An organic compound has an empirical formula ${\rm{(C}}{{\rm{H}}_2}{\rm{O)}}$ its vapour density is 45. The molecular formula of the compound is

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

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{O}}$

${{\rm{C}}_2}{{\rm{H}}_2}{\rm{O}}$

Which is correct for ${\rm{N}}{{\rm{a}}_2}{\rm{HP}}{{\rm{O}}_3}$?

${\rm{Eq}}{\rm{. wt}}{\rm{.}} = M/2$

The electrochemical equivalent of a metal is $'x'$ g ${\rm{coulom}}{{\rm{b}}^{ - 1}}{\rm{.}}$ The equivalent weight of metal is

$1.6 \times {10^{ - 19}} \times x$

SOME BASIC CONCEPTS OF CHEMISTRY MCQs for NEET UG CHEMISTRY — Medium, Page 3

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Application +4 | -1

A bivalent metal has an equivalent mass of 32. The molecular mass of the metal nitrate is

View answer and explanation

Correct option: D

188

Option D (188) is the correct answer. The equivalent mass is related to the atomic mass and valency, and this allows us to calculate the molecular mass of the metal nitrate.

Solution approach

Correct Answer + Core Reason: 188. The molecular mass of the metal nitrate is calculated using the equivalent mass and valency of the metal.

Step-by-Step Solution:

Step 1: Determine the atomic mass of the metal.\nEquivalent mass = $\frac{{\text{{Atomic mass}}}}{{\text{{Valency}}}}$ \nAtomic mass = Equivalent mass $\times$ Valency\nAtomic mass = $32 \times 2 = 64$

Step 2: Determine the formula of the metal nitrate.\nSince the metal is bivalent (valency 2), the formula of the metal nitrate is $M(NO_3)_2$ .

Step 3: Calculate the molecular mass of the metal nitrate.\nMolecular mass of $M(NO_3)_2$ = Atomic mass of M + 2 $\times$ (Atomic mass of N + 3 $\times$ Atomic mass of O)\nMolecular mass of $M(NO_3)_2$ = $64 + 2 \times (14 + 3 \times 16)$ \nMolecular mass of $M(NO_3)_2$ = $64 + 2 \times (14 + 48)$ \nMolecular mass of $M(NO_3)_2$ = $64 + 2 \times 62$ \nMolecular mass of $M(NO_3)_2$ = $64 + 124$ \nMolecular mass of $M(NO_3)_2$ = $188$

Why other options are wrong:
- Option A: Incorrect. This value does not correspond to the correct calculation.
- Option B: Incorrect. This value does not correspond to the correct calculation.
- Option C: Incorrect. This value does not correspond to the correct calculation.

Key Concepts:
- Equivalent mass
- Valency
- Molecular mass

Memory Aids:
- Equivalent mass = Atomic mass / Valency

Quick Tips:
- Always determine the formula of the compound before calculating the molecular mass.

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Understanding +4 | -1

A solution of known normality is diluted to two times. Which of the following changes during dilution?

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because dilution only changes the concentration (normality or molarity), not the number of equivalents or moles of solute.

Solution approach

Correct Answer + Core Reason: None of the above. Dilution changes concentration, not the amount of solute.

Step-by-Step Solution:

Step 1: Understand dilution: Dilution involves adding more solvent to a solution.

Step 2: Consider the effect on solute: The amount of solute remains the same during dilution.

Step 3: Analyze the options:
- Equivalent of solute: Remains constant.
- Moles of solute: Remains constant.
- Volume of 1 g solution: Changes, but this is not directly related to the amount of solute.

Step 4: Final Answer: None of the above

Why other options are wrong:
- Option A: The number of equivalents of solute remains constant during dilution.
- Option B: The number of moles of solute remains constant during dilution.
- Option C: While the volume of 1 g solution might change, it's not the primary change related to the amount of solute.

Key Concepts:
- Dilution
- Molarity and Normality

Memory Aids:
- M1V1 = M2V2 (Molarity)
- N1V1 = N2V2 (Normality)

Quick Tips:
- Remember that dilution only changes the concentration, not the amount of solute.

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Application +4 | -1

The equivalent weight of iron in

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}

would be :

View answer and explanation

Correct option: A

18.6

Option A (18.6) is correct because the equivalent weight is calculated by dividing the molar mass of iron by its valency change.

Solution approach

Correct Answer + Core Reason: 18.6. The equivalent weight of iron is calculated by dividing its atomic weight by the number of electrons involved in the redox process.

Step-by-Step Solution:

Step 1: Determine the oxidation state of iron in $Fe_2O_3$ .
$Fe_2O_3$ : Oxygen has an oxidation state of -2, so 3 oxygen atoms have a total oxidation state of -6.\nSince the compound is neutral, 2 iron atoms must have a total oxidation state of +6.\nTherefore, each iron atom has an oxidation state of +3.

Step 2: Determine the valency of iron.\nThe valency of iron is the change in oxidation state, which is 3.

Step 3: Calculate the equivalent weight of iron.\nAtomic weight of iron (Fe) = 56 g/mol\nEquivalent weight of iron = $\frac{{\text{{Atomic weight}}}}{{\text{{Valency}}}} = \frac{{56}}{3} \approx 18.6$

Why other options are wrong:
- Option B: 28: This is incorrect calculation.
- Option C: 56: This is the atomic weight of iron, not the equivalent weight.
- Option D: 11: This is incorrect calculation.

Key Concepts:
- Equivalent weight
- Atomic weight
- Oxidation state
- Valency

Memory Aids:
- Equivalent weight = Atomic weight / Valency

Quick Tips:
- Determine the oxidation state of the element in the compound.

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Application +4 | -1

An organic compound has an empirical formula

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

its vapour density is 45. The molecular formula of the compound is

View answer and explanation

Correct option: D

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

Option D (

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

) is correct. The molecular formula is a multiple of the empirical formula, and this multiple is determined by the vapor density.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_3}{\mathrm{H}_6}{\mathrm{O}_3}$ . The molecular formula is determined by the empirical formula and the vapor density.

