the equivalent weight of Br2 is calculated based on the change in oxidation state during the reaction.

Solution approach

Correct Answer + Core Reason: $\frac{3 M}{5}$ . The equivalent weight of Br2 is calculated based on the change in oxidation state.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation.
$2 \mathrm{BrO}_3^{-} + 12 \mathrm{H}^{+} + 10 \mathrm{Br}^{-} \longrightarrow 6 \mathrm{Br}_2 + 6 \mathrm{H}_2\mathrm{O}$

Step 2: Determine the change in oxidation state of bromine.\nIn $\mathrm{BrO}_3^{-}$ , the oxidation state of Br is +5.\nIn $\mathrm{Br}^{-}$ , the oxidation state of Br is -1.\nIn $\mathrm{Br}_2$ , the oxidation state of Br is 0.

Step 3: Determine the n-factor for $\mathrm{Br}_2$ .
10 moles of electrons are required for the formation of 6 moles of $\mathrm{Br}_2$ .\nTherefore, the n-factor for $\mathrm{Br}_2$ is $\frac{10}{6} = \frac{5}{3}$ .\nAlternatively, consider the change in oxidation state for each Br in $\mathrm{Br}_2$ formed from $\mathrm{Br}^-$ : $0 - (-1) = 1$ . Since there are 2 Br atoms in $\mathrm{Br}_2$ , the change is 2. For $\mathrm{Br}_2$ formed from $\mathrm{BrO}_3^-$ , the change in oxidation state is $5 - 0 = 5$ . So, the total change is $5+1+1+1+1+1+1+1+1+1 = 10$ for 6 moles of $\mathrm{Br}_2$ . Thus, for 1 mole of $\mathrm{Br}_2$ , the n-factor is $\frac{10}{6} = \frac{5}{3}$ .

Step 4: Calculate the equivalent weight of $\mathrm{Br}_2$ .\nEquivalent weight = $\frac{\text{Molar mass}}{\text{n-factor}} = \frac{M}{5/3} = \frac{3M}{5}$

Why other options are wrong:
- **Option A ( $\frac{5 M}{8}$ ):** This is incorrect due to an error in calculating the n-factor for $\mathrm{Br}_2$ .
- **Option B ( $\frac{5 M}{3}$ ): This is incorrect because it uses the reciprocal of the correct n-factor in the equivalent weight calculation.
- Option D ( $\frac{4 M}{6}$ ):** This is incorrect due to an error in calculating the n-factor for $\mathrm{Br}_2$ .

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

Memory Aids:
- Equivalent weight = $\frac{\text{Molar mass}}{\text{n-factor}}$

Quick Tips:
- Always balance the chemical equation before determining the n-factor.
- Identify the change in oxidation state correctly to determine the n-factor accurately.

258 Medium

258 of 272

Application +4 | -1

30 \mathrm{~mL}

of

0.2 \mathrm{N} \mathrm{~BaC}{l}

is mixed with

40 \mathrm{~mL}

of

0.3 \mathrm{N} \mathrm{~A}{l}_2\left(\mathrm{SO}_4\right)_3

. How many

\mathrm{g}

of

\mathrm{BaSO}_4

are formed?

View answer and explanation

Correct option: D

0.70 \mathrm{~g}

Option D (

0.70 \mathrm{~g}

) is correct because the amount of BaSO4 formed is determined by the limiting reactant, which in this case is BaCl2.

Solution approach

Correct Answer + Core Reason: $0.70 \mathrm{~g}$ . The amount of BaSO4 formed is determined by the limiting reactant.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the reaction:
$3 \mathrm{BaCl}_2 + \mathrm{Al}_2(\mathrm{SO}_4)_3 \rightarrow 3 \mathrm{BaSO}_4 + 2 \mathrm{AlCl}_3$

Step 2: Calculate the milliequivalents (meq) of $\mathrm{BaCl}_2$ :
$\text{meq of } \mathrm{BaCl}_2 = \text{Volume (mL)} \times \text{Normality (N)} = 30 \text{ mL} \times 0.2 \text{ N} = 6 \text{ meq}$

Step 3: Calculate the milliequivalents (meq) of $\mathrm{Al}_2(\mathrm{SO}_4)_3$ :
$\text{meq of } \mathrm{Al}_2(\mathrm{SO}_4)_3 = \text{Volume (mL)} \times \text{Normality (N)} = 40 \text{ mL} \times 0.3 \text{ N} = 12 \text{ meq}$

Step 4: Determine the limiting reactant. From the balanced equation, 3 moles of $\mathrm{BaCl}_2$ react with 1 mole of $\mathrm{Al}_2(\mathrm{SO}_4)_3$ . Therefore, 6 meq of $\mathrm{BaCl}_2$ will react with $\frac{6}{3} = 2$ meq "units" of $\mathrm{Al}_2(\mathrm{SO}_4)_3$ . Since we have 12 meq of $\mathrm{Al}_2(\mathrm{SO}_4)_3$ , $\mathrm{BaCl}_2$ is the limiting reactant.

Step 5: Calculate the mass of $\mathrm{BaSO}_4$ formed. Since 3 moles of $\mathrm{BaSO}_4$ are formed from 3 moles of $\mathrm{BaCl}_2$ , the number of meq of $\mathrm{BaSO}_4$ formed is equal to the number of meq of $\mathrm{BaCl}_2$ used, which is 6 meq.

Step 6: Calculate the equivalent weight of $\mathrm{BaSO}_4$ . The molar mass of $\mathrm{BaSO}_4$ is 233 g/mol. Since the valency of BaSO4 is 2, the equivalent weight is $\frac{233}{2} = 116.5 \text{ g/eq}$

Step 7: Calculate the mass of $\mathrm{BaSO}_4$ formed: $\text{Mass} = \frac{\text{meq} \times \text{Equivalent weight}}{1000} = \frac{6 \times 116.5}{1000} = 0.699 \approx 0.70 \text{ g}$

Why other options are wrong:
- **Option A ( $0.10 \mathrm{~g}$ ): Incorrect. This is likely a calculation error or misunderstanding of stoichiometry.
- Option B ( $0.60 \mathrm{~g}$ ): Incorrect. This is likely a calculation error or misunderstanding of stoichiometry.
- Option C ( $0.90 \mathrm{~g}$ ): Incorrect. This is likely a calculation error or misunderstanding of stoichiometry.

Key Concepts:
- Stoichiometry
- Limiting reactant
- Equivalent weight
- Milliequivalents

Memory Aids:**
- $\text{meq} = \text{Volume (mL)} \times \text{Normality (N)}$
- $\text{Mass} = \frac{\text{meq} \times \text{Equivalent weight}}{1000}$

Quick Tips:
- Always balance the chemical equation first.
- Identify the limiting reactant.
- Use milliequivalents for easier calculations.

259 Medium

259 of 272

Application +4 | -1

The weight of

1 \times 10^{22}

molecules of

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

is

View answer and explanation

Correct option: C

4.159 \mathrm{~g}

Option C (

4.159 \mathrm{~g}

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

Solution approach

Correct Answer + Core Reason: $4.159 \mathrm{~g}$ . This is calculated using the mole concept and molar mass.

Step-by-Step Solution:

Step 1: Calculate the molar mass of $CuSO_4 \cdot 5H_2O$ .\nMolar mass = 63.5 (Cu) + 32 (S) + 4(16) (O) + 5[2(1) (H) + 16 (O)]\nMolar mass = 63.5 + 32 + 64 + 5(2 + 16) = 63.5 + 32 + 64 + 5(18) = 63.5 + 32 + 64 + 90 = 249.5 g/mol

Step 2: Calculate the number of moles of $CuSO_4 \cdot 5H_2O$ .\nMoles = $\frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{1 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0166 moles$

Step 3: Calculate the mass of $CuSO_4 \cdot 5H_2O$ .\nMass = Moles x Molar mass = 0.0166 mol x 249.5 g/mol ≈ 4.14 g

Why other options are wrong:
- Option A: Incorrect. This value is too high.
- Option B: Incorrect. This value is much too high.
- Option D: Incorrect. Option C is the correct answer.