Step-by-Step Solution:

Step 1: Calculate the empirical formula mass.\nEmpirical formula is $CH_2O$ .\nEmpirical formula mass = 12 + 2(1) + 16 = 30

Step 2: Calculate the molecular mass.\nVapor density = $\frac{{\text{{Molecular mass}}}}{2}$ \nMolecular mass = 2 $\times$ Vapor density = 2 $\times$ 45 = 90

Step 3: Determine the multiplier (n).\nn = $\frac{{\text{{Molecular mass}}}}{{\text{{Empirical formula mass}}}} = \frac{{90}}{{30}} = 3$

Step 4: Calculate the molecular formula.\nMolecular formula = $(CH_2O)_n = (CH_2O)_3 = C_3H_6O_3$

Why other options are wrong:
- Option A: Incorrect. This is the empirical formula, not the molecular formula.
- Option B: Incorrect. This formula does not match the empirical formula.
- Option C: Incorrect. This formula does not match the empirical formula.

Key Concepts:
- Empirical formula
- Molecular formula
- Vapor density

Memory Aids:
- Molecular mass = 2 $\times$ Vapor density

Quick Tips:
- Always calculate the empirical formula mass first.

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Application +4 | -1

Mass ratio of

N{H_3}\,and\mathrm{ }C{O_2}

for maximum product formation as per reaction :

2N{H_3} + C{O_2} \to N{H_2}COON{H_4}

View answer and explanation

Correct option: A

\frac{{17}}{{22}}

Option A (17/22) is correct because it represents the stoichiometric mass ratio of NH3 to CO2 required for complete reaction.

Solution approach

Correct Answer + Core Reason: $\frac{{17}}{{22}}$ . The mass ratio is determined by the stoichiometry of the balanced chemical equation.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2N{H_3} + C{O_2} \to N{H_2}COON{H_4}$

Step 2: Determine the molar masses of $NH_3$ and $CO_2$ .\nMolar mass of $NH_3 = 14 + 3(1) = 17 \text{ g/mol}$ \nMolar mass of $CO_2 = 12 + 2(16) = 44 \text{ g/mol}$

Step 3: Determine the mass ratio based on the stoichiometry.\nFrom the balanced equation, 2 moles of $NH_3$ react with 1 mole of $CO_2$ .\nMass of 2 moles of $NH_3 = 2 \times 17 = 34 \text{ g}$ \nMass of 1 mole of $CO_2 = 44 \text{ g}$

Step 4: Calculate the mass ratio of $NH_3$ to $CO_2$ .\nRatio = $\frac{{34}}{{44}} = \frac{{17}}{{22}}$

Why other options are wrong:
- Option B (17/44): This is the inverse of the correct ratio or uses only 1 mole of NH3 instead of 2.
- Option C (22/17): This is the inverse of the correct ratio.
- Option D (44/17): This is the inverse of the correct ratio and also uses incorrect mole ratios.

Key Concepts:
- Stoichiometry
- Molar mass
- Balanced chemical equations

Memory Aids:
- Remember to balance the chemical equation before calculating mole ratios.

Quick Tips:
- Always double-check the balanced equation to ensure correct mole ratios.

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Remembering +4 | -1

The formula of a compound which gives simple whole number atomic ratio in one molecule of a compound is called:

View answer and explanation

Correct option: C

Empirical formula

Option C is correct because the empirical formula represents the simplest whole-number ratio of atoms in a compound.

Solution approach

Correct Answer + Core Reason: Empirical formula. The empirical formula represents the simplest whole number ratio of atoms in a compound.

Step-by-Step Solution:

Step 1: Define empirical formula. The empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

Step 2: Consider the other options:
A. Structure formula: Shows the arrangement of atoms in a molecule.
B. Molecular formula: Shows the actual number of atoms of each element in a molecule.
D. Projection formula: Used to represent three-dimensional structures in two dimensions.

Step 3: Conclusion: The correct answer is empirical formula.

Why other options are wrong:
- Option A: Structural formula shows the arrangement of atoms, not the simplest ratio.
- Option B: Molecular formula shows the actual number of atoms, not the simplest ratio.
- Option D: Projection formula is a way to represent 3D structures, not a chemical formula.

Key Concepts:
- Empirical formula
- Molecular formula
- Structural formula

Memory Aids:
- Empirical = Simplest

Quick Tips:
- Remember the definitions of different types of chemical formulas.

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Remembering +4 | -1

Specific gravity of solution is given by:

View answer and explanation

Correct option: A

Weight of 1 mL solution

Option A is correct because specific gravity is the ratio of the density of a substance to the density of a reference substance (usually water), which can be represented as the weight of 1 mL of the solution.

Solution approach

Correct Answer + Core Reason: Weight of 1 mL solution. Specific gravity is related to the density, which is mass per unit volume.

Step-by-Step Solution:

Step 1: Define specific gravity. Specific gravity = (Density of substance) / (Density of reference substance).

Step 2: Density can be expressed as mass/volume or weight/volume.

Step 3: For a solution, the weight of 1 mL represents its density relative to water.

Step 4: Final Answer: Weight of 1 mL solution.

Why other options are wrong:
- Option B: Mole present in 1 mL solution represents molarity, not specific gravity.
- Option C: Volume of 1 g solution represents the inverse of density, not specific gravity.
- Option D: The correct answer is among the options.

Key Concepts:
- Specific gravity
- Density

Memory Aids:
- Specific gravity is a dimensionless quantity.

Quick Tips:
- Remember that specific gravity is a ratio.

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Application +4 | -1

Weight of

\mathrm{FeS}{\mathrm{O}_4}\left( {\mathrm{mol}\mathrm{.wt}\mathrm{.} = 152} \right)

oxidized by 200 mL of 1

N

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

solution is :

View answer and explanation

Correct option: A

30.4 g

Option A (30.4 g) is correct because it is the correct mass of FeSO4 oxidized based on the stoichiometry of the reaction with KMnO4.

Solution approach

Correct Answer + Core Reason: 30.4 g. The mass of $FeSO_4$ oxidized is calculated based on the stoichiometry of the reaction with $KMnO_4$ .

Step-by-Step Solution:

Step 1: Write the balanced redox reaction.
$MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

Step 2: Calculate the milliequivalents of $KMnO_4$ .\nMilliequivalents of $KMnO_4 = Normality \times Volume (in mL) = 1 N \times 200 mL = 200$ meq

Step 3: Calculate the milliequivalents of $FeSO_4$ .\nMilliequivalents of $FeSO_4$ = Milliequivalents of $KMnO_4 = 200$ meq

Step 4: Calculate the mass of $FeSO_4$ .\nEquivalent weight of $FeSO_4 = \frac{{Molar mass}}{{n - factor}} = \frac{{152}}{1} = 152$ g/eq\nMilliequivalent weight of $FeSO_4 = 152$ mg/meq\nMass of $FeSO_4 = Milliequivalents \times Milliequivalent weight =$ 200 meq × 152 mg/meq = 30400 mg = 30.4 g $\n\n**Why other options are wrong:**\n- **Option B:** 15.2 g: This is incorrect because it divides the correct answer by 2, possibly due to a miscalculation in the mole ratio or equivalent weight.\n- **Option C:** 60.8 g: This is incorrect because it multiplies the correct answer by 2, possibly due to a miscalculation in the mole ratio or equivalent weight.\n- **Option D:** 158 g: This is incorrect because it uses an incorrect molar mass or performs an incorrect calculation.\n\n**Key Concepts:**\n- Redox titrations\n- Normality\n- Equivalent weight\n- Stoichiometry\n\n**Memory Aids:**\n- Milliequivalents = Normality x Volume (in mL)\n\n**Quick Tips:**\n- Always balance the redox reaction before performing stoichiometric calculations.\n- Ensure correct units are used in calculations.$

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Application +4 | -1

10 mL of concentrated

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

(18 M)is diluted to one litre. The approximate molecular of the dilute acid is:

View answer and explanation

Correct option: C

0.18 M

Option C (0.18 M) is correct. The molarity of the diluted solution is calculated using the dilution formula:

M_1V_1 = M_2V_2

.

Solution approach

Correct Answer + Core Reason: 0.18 M. The molarity of the diluted solution is calculated using the dilution formula.

Step-by-Step Solution:

Step 1: Identify the given values.
$M_1 = 18 \text{{ M}}$
$V_1 = 10 \text{{ mL}}$
$V_2 = 1 \text{{ L}} = 1000 \text{{ mL}}$

Step 2: Apply the dilution formula.
$M_1V_1 = M_2V_2$
$18 \times 10 = M_2 \times 1000$

Step 3: Solve for $M_2$ .
$M_2 = \frac{{18 \times 10}}{{1000}} = \frac{{180}}{{1000}} = 0.18 \text{{ M}}$

Why other options are wrong:
- Option A: Incorrect. This is the initial molarity before dilution.
- Option B: Incorrect. This value is obtained by multiplying the initial molarity and volume without dividing by the final volume.
- Option D: Incorrect. This value is obtained by dividing the initial molarity by 10 instead of 100.

Key Concepts:
- Molarity
- Dilution formula: $M_1V_1 = M_2V_2$

Memory Aids:
- Dilution means decreasing the concentration.

Quick Tips:
- Always convert volumes to the same units before applying the dilution formula.

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Application +4 | -1

Mole fraction of the solute in a 1.00 molal aqueous solution is:

View answer and explanation

Correct option: C

0.0177

Option C (0.0177) is correct. The mole fraction of the solute in a 1.00 molal aqueous solution is calculated by considering the moles of solute and solvent.

Solution approach

Correct Answer + Core Reason: 0.0177. The mole fraction is calculated by dividing moles of solute by total moles.

Step-by-Step Solution:

Step 1: Define molality.
1.00 molal aqueous solution means 1 mole of solute in 1 kg of water.

Step 2: Calculate moles of water.\nMoles of water = $\frac{{1000 \text{{ g}}}}{{18 \text{{ g/mol}}}} = 55.55 \text{{ moles}}$

Step 3: Calculate the mole fraction of the solute.\nMole fraction of solute = $\frac{{\text{{Moles of solute}}}}{{\text{{Moles of solute}} + \text{{Moles of solvent}}}}$ \nMole fraction of solute = $\frac{{1}}{{1 + 55.55}} = \frac{{1}}{{56.55}} = 0.0177$

Why other options are wrong:
- Option A: Incorrect. This value is significantly larger than the expected mole fraction for a dilute solution.
- Option B: Incorrect. This value is larger than the correct mole fraction.
- Option D: Incorrect. This value is not the correct mole fraction based on the calculation.

Key Concepts:
- Molality
- Mole fraction

Memory Aids:
- Molality = Moles of solute / kg of solvent
- Mole fraction = Moles of component / Total moles

Quick Tips:
- Always remember that molality is defined per kg of solvent.

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Remembering +4 | -1

Who among the following was the first proponent of atomic theory?

View answer and explanation

Correct option: A

Acharya Kanda

Option A (Acharya Kanda) is correct because Acharya Kanda was one of the earliest proponents of atomic theory in ancient India.

Solution approach

Correct Answer + Core Reason: Acharya Kanda. He was an ancient Indian philosopher who proposed atomic theory.

Step-by-Step Solution:

Step 1: Recall the history of atomic theory.

Step 2: Acharya Kanda, an ancient Indian philosopher, is credited with proposing an early form of atomic theory, known as Vaisheshika.

Step 3: His ideas predate Dalton's atomic theory by centuries.

Why other options are wrong:
- Option B: Chakrapani is not primarily known for atomic theory.
- Option C: Nagarjuna is known for his contributions to Buddhist philosophy and alchemy, not primarily atomic theory.
- Option D: None of the above is incorrect because option A is correct.

Key Concepts:
- History of atomic theory
- Ancient Indian philosophy
- Acharya Kanda

Memory Aids:
- Kanda = Early atomic theory.

Quick Tips:
- Be aware of the historical context of scientific discoveries.

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Understanding +4 | -1

The weights of two elements which combine with one another are in the ratio of their:

View answer and explanation

Correct option: C

Eq. wt.

The correct answer is C (Eq. wt.) because the weights of two elements that combine with each other are in the ratio of their equivalent weights.

Solution approach

Correct Answer + Core Reason: Eq. wt. The weights of two elements that combine with each other are in the ratio of their equivalent weights.

Step-by-Step Solution:

Step 1: Define Equivalent Weight: The equivalent weight of an element is the mass of the element that combines with or displaces 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen, or 35.5 parts by mass of chlorine.

Step 2: The weights of two elements that combine with each other are in the ratio of their equivalent weights.

Why other options are wrong:
- Option A: At. wt. (Atomic weight) is the average mass of atoms of an element, but doesn't directly relate to the combining ratios.
- Option B: Mol. wt (Molecular weight) is the sum of the atomic weights of the atoms in a molecule, but doesn't directly relate to the combining ratios of elements.
- Option D: None of these is incorrect because equivalent weight is the correct answer.