Key Concepts:
- Molar Mass Calculation
- Mole Concept
- Avogadro's Number

Memory Aids:
- Remember: Moles = Mass / Molar mass
- Remember: 1 mole = $6.022 \times 10^{23}$ particles

Quick Tips:
- Always calculate the molar mass correctly before calculating the number of moles or mass.

260 Medium

260 of 272

Application +4 | -1

Volume of

\mathrm{N}_2

required at STP to produce 10 moles of

\mathrm{NH}_3

through Haber's process is

View answer and explanation

Correct option: A

112 \mathrm{~L}

The correct answer is A (

112 \mathrm{~L}

) because 5 moles of N2 are required to produce 10 moles of NH3, and the volume of 5 moles of any gas at STP is 5 * 22.4 L = 112 L.

Solution approach

Correct Answer + Core Reason: $112 \mathrm{~L}$ . The stoichiometry of the Haber process dictates the volume of N2 required.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the Haber's process: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

Step 2: Determine the mole ratio between $N_2$ and $NH_3$ . From the balanced equation, 1 mole of $N_2$ produces 2 moles of $NH_3$ .

Step 3: Calculate the moles of $N_2$ required to produce 10 moles of $NH_3$ . Since the ratio is 1:2, we need $\frac{10}{2} = 5$ moles of $N_2$ .

Step 4: Calculate the volume of 5 moles of $N_2$ at STP. At STP, 1 mole of any gas occupies 22.4 L. Therefore, 5 moles of $N_2$ occupy $5 \times 22.4 \mathrm{~L} = 112 \mathrm{~L}$ .

Why other options are wrong:
- **Option B ( $224 \mathrm{~L}$ ): This is incorrect because it calculates the volume of 10 moles of a gas at STP, instead of the volume of N2 required based on the stoichiometry.
- Option C ( $145 \mathrm{~L}$ ): This value doesn't correspond to any logical calculation based on the stoichiometry or molar volume at STP.
- Option D ( $184 \mathrm{~L}$ ): This value also doesn't correspond to any logical calculation based on the stoichiometry or molar volume at STP.

Key Concepts:
- Stoichiometry
- Haber's process
- Molar volume at STP (22.4 L/mol)

Memory Aids:**
- Remember the balanced equation for the Haber process: $N_2 + 3H_2 \rightarrow 2NH_3$
- Remember the molar volume at STP: 22.4 L/mol

Quick Tips:
- Always start with a balanced chemical equation.
- Use the mole ratio from the balanced equation to determine the amount of reactants needed or products formed.

261 Medium

261 of 272

Application +4 | -1

Volume of one molecule of oxygen gas at S.T.P. is

View answer and explanation

Correct option: A

3.7 \times 10^{-20} {~mL}

Option A (

3.7 \times 10^{-20} {~mL}

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

Solution approach

Correct Answer + Core Reason: $3.7 \times 10^{-20} {~mL}$ . The volume is calculated by dividing the molar volume at STP by Avogadro's number.

Step-by-Step Solution:

Step 1: Volume of one mole of any gas at STP = 22400 mL

Step 2: Number of molecules in one mole = $6.02 \times 10^{23}$

Step 3: Volume of one molecule of $O_2$ at STP = $\frac{{22400}}{{6.02 \times 10^{23}}} mL$

Step 4: Calculate: Volume = $3.7 \times 10^{-20} mL$

Why other options are wrong:
- **Option B ( $2.5 \times 10^{-21} {~mL}$ ): This value is incorrect due to a miscalculation or incorrect application of Avogadro's number.
- Option C ( $3.1 \times 10^{-22} {~mL}$ ): This value is incorrect due to a miscalculation or incorrect application of Avogadro's number.
- Option D ( $6.1 \times 10^{-23} {~mL}$ ): This value is incorrect due to a miscalculation or incorrect application of Avogadro's number.