Key Concepts:
- Equivalent weight

Memory Aids:
- Equivalent weight is related to combining capacity.

Quick Tips:
- Remember the definition of equivalent weight.

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Application +4 | -1

Which among the following has maximum number of atoms among the following?

View answer and explanation

Correct option: A

22\mathrm{ }g\mathrm{ }C{O_2}

Option A (22 g CO2) is correct because it contains the highest number of moles of molecules, and each molecule has 3 atoms.

Solution approach

Correct Answer + Core Reason: $22\mathrm{ }g\mathrm{ }C{O_2}$ . The number of atoms is proportional to the number of moles multiplied by the number of atoms per molecule.

Step-by-Step Solution:

Step 1: Calculate the number of moles for each option.
- A. $22\mathrm{ }g\mathrm{ }C{O_2}$ : Moles = $\frac{{22}}{{44}} = 0.5$ moles. Atoms per molecule = 3. Total atoms = $0.5 \times 3 \times N_A = 1.5 N_A$
- B. $16\mathrm{ }g\mathrm{ }{O_3}$ : Moles = $\frac{{16}}{{48}} = \frac{{1}}{{3}}$ moles. Atoms per molecule = 3. Total atoms = $\frac{{1}}{{3}} \times 3 \times N_A = N_A$
- C. $14\mathrm{ }g\mathrm{ }{N_2}$ : Moles = $\frac{{14}}{{28}} = 0.5$ moles. Atoms per molecule = 2. Total atoms = $0.5 \times 2 \times N_A = N_A$
- D. $24\mathrm{ }g\mathrm{ }Mg$ : Moles = $\frac{{24}}{{24}} = 1$ mole. Atoms per molecule = 1. Total atoms = $1 \times 1 \times N_A = N_A$

Step 2: Compare the total number of atoms for each option.\nOption A has the highest number of atoms ( $1.5 N_A$ ).

Why other options are wrong:
- Option B (16 g O3): Contains fewer moles of molecules than CO2, resulting in fewer total atoms.
- Option C (14 g N2): Contains the same number of moles as CO2, but fewer atoms per molecule, resulting in fewer total atoms.
- Option D (24 g Mg): Contains more moles than CO2, but only one atom per \

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Analyzing +4 | -1

Which is correct for

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

?

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because Na2HPO3 is an acid salt, has an equivalent weight of M/2, and the oxidation number of P is +3.

Solution approach

Correct Answer + Core Reason: All of these. Na2HPO3 satisfies all the given conditions.

Step-by-Step Solution:

Step 1: Analyze the given compound $\mathrm{N}{\mathrm{a}_2}\mathrm{HP}{\mathrm{O}_3}$ .

Step 2: Check if it is an acid salt.
$\mathrm{N}{\mathrm{a}_2}\mathrm{HP}{\mathrm{O}_3}$ is formed by replacing two hydrogen atoms of ${\mathrm{H}_3}\mathrm{P}{\mathrm{O}_3}$ with sodium atoms. It still has one replaceable hydrogen atom, so it is an acid salt. Therefore, option A is incorrect.

Step 3: Determine the basicity of the acid salt.\nSince it has one replaceable hydrogen atom, its basicity is 2. Thus, $\mathrm{Eq}\mathrm{. wt}\mathrm{.} = M/2$ .

Step 4: Calculate the oxidation number of P.\nLet the oxidation number of P be x.
$2(+1) + (+1) + x + 3(-2) = 0$
$2 + 1 + x - 6 = 0$ \nx = +3

Step 5: Conclusion\nSince $\mathrm{N}{\mathrm{a}_2}\mathrm{HP}{\mathrm{O}_3}$ is an acid salt, $\mathrm{Eq}\mathrm{. wt}\mathrm{.} = M/2$ , and the oxidation number of P is +3, option D is correct.

Why other options are wrong:
- Option A: Incorrect because it is an acid salt.
- Option B: Correct statement.
- Option C: Correct statement.

Key Concepts:
- Acid salts
- Equivalent weight
- Oxidation number

Memory Aids:
- Acid salts contain replaceable hydrogen atoms.

Quick Tips:
- Determine the number of replaceable hydrogen atoms in the acid salt to find its basicity.

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Analyzing +4 | -1

0.75 moles of a solid

{A_4}

and 2 moles of

{O_2}(g)

are heated in a sealed vessel, completely using up the reactants and produces only one compound. It is found that when the temperature is used to initial temperature, the contents of the vessel exhibit a pressure equal to

\frac{1}{2}

of the original pressure. The formula of the product will be

View answer and explanation

Correct option: C

{A_3}{O_4}

Option C (

{A_3}{O_4}

) is correct because the stoichiometry of the reaction and the pressure change indicate that 3 moles of

{A_4}

react with 8 moles of

{O_2}

to produce 4 moles of

{A_3}{O_4}

.

Solution approach

Correct Answer + Core Reason: ${A_3}{O_4}$ is the correct formula because the pressure drops to half, implying a decrease in the number of moles of gas by half after the reaction.

Step-by-Step Solution:

Step 1: Write the general reaction: ${A_4} + {O_2} \rightarrow {A_x}{O_y}$

Step 2: Balance the reaction based on the given information. 0.75 moles of ${A_4}$ and 2 moles of ${O_2}$ react completely.

Step 3: Use the pressure information. Since the pressure is halved at the same temperature, the number of moles of gaseous product is half the number of moles of gaseous reactant.\nInitial moles of gas ( ${O_2}$ ) = 2 moles.\nFinal moles of gas (product) = 2/2 = 1 mole.

Step 4: Let's consider the reaction: $a{A_4} + b{O_2} \rightarrow c{A_x}{O_y}$ \nGiven: $a = 0.75$ moles, $b = 2$ moles, and the product is 1 mole.\nSo, $0.75{A_4} + 2{O_2} \rightarrow 1{A_x}{O_y}$ \nMultiply by 4 to get whole numbers: $3{A_4} + 8{O_2} \rightarrow 4{A_x}{O_y}$

Step 5: Balance the number of A and O atoms:
$3 \times 4 = 12$ A atoms on the left, so $4x = 12$ , which means $x = 3$ .
$8 \times 2 = 16$ O atoms on the left, so $4y = 16$ , which means $y = 4$ .\nTherefore, the formula of the product is ${A_3}{O_4}$ .