Key Concepts:
- Molar volume at STP
- Avogadro's number

Memory Aids:
- At STP, 1 mole of any gas occupies 22.4 L or 22400 mL.

Quick Tips:**
- Remember the standard molar volume at STP.

262 Medium

262 of 272

Analyzing +4 | -1

Select the option having maximum number of carbon atoms among the following

View answer and explanation

Correct option: D

0.5 \,{~g}\, {C}_5 {H}_{10}

Option D (0.5 g C5H10) is correct because it contains the maximum number of carbon atoms compared to the other options when calculated based on their molar masses.

Solution approach

Correct Answer + Core Reason: $0.5 \{~g}\, {C}_5 {H}_{10}$ . This compound has the highest number of carbon atoms for the same mass due to its lower molar mass and higher carbon content.

Step-by-Step Solution:

Step 1: Calculate the number of moles of each compound.

Step 2: Molar mass of $\mathrm{CO}_2 = 44 \mathrm{g/mol}$ . Moles of $\mathrm{CO}_2 = \frac{{0.5}}{{44}} = 0.0114 \mathrm{mol}$ . Number of C atoms = $0.0114 \times N_A$

Step 3: Molar mass of $\mathrm{C}_3\mathrm{O}_2 = 68 \mathrm{g/mol}$ . Moles of $\mathrm{C}_3\mathrm{O}_2 = \frac{{0.5}}{{68}} = 0.00735 \mathrm{mol}$ . Number of C atoms = $3 \times 0.00735 \times N_A = 0.022 \times N_A$

Step 4: Molar mass of $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 = 180 \mathrm{g/mol}$ . Moles of $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 = \frac{{0.5}}{{180}} = 0.00278 \mathrm{mol}$ . Number of C atoms = $6 \times 0.00278 \times N_A = 0.0167 \times N_A$

Step 5: Molar mass of $\mathrm{C}_5\mathrm{H}_{10} = 70 \mathrm{g/mol}$ . Moles of $\mathrm{C}_5\mathrm{H}_{10} = \frac{{0.5}}{{70}} = 0.00714 \mathrm{mol}$ . Number of C atoms = $5 \times 0.00714 \times N_A = 0.0357 \times N_A$

Step 6: Compare the number of carbon atoms.

Why other options are wrong:
- Option A: $0.5 \{~g} \, {CO}_2$ has fewer carbon atoms than $0.5 \{~g}\, {C}_5 {H}_{10}$ .
- Option B: $0.5 \{~g} \, {C}_3 {O}_2$ has fewer carbon atoms than $0.5 \{~g}\, {C}_5 {H}_{10}$ .
- Option C: $0.5 \{~g} \, {C}_6 {H}_{12} {O}_6$ has fewer carbon atoms than $0.5 \{~g}\, {C}_5 {H}_{10}$ .

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

Memory Aids:
- Moles = Mass / Molar mass

Quick Tips:
- Calculate the number of moles and then multiply by the number of carbon atoms in each molecule.

263 Medium

263 of 272

Application +4 | -1

If 6.3 g of

NaHC{O_3}

are added to 15.0 g

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

solution, the residue is found to weight 18.0 g. what is the mass of

C{O_2}

released in the reaction?

View answer and explanation

Correct option: B

3.3 g

Option B (3.3 g) is correct because the mass of CO2 released is calculated using the law of conservation of mass.

Solution approach

Correct Answer + Core Reason: 3.3 g. The mass of $CO_2$ released is determined by applying the law of conservation of mass.