Why other options are wrong:
- Option A: ${A_2}{O_3}$ would require a different stoichiometry and wouldn't result in the observed pressure change.
- Option B: ${A_3}{O_8}$ would require more oxygen than available and wouldn't result in the observed pressure change.
- Option D: $A{O_2}$ would require a different stoichiometry and wouldn't result in the observed pressure change.

Key Concepts:
- Stoichiometry
- Ideal Gas Law (Pressure and Moles Relationship)
- Balancing Chemical Equations

Memory Aids:
- PV = nRT (At constant V and T, P is proportional to n)

Quick Tips:
- Use the pressure change to determine the change in the number of moles of gas.
- Balance the equation to find the correct stoichiometry.

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Understanding +4 | -1

In the reaction

3Mg + {N_2} \to M{g_3}{N_2}

View answer and explanation

Correct option: B

Mg

is oxidised

Option B (

Mg

is oxidised) is correct because Magnesium loses electrons (oxidation) to form

Mg_3N_2

.

Solution approach

Correct Answer + Core Reason: $Mg$ is oxidised. Magnesium loses electrons, hence it is oxidised.

Step-by-Step Solution:

Step 1: Write the reaction: $3Mg + N_2 \to Mg_3N_2$

Step 2: Determine oxidation states: In $Mg$ , the oxidation state is 0. In $Mg_3N_2$ , the oxidation state of $Mg$ is +2.\nIn $N_2$ , the oxidation state is 0. In $Mg_3N_2$ , the oxidation state of $N$ is -3.

Step 3: Identify oxidation and reduction: $Mg$ goes from 0 to +2 (oxidation). $N$ goes from 0 to -3 (reduction).

Step 4: Final Answer: $Mg$ is oxidised

Why other options are wrong:
- Option A: $Mg$ is not reduced; it is oxidized.
- Option C: $N$ is reduced, not oxidized.
- Option D: $N$ is an oxidizing agent, not a reductant.

Key Concepts:
- Oxidation and Reduction
- Oxidation states

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

Quick Tips:
- Determine the oxidation states of each element before and after the reaction.

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Analyzing +4 | -1

Which of the following has the smallest number of molecules?

View answer and explanation

Correct option: A

0.1 mole of

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

gas

Option A (0.1 mole of

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

gas) is correct because it represents the smallest number of molecules compared to the other options when converted to moles.

Solution approach

Correct Answer + Core Reason: 0.1 mole of $\mathrm{C}{\mathrm{O}_2}$ gas has the smallest number of molecules because it represents the smallest number of moles.

Step-by-Step Solution:

Step 1: Convert all options to moles:
- A. 0.1 mole of $\mathrm{C}{\mathrm{O}_2}$ gas: 0.1 mole
- B. 11.2 L of $\mathrm{C}{\mathrm{O}_2}$ gas at STP: At STP, 1 mole occupies 22.4 L. So, 11.2 L is $\frac{{11.2}}{{22.4}} = 0.5$ moles.
- C. 22 g of $\mathrm{C}{\mathrm{O}_2}$ gas: The molar mass of $\mathrm{C}{\mathrm{O}_2}$ is 12 + 2(16) = 44 g/mol. So, 22 g is $\frac{{22}}{{44}} = 0.5$ moles.
- D. $22.4 \times {10^3}$ mL of $\mathrm{C}{\mathrm{O}_2}$ gas at STP: $22.4 \times {10^3}$ mL = 22.4 L. At STP, 1 mole occupies 22.4 L. So, 22.4 L is 1 mole.

Step 2: Compare the number of moles: 0.1 < 0.5 < 0.5 < 1

Step 3: Conclusion: The smallest number of moles corresponds to the smallest number of molecules.

Why other options are wrong:
- Option B: 11.2 L of $\mathrm{C}{\mathrm{O}_2}$ gas at STP is 0.5 moles, which is more than 0.1 mole.
- Option C: 22 g of $\mathrm{C}{\mathrm{O}_2}$ gas is 0.5 moles, which is more than 0.1 mole.
- Option D: $22.4 \times {10^3}$ mL of $\mathrm{C}{\mathrm{O}_2}$ gas at STP is 1 mole, which is more than 0.1 mole.

Key Concepts:
- Mole Concept
- STP (Standard Temperature and Pressure)
- Molar Mass

Memory Aids:
- At STP, 1 mole of any gas occupies 22.4 L.
- Number of molecules = number of moles x Avogadro's number

Quick Tips:
- Convert all quantities to moles before comparing the number of molecules.
- Remember the molar volume of a gas at STP.

78 Medium

78 of 116Question permalink

Understanding +4 | -1

Reactants react in the equal number of …….. to give products.

View answer and explanation

Correct option: C

Equivalent

Option C (Equivalent) is correct because reactants react in equal numbers of equivalents to give products, based on the concept of equivalent weight.

Solution approach

Correct Answer + Core Reason: Equivalent. Reactants react in equal numbers of equivalents.

Step-by-Step Solution:

Step 1: Recall the concept of equivalents in chemical reactions.

Step 2: The law of equivalence states that substances react in the ratio of their equivalent weights.

Step 3: Therefore, reactants react in equal numbers of equivalents to give products.

Why other options are wrong:
- Option A: Mole is incorrect. Reactants do not necessarily react in equal numbers of moles.
- Option B: Weights is incorrect. Reactants do not necessarily react in equal weights.
- Option D: All of these is incorrect because only equivalents is correct.

Key Concepts:
- Equivalent weight
- Law of equivalence
- Stoichiometry

Memory Aids:
- Remember that equivalents are related to the combining capacity of a substance.

Quick Tips:
- Use the concept of equivalents to simplify stoichiometric calculations.

79 Medium

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Understanding +4 | -1

Molecular weight of tribasic acid is W. Its equivalent weight will be:

View answer and explanation

Correct option: B

\frac{W}{3}

Option B (W/3) is correct because equivalent weight of an acid is its molecular weight divided by its basicity, which is 3 for a tribasic acid.

Solution approach

Correct Answer + Core Reason: $\frac{W}{3}$ . Equivalent weight is molecular weight divided by basicity.

Step-by-Step Solution:

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

Step 2: Identify the basicity: Tribasic acid has a basicity of 3.

Step 3: Apply the formula: Equivalent weight = $\frac{W}{3}$

Step 4: Final Answer: $\frac{W}{3}$

Why other options are wrong:
- Option A: This is incorrect because it divides the molecular weight by 2, which would be correct for a dibasic acid, not a tribasic acid.
- Option C: This is incorrect because it states the equivalent weight is equal to the molecular weight, which is only true for monobasic acids.
- Option D: This is incorrect because it multiplies the molecular weight by 3, which is the opposite of what should be done to find the equivalent weight of a tribasic acid.

Key Concepts:
- Equivalent weight
- Basicity of an acid

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

Quick Tips:
- Remember to divide by the correct basicity (number of replaceable H+ ions).

80 Medium

80 of 116Question permalink

Understanding +4 | -1

The stoichiometry of the following reaction is

{\mathrm{K}_2}{\mathrm{S}_2}{\mathrm{O}_8}\left( {aq} \right) + 2\mathrm{KI}\left( {aq} \right) \to 2{\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4}\left( {aq} \right) + {\mathrm{I}_2}\left( {aq} \right)

View answer and explanation

Correct option: C

1\mathrm{ }:\mathrm{ }2

Option C is correct because the stoichiometry of

K_2S_2O_8

to KI is 1:2 as indicated by the balanced chemical equation.

Solution approach

Correct Answer + Core Reason: $1\mathrm{ }:\mathrm{ }2$ . The balanced equation shows 1 mole of ${\mathrm{K}_2}{\mathrm{S}_2}{\mathrm{O}_8}$ reacts with 2 moles of KI.

Step-by-Step Solution:

Step 1: Examine the balanced chemical equation: ${\mathrm{K}_2}{\mathrm{S}_2}{\mathrm{O}_8}\left( {aq} \right) + 2\mathrm{KI}\left( {aq} \right) \to 2{\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4}\left( {aq} \right) + {\mathrm{I}_2}\left( {aq} \right)$ .

Step 2: Identify the stoichiometric coefficients of ${\mathrm{K}_2}{\mathrm{S}_2}{\mathrm{O}_8}$ and KI.

Step 3: The coefficient of ${\mathrm{K}_2}{\mathrm{S}_2}{\mathrm{O}_8}$ is 1, and the coefficient of KI is 2.

Step 4: Therefore, the stoichiometry is 1:2.

Step 5: Final Answer: $1\mathrm{ }:\mathrm{ }2$ .

Why other options are wrong:
- Option A: $2\mathrm{ }:\mathrm{ }2$ is incorrect.
- Option B: $1\mathrm{ }:\mathrm{ }1$ is incorrect.
- Option D: $2\mathrm{ }:\mathrm{ }1$ is incorrect.

Key Concepts:
- Stoichiometry
- Balanced chemical equations

Memory Aids:
- The coefficients in a balanced equation represent the mole ratios.

Quick Tips:
- Always make sure the chemical equation is balanced before determining stoichiometry.

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Understanding +4 | -1

The electrochemical equivalent of a metal is

'x'

g

\mathrm{coulom}{\mathrm{b}^{ - 1}}\mathrm{.}

The equivalent weight of metal is

View answer and explanation

Correct option: B

x \times 96500

Option B (

x \times 96500

) is correct because the equivalent weight is the product of the electrochemical equivalent and Faraday's constant (96500 C/mol).

Solution approach

Correct Answer + Core Reason: $x \times 96500$ because Equivalent weight = Electrochemical equivalent $\times$ Faraday's constant.

Step-by-Step Solution:

Step 1: Define electrochemical equivalent (ECE): The electrochemical equivalent is the mass of a substance deposited or liberated at an electrode when 1 coulomb of charge is passed through the electrolyte.

Step 2: Define equivalent weight: The equivalent weight of a metal is the mass of the metal that combines with or displaces 1.008 g of hydrogen or 8 g of oxygen.

Step 3: Relate ECE and equivalent weight: Equivalent weight = ECE $\times$ Faraday's constant (F). Faraday's constant is approximately 96500 Coulombs per mole.

Step 4: Substitute the given value: Given ECE = $x$ g/C. Therefore, Equivalent weight = $x \times 96500$ g/mol.

Why other options are wrong:
- Option A: $x$ : This is incorrect because it only represents the electrochemical equivalent and not the equivalent weight.
- Option C: $\frac{x}{{96500}}$ : This is incorrect because it divides the electrochemical equivalent by Faraday's constant, which is the inverse of the correct relationship.
- Option D: $1.6 \times {10^{ - 19}} \times x$ : This is incorrect because it uses the charge of an electron instead of Faraday's constant.

Key Concepts:
- Electrochemical equivalent
- Equivalent weight
- Faraday's constant

Memory Aids:
- Equivalent weight = ECE $\times$ Faraday's constant

Quick Tips:
- Remember the relationship between electrochemical equivalent, equivalent weight, and Faraday's constant.

82 Medium

82 of 116Question permalink

Understanding +4 | -1

A: Ratio of empirical formula mass and molecular formula mass may be a whole number.

R: Molecular formula mass

=\mathrm{n} \times

empirical formula mass, where

\mathrm{n}

is the simplest whole number.

View answer and explanation

Correct option: B

Both A and R are correct, but R is not the correct explanation of A.

Option B (Both A and R are correct, but R is not the correct explanation of A) is correct because both the assertion and the reason are correct statements, but the reason doesn't directly explain why the ratio in the assertion is a whole number.

Solution approach

Correct Answer + Core Reason: Both A and R are correct, but R is not the correct explanation of A. Both statements are true, but the reason doesn't directly explain the assertion.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Ratio of empirical formula mass and molecular formula mass may be a whole number.

Step 2: This is correct. The molecular formula is a whole number multiple of the empirical formula. Therefore, the ratio of their masses can be a whole number (or the inverse of a whole number).

Step 3: Analyze Reason (R): Molecular formula mass = n × empirical formula mass, where n is the simplest whole number.

Step 4: This is also correct. The molecular formula is a whole number multiple (n) of the empirical formula.

Step 5: However, the Reason doesn't directly explain *why* the ratio *may* be a whole number. The reason states the relationship, but the assertion is about the ratio itself being a whole number.

Step 6: Final Answer: Both A and R are correct, but R is not the correct explanation of A.

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

Key Concepts:
- Empirical formula
- Molecular formula
- Relationship between empirical and molecular formulas

Memory Aids:
- Molecular Formula = n * Empirical Formula

Quick Tips:
- Understand the definitions of empirical and molecular formulas and their relationship.