Step-by-Step Solution:

Step 1: State the law of conservation of mass.\nMass of reactants = Mass of products

Step 2: Identify the reactants and products.\nReactants: $NaHC{O_3}$ and $CH_3COOH$ \nProducts: Residue and $CO_2$

Step 3: Write the equation representing the mass balance.\nMass of $NaHC{O_3}$ + Mass of $CH_3COOH$ = Mass of residue + Mass of $CO_2$

Step 4: Substitute the given values.
6.3 g + 15.0 g = 18.0 g + Mass of $CO_2$

Step 5: Solve for the mass of $CO_2$ .\nMass of $CO_2$ = 6.3 g + 15.0 g - 18.0 g = 21.3 g - 18.0 g = 3.3 g

Why other options are wrong:
- Option A: This is incorrect. It's not the correct mass difference.
- Option C: This is incorrect. It's not the correct mass difference.
- Option D: This is incorrect. It's not the correct mass difference.

Key Concepts:
- Law of conservation of mass

Memory Aids:
- Mass is conserved in chemical reactions.

Quick Tips:
- Ensure the equation is balanced before applying the law of conservation of mass.

264 Medium

264 of 272

Analyzing +4 | -1

Statement-I: Molarity of a solution does not depend upon temperature whereas molality depends.

Statement-II: Molarity and molality both depend only on the number of moles of solute particles.

View answer and explanation

Correct option: D

Statement II is correct but statement I is incorrect

Option D is correct because molarity depends on volume, which changes with temperature, while molality depends on mass, which is independent of temperature. Also, both depend on the number of moles of solute.

Solution approach

Correct Answer + Core Reason: Statement II is correct but statement I is incorrect. Molarity is temperature-dependent, while molality is not.

Step-by-Step Solution:

Step 1: Define molarity and molality.\nMolarity (M) = moles of solute / liters of solution\nMolality (m) = moles of solute / kilograms of solvent

Step 2: Analyze the temperature dependence.\nVolume changes with temperature, so molarity changes with temperature.\nMass does not change with temperature, so molality does not change with temperature.

Step 3: Analyze the dependence on moles of solute.\nBoth molarity and molality depend on the number of moles of solute.

Step 4: Evaluate the statements.\nStatement I: Molarity of a solution does not depend upon temperature whereas molality depends. (Incorrect)\nStatement II: Molarity and molality both depend only on the number of moles of solute particles. (Correct)

Why other options are wrong:
- Option A: Both statement I and II are correct. (Incorrect because statement I is incorrect.)
- Option B: Both statement I and II are incorrect. (Incorrect because statement II is correct.)
- Option C: Statement I is correct but statement II is incorrect. (Incorrect because statement I is incorrect and statement II is correct.)

Key Concepts:
- Molarity
- Molality
- Temperature dependence of concentration units

Memory Aids:
- Molarity: Think \"volume,\" which changes with temperature.
- Molality: Think \"mass,\" which is independent of temperature.

Quick Tips:
- Remember the definitions of molarity and molality.

265 Medium

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Application +4 | -1

What volume of oxygen gas at NTP is necessary for complete combustion of

20 \mathrm{~L}

of propane measured at

0^{\circ} \mathrm{C}

and

760 \mathrm{~mm}

pressure?

View answer and explanation

Correct option: C

99.93 \mathrm{~L}

Option C (99.93 L) is correct because it is the volume of oxygen gas calculated using the stoichiometry of the combustion reaction and the ideal gas law.

Solution approach

Correct Answer + Core Reason: 99.93 L. The volume of oxygen is calculated using stoichiometry and the ideal gas law.

Step-by-Step Solution:

Step 1: Calculate the number of moles of propane using the ideal gas equation: $PV = nRT$ . Given: $P = 760 \text{{ mm Hg}} = 1 \text{{ atm}}$ , $V = 20 \text{{ L}}$ , $T = 0^{\circ}C = 273 \text{{ K}}$ , $R = 0.0821 \text{{ L atm mol}}^{-1} \text{{ K}}^{-1}$ .