83 Medium

83 of 116Question permalink

Understanding +4 | -1

Mole and millimole of reactants react in the ……….as represented by balanced stoichiometric equation.

View answer and explanation

Correct option: A

Molar ratio

Option A (Molar ratio) is correct because reactants react in the molar ratio as represented by the balanced stoichiometric equation.

Solution approach

Correct Answer + Core Reason: Molar ratio. Reactants react according to the molar ratio defined by the balanced equation.

Step-by-Step Solution:

Step 1: Understand stoichiometry.\nStoichiometry deals with the quantitative relationships between reactants and products in chemical reactions.

Step 2: Define molar ratio.\nThe molar ratio is the ratio of moles of reactants and products as indicated by the coefficients in the balanced chemical equation.

Step 3: Conclusion.\nTherefore, moles and millimoles of reactants react in the molar ratio as represented by the balanced stoichiometric equation.

Why other options are wrong:
- Option B: Equal amount is incorrect because reactants do not necessarily react in equal amounts; they react according to the molar ratio.
- Option C: Both (a) and (b) is incorrect because only the molar ratio is correct.
- Option D: None of these is incorrect because option A is correct.

Key Concepts:
- Stoichiometry
- Molar ratio
- Balanced chemical equation

Memory Aids:
- Remember that the coefficients in a balanced equation represent the molar ratio.

Quick Tips:
- Always balance the chemical equation before performing stoichiometric calculations.

84 Medium

84 of 116Question permalink

Application +4 | -1

Which has maximum number of atoms ?

View answer and explanation

Correct option: B

44\mathrm{ }g\mathrm{ }of\mathrm{ }C{O_2}

Option B (

44\mathrm{ }g\mathrm{ }of\mathrm{ }C{O_2}

) is correct because it contains the highest number of moles of molecules and the highest number of atoms per molecule compared to the other options after correcting the calculation.

Solution approach

Correct Answer + Core Reason: $44\mathrm{ }g\mathrm{ }of\mathrm{ }C{O_2}$ . It has the highest number of moles and atoms per molecule after correction.

Step-by-Step Solution:

Step 1: Calculate moles for each option:
- A: $16\mathrm{ }g\mathrm{ }of\mathrm{ }{O_3} = \frac{16}{48} = \frac{1}{3}$ moles
- B: $44\mathrm{ }g\mathrm{ }of\mathrm{ }C{O_2} = \frac{44}{44} = 1$ mole
- C: $1g\mathrm{ }of\mathrm{ }{H_2} = \frac{1}{2} = 0.5$ moles
- D: $23\mathrm{ }g\mathrm{ }of\mathrm{ }N{O_2} = \frac{23}{46} = 0.5$ moles

Step 2: Calculate the number of atoms per molecule:
- A: $O_3$ has 3 atoms
- B: $CO_2$ has 3 atoms
- C: $H_2$ has 2 atoms
- D: $NO_2$ has 3 atoms

Step 3: Calculate the total number of atoms:
- A: $\frac{1}{3} \times 3 \times N_A = 1 N_A$
- B: $1 \times 3 \times N_A = 3 N_A$
- C: $0.5 \times 2 \times N_A = 1 N_A$
- D: $0.5 \times 3 \times N_A = 1.5 N_A$

Step 4: Final Answer: $44\mathrm{ }g\mathrm{ }of\mathrm{ }C{O_2}$

Why other options are wrong:
- Option A: Although $O_3$ has 3 atoms per molecule, the number of moles is low.
- Option C: The number of moles is 0.5 and the number of atoms per molecule is 2, resulting in fewer total atoms.
- Option D: The number of moles is 0.5 and the number of atoms per molecule is 3, resulting in fewer total atoms than option B.

Key Concepts:
- Moles
- Avogadro's Number

Memory Aids:
- Moles = Mass / Molar Mass

Quick Tips:
- Calculate the number of moles first, then multiply by the number of atoms per molecule.

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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 it calculates the volume of Na2S2O3 required based on the stoichiometry of the reactions.

Solution approach

Correct Answer + Core Reason: 25 mL. The volume of $Na_2S_2O_3$ required is calculated based on the stoichiometry of the reactions.

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: Calculate the moles of $CuSO_4$ .\nMoles of $CuSO_4 = Molarity \times Volume (in L) = 0.20 M \times 0.050 L = 0.01$ mol

Step 3: Determine the mole ratio of $CuSO_4$ to $I_2$ .\nFrom the first equation, 2 moles of $CuSO_4$ produce 1 mole of $I_2$ . Therefore, 0.01 mol of $CuSO_4$ produces 0.005 mol of $I_2$ .

Step 4: Determine the mole ratio of $I_2$ to $Na_2S_2O_3$ .\nFrom the second equation, 1 mole of $I_2$ reacts with 2 moles of $Na_2S_2O_3$ . Therefore, 0.005 mol of $I_2$ reacts with $2 \times 0.005 = 0.01$ mol of $Na_2S_2O_3$ .

Step 5: Calculate the volume of $Na_2S_2O_3$ .\nVolume of $Na_2S_2O_3 = \frac{{Moles}}{{Molarity}} = \frac{{0.01 mol}}{{0.40 M}} = 0.025 L = 25 mL$

Why other options are wrong:
- Option A: 12.5 mL: This is incorrect because it divides the correct answer by 2, possibly due to a miscalculation in the mole ratio or volume calculation.
- Option C: 50 mL: This is incorrect because it multiplies the correct answer by 2, possibly due to a miscalculation in the mole ratio or volume calculation.
- Option D: 2.5 mL: This is incorrect because it divides the correct answer by 10, possibly due to a miscalculation in the mole ratio or volume calculation.

Key Concepts:
- Stoichiometry
- Molarity
- Redox reactions

Memory Aids:
- Molarity = Moles / Volume (in L)

Quick Tips:
- Always balance the chemical equations before performing stoichiometric calculations.
- Ensure correct units are used in calculations.

86 Medium

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Understanding +4 | -1

A mixture contains two or more substances in ________ which are called its ________.

View answer and explanation

Correct option: C

any ratio, components

Option C (any ratio, components) is correct because mixtures can have variable composition, and the substances within are called components.

Solution approach

Correct Answer + Core Reason: any ratio, components. Mixtures are defined by variable composition and constituent components.