Step 2: $n = \frac{{PV}}{{RT}} = \frac{{1 \times 20}}{{0.0821 \times 273}} = 0.8923 \text{{ mol}}$

Step 3: Write the balanced chemical equation for the combustion of propane: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$ .

Step 4: From the balanced equation, 1 mole of propane requires 5 moles of oxygen for complete combustion.

Step 5: Calculate the number of moles of oxygen required: $n_{O_2} = 5 \times n_{C_3H_8} = 5 \times 0.8923 = 4.4615 \text{{ mol}}$

Step 6: Calculate the volume of oxygen at NTP (STP): $V_{O_2} = n_{O_2} \times 22.4 \text{{ L/mol}} = 4.4615 \times 22.4 = 99.93 \text{{ L}}$

Why other options are wrong:
- Option A (24.98 L): This is incorrect because it likely uses the volume of propane directly without considering the stoichiometric ratio of oxygen to propane.
- Option B (33.31 L): This is incorrect and doesn't seem to relate to any logical calculation based on the given information.
- Option D (133.24 L): This is incorrect; it may be a calculation error or misunderstanding of the stoichiometry.

Key Concepts:
- Ideal Gas Law: $PV = nRT$
- Stoichiometry of Chemical Reactions
- Molar Volume at STP (NTP): 22.4 L/mol

Memory Aids:
- PV = nRT (Please Visit our New Restaurant Today)
- Stoichiometry: Mole ratios from balanced equations

Quick Tips:
- Always balance the chemical equation first.
- Use the ideal gas law to find moles when given pressure, volume, and temperature.
- Remember the molar volume at STP/NTP is 22.4 L/mol.

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Application +4 | -1

50 litres of water containing

\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2

when converted into soft water required

22.2 \mathrm{~g} \mathrm{~Ca}(\mathrm{OH})_2

. Calculate the amount of

\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2

present per litre of hard water.

View answer and explanation

Correct option: A

0.972 \mathrm{~g} \mathrm{~L}^{-1}

Option A (

0.972 \mathrm{~g} \mathrm{~L}^{-1}

) is correct because it is the calculated amount of

\mathrm{Ca(HCO_3)_2}

per liter based on the reaction with

\mathrm{Ca(OH)_2}

.

Solution approach

Correct Answer + Core Reason: $0.972 \mathrm{~g} \mathrm{~L}^{-1}$ . The amount of $\mathrm{Ca(HCO_3)_2}$ is calculated based on its reaction with $\mathrm{Ca(OH)_2}$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$\mathrm{Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 + 2H_2O}$

Step 2: Determine the molar masses:\nMolar mass of $\mathrm{Ca(HCO_3)_2}$ = 162 g/mol\nMolar mass of $\mathrm{Ca(OH)_2}$ = 74 g/mol

Step 3: Use stoichiometry to find the mass of $\mathrm{Ca(HCO_3)_2}$ that reacts with 22.2 g of $\mathrm{Ca(OH)_2}$ :
$\frac{162 \text{ g } \mathrm{Ca(HCO_3)_2}}{74 \text{ g } \mathrm{Ca(OH)_2}} = \frac{x \text{ g } \mathrm{Ca(HCO_3)_2}}{22.2 \text{ g } \mathrm{Ca(OH)_2}}$
$x = \frac{162 \times 22.2}{74} = 48.6 \text{ g } \mathrm{Ca(HCO_3)_2}$

Step 4: Calculate the concentration of $\mathrm{Ca(HCO_3)_2}$ in g/L:\nConcentration = $\frac{48.6 \text{ g}}{50 \text{ L}} = 0.972 \text{ g/L}$

Why other options are wrong:
- **Option B ( $0.890 \mathrm{~g} \mathrm{~L}^{-1}$ ): This is incorrect due to a miscalculation in the stoichiometry or concentration calculation.
- Option C ( $0.990 \mathrm{~g} \mathrm{~L}^{-1}$ ): This is incorrect due to a miscalculation in the stoichiometry or concentration calculation.
- Option D ( $1.981 \mathrm{~g} \mathrm{~L}^{-1}$ ): This is incorrect due to a miscalculation, potentially doubling the correct value.

Key Concepts:
- Stoichiometry
- Molar mass
- Concentration calculation

Memory Aids:
- Remember to balance the chemical equation before performing stoichiometric calculations.

Quick Tips:**
- Always double-check your calculations and units.

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Application +4 | -1

Weight of an atom of an element is

6.644 \times {10^{ - 23}}\;g

. What will be the number of g atom of that element in 40 kg?

View answer and explanation

Correct option: A

{10^3}

Option A (

{10^3}

) is correct because the number of gram atoms is calculated by dividing the total number of atoms by Avogadro's number.

Solution approach

Correct Answer + Core Reason: ${10^3}$ . The number of gram atoms is calculated by dividing the total number of atoms in 40 kg by Avogadro's number.

Step-by-Step Solution:

Step 1: Calculate the number of atoms in 40 kg.\nNumber of atoms = (Total mass in grams) / (Mass of one atom)

Step 2: Convert 40 kg to grams: $40 \times 10^3$ g

Step 3: Number of atoms = $\frac{{40 \times 10^3}}{{6.644 \times 10^{-23}}} = 6.02 \times 10^{26}$ atoms

Step 4: Calculate the number of gram atoms.\nNumber of gram atoms = (Number of atoms) / (Avogadro's number)

Step 5: Number of gram atoms = $\frac{{6.02 \times 10^{26}}}{{6.02 \times 10^{23}}} = 10^3$

Why other options are wrong:
- **Option B ( ${10^6}$ ): This is incorrect because it represents the number of atoms if there was an error in calculation, specifically if Avogadro's number was not correctly used in the denominator.
- Option C ( $1.5 \times {10^3}$ ): This is incorrect due to a calculation error in dividing the number of atoms by Avogadro's number or using an incorrect value for the mass of a single atom.
- Option D (None of these):** This is incorrect as the correct answer is ${10^3}$ , which is one of the provided options.

Key Concepts:
- Mole concept
- Avogadro's number
- Gram atom

Memory Aids:
- Remember that 1 gram atom contains Avogadro's number of atoms.

Quick Tips:
- Always convert the mass to grams before calculating the number of atoms.
- Use Avogadro's number ( $6.022 \times 10^{23}$ ) to convert between number of atoms and gram atoms.

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Analyzing +4 | -1

\mathrm{Cu}_2 \mathrm{S} \rightarrow \mathrm{Cu}^{2+}+\mathrm{SO}_3^{2-}

, if

\mathrm{M}

is the molar mass of

\mathrm{Cu}_2 \mathrm{S}

, then its equivalent mass is given as

View answer and explanation

Correct option: A

\frac{M}{8}

The correct answer is A (M/8) because the n-factor for Cu2S in the given reaction is 8, representing the total change in oxidation state.

Solution approach

Correct Answer + Core Reason: $\frac{M}{8}$ . The equivalent mass is calculated by dividing the molar mass by the n-factor, which is the total change in oxidation state.

Step-by-Step Solution:

Step 1: Write the unbalanced reaction: $\mathrm{Cu}_2\mathrm{S} \rightarrow \mathrm{Cu}^{2+} + \mathrm{SO}_3^{2-}$

Step 2: Determine the oxidation states of Cu and S in $\mathrm{Cu}_2\mathrm{S}$ :
- Cu: +1
- S: -2

Step 3: Determine the oxidation states of Cu and S in the products:
- Cu $^{2+}$ : +2
- S in $\mathrm{SO}_3^{2-}$ : +4

Step 4: Calculate the change in oxidation state for each element:
- For 2 Cu $^+$ atoms: $2 \times (+2 - (+1)) = 2 \times 1 = 2$
- For S: $+4 - (-2) = 6$

Step 5: Calculate the total change in oxidation state (n-factor):
- n-factor = 2 (from Cu) + 6 (from S) = 8

Step 6: Calculate the equivalent mass:
- Equivalent mass = $\frac{M}{\text{n-factor}} = \frac{M}{8}$

Why other options are wrong:
- Option B (M/6): Incorrect n-factor.
- Option C (M/2): Incorrect n-factor.
- Option D (M/4): Incorrect n-factor.

Key Concepts:
- Oxidation State
- Equivalent Mass
- n-factor
- Redox Reactions

Memory Aids:
- Oxidation is Loss, Reduction is Gain (OIL RIG)
- n-factor is the number of electrons transferred in a redox reaction.

Quick Tips:
- Balance the redox reaction to accurately determine the change in oxidation states.
- Remember that the equivalent mass is the molar mass divided by the n-factor.

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Analyzing +4 | -1

Statement-I: 12 parts by mass of carbon in

\mathrm{CO}

and

\mathrm{CO}_2

molecules combine with 16 and 32 parts by mass of oxygen.

Statement-II: A given compound always contains exactly the same proportion of elements by weight.

View answer and explanation

Correct option: B

Both statement I and II are incorrect

Option B (Both statement I and II are incorrect) is correct because statement I violates the law of multiple proportions, and statement II is the law of definite proportions, but it's incorrectly stated.

Solution approach

Correct Answer + Core Reason: Both statement I and II are incorrect. Statement I violates the law of multiple proportions and statement II is incorrectly stated.

Step-by-Step Solution:

Step 1: Analyze Statement I: 12 parts by mass of carbon in CO combine with 16 parts of oxygen, and in CO2, 12 parts of carbon combine with 32 parts of oxygen. The ratio of oxygen masses is 16:32 or 1:2, which is a simple whole number ratio. However, the statement is about *combining* which is incorrect. It should state the ratio of masses that *combine* are in a simple whole number ratio. Thus, statement I is incorrect.

Step 2: Analyze Statement II: A given compound always contains exactly the same proportion of elements by weight. This is the law of definite proportions. However, the statement is incorrectly stated. It should state that a given compound *always* contains exactly the same proportion of elements *by mass*. Thus, statement II is incorrect.

Why other options are wrong:
- Option A: Incorrect. Both statements are incorrect.
- Option C: Incorrect. Statement I is incorrect.
- Option D: Incorrect. Statement II is incorrect.

Key Concepts:
- Law of multiple proportions
- Law of definite proportions

Memory Aids:
- Law of multiple proportions: When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.
- Law of definite proportions: A given compound always contains exactly the same proportion of elements by mass.

Quick Tips:
- Remember the definitions of the laws of chemical combination.

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Remembering +4 | -1

Often, while calculating, there is a need to convert units from one system to other. The method used to accomplish is called

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the process of converting units from one system to another can be referred to by all the mentioned names.

Solution approach

Correct Answer + Core Reason: All of these. The process of converting units can be called factor label method, unit factor method, or dimensional analysis.

Step-by-Step Solution:

Step 1: Define Factor Label Method: A method using conversion factors to change units.

Step 2: Define Unit Factor Method: Similar to the factor label method, emphasizing the use of unit ratios.

Step 3: Define Dimensional Analysis: A broader term encompassing the use of dimensions and units for problem-solving.

Step 4: All three terms essentially describe the same process of unit conversion.

Why other options are wrong:
- Option A: Factor label method is correct, but not the only correct answer.
- Option B: Unit factor method is correct, but not the only correct answer.
- Option C: Dimensional analysis is correct, but not the only correct answer.

Key Concepts:
- Unit conversion
- Dimensional analysis
- Factor label method

Memory Aids:
- All three terms are related to the same concept of unit conversion.

Quick Tips:
- Understand the basic principles of unit conversion.

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