Step-by-Step Solution:

Step 1: Define a mixture. A mixture is a substance comprising two or more components not chemically bonded.

Step 2: Consider the ratio of components. The components of a mixture can be present in any ratio.

Step 3: Define the constituents. The individual substances making up the mixture are called components.

Step 4: Conclusion: A mixture contains two or more substances in any ratio, which are called its components.

Why other options are wrong:
- Option A: Fixed ratio, compounds: Mixtures do not have a fixed ratio, and the constituents are not necessarily compounds.
- Option B: Fixed ratio, elements: Mixtures do not have a fixed ratio, and the constituents are not necessarily elements.
- Option D: Any ratio, elements: While the ratio is correct, the constituents are not necessarily elements; they can be compounds as well.

Key Concepts:
- Definition of a mixture
- Composition of mixtures

Memory Aids:
- Mixtures = variable composition

Quick Tips:
- Remember the difference between mixtures and compounds.

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Application +4 | -1

Oxidation number of carbon in carbon suboxide is :

View answer and explanation

Correct option: B

+ \frac{4}{3}

Option B (

+ \frac{4}{3}

) is correct because the oxidation number of carbon in carbon suboxide (

C_3O_2

) is calculated based on the oxidation state of oxygen being -2.

Solution approach

Correct Answer + Core Reason: $+ \frac{4}{3}$ . The oxidation number is calculated by considering the overall charge of the molecule being zero.

Step-by-Step Solution:

Step 1: Determine the oxidation state of oxygen. Oxygen usually has an oxidation state of -2.

Step 2: Let the oxidation state of carbon be x. The formula for carbon suboxide is $C_3O_2$ .

Step 3: Set up the equation: $3x + 2(-2) = 0$

Step 4: Solve for x: $3x - 4 = 0 \Rightarrow 3x = 4 \Rightarrow x = \frac{4}{3}$

Step 5: The oxidation number of carbon is $+ \frac{4}{3}$ .

Why other options are wrong:
- Option A: This is incorrect because the calculation doesn't correctly account for the number of carbon and oxygen atoms in the molecule.
- Option C: This is incorrect as it overestimates the oxidation state of carbon in carbon suboxide.
- Option D: This is incorrect because oxidation numbers are usually positive for carbon when bonded to oxygen, and the magnitude is also incorrect.

Key Concepts:
- Oxidation number rules
- Balancing oxidation states in a compound

Memory Aids:
- Oxygen usually has an oxidation state of -2.

Quick Tips:
- Remember to set the sum of oxidation states to zero for neutral molecules.

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Remembering +4 | -1

The gram molecular weight of hydrogen peroxide is 34. What is the unit of gram molecular weight?

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because gram molecular weight is expressed as grams per mole.

Solution approach

Correct Answer + Core Reason: $g\;\mathrm{mo}{\mathrm{l}^{ - 1}}$ . Gram molecular weight is mass per mole.

Step-by-Step Solution:

Step 1: Define gram molecular weight. Gram molecular weight is the mass of one mole of a substance.

Step 2: Determine the units. Mass is measured in grams (g), and the amount of substance is measured in moles (mol). Therefore, the unit of gram molecular weight is grams per mole.

Step 3: Express the units. Grams per mole is written as $g/mol$ or $g \;mol^{-1}$ .

Why other options are wrong:
- Option A: g is incorrect because it only represents mass, not mass per mole.
- Option B: mole is incorrect because it represents the amount of substance, not mass per mole.
- Option D: $\mathrm{mol }\;\mathrm{g}$ is incorrect because it represents moles per gram, which is the inverse of gram molecular weight.

Key Concepts:
- Gram molecular weight
- Mole concept
- Units of measurement

Memory Aids:
- Gram molecular weight = grams / mole

Quick Tips:
- Remember that gram molecular weight is a ratio of mass to the amount of substance.

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Application +4 | -1

If

30 \mathrm{~mL}

of

\mathrm{H}_2

and

20 \mathrm{~mL}

of

\mathrm{O}_2

react to form water, what is left at the end of the reaction?

View answer and explanation

Correct option: D

5 \mathrm{~mL}

of

\mathrm{O}_2

Option D (

5 \mathrm{~mL}

of

\mathrm{O}_2

) is correct because the reaction consumes twice the volume of H2 compared to O2, leaving excess O2.

Solution approach

Correct Answer + Core Reason: $5 \mathrm{~mL}$ of $\mathrm{O}_2$ . The reaction consumes twice the volume of H2 compared to O2, leaving excess O2.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the formation of water: $2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$ .

Step 2: Determine the volume ratio of $\mathrm{H}_2$ to $\mathrm{O}_2$ from the balanced equation: 2 volumes of $\mathrm{H}_2$ react with 1 volume of $\mathrm{O}_2$ .

Step 3: Calculate the volume of $\mathrm{O}_2$ required to react with $30 \mathrm{~mL}$ of $\mathrm{H}_2$ : $\frac{{30 \mathrm{~mL} \; \mathrm{H}_2}}{{2}} = 15 \mathrm{~mL} \; \mathrm{O}_2$ .

Step 4: Calculate the amount of $\mathrm{O}_2$ left after the reaction: $20 \mathrm{~mL} \; \mathrm{O}_2 - 15 \mathrm{~mL} \; \mathrm{O}_2 = 5 \mathrm{~mL} \; \mathrm{O}_2$ .

Step 5: Therefore, $5 \mathrm{~mL}$ of $\mathrm{O}_2$ is left at the end of the reaction.

Why other options are wrong:
- Option A: $10 \mathrm{~mL}$ of $\mathrm{H}_2$ : This is incorrect because all the $\mathrm{H}_2$ is consumed in the reaction.
- Option B: $5 \mathrm{~mL}$ of $\mathrm{H}_2$ : This is incorrect because all the $\mathrm{H}_2$ is consumed in the reaction.
- Option C: $10 \mathrm{~mL}$ of $\mathrm{O}_2$ : This is incorrect because some of the $\mathrm{O}_2$ is consumed in the reaction.

Key Concepts:
- Stoichiometry
- Balanced chemical equations
- Volume ratios of gases

Memory Aids:
- Remember to balance the chemical equation before determining the volume ratios.

Quick Tips:
- Always write the balanced chemical equation first to determine the correct stoichiometric ratios.

90 Medium

90 of 116Question permalink

Analyzing +4 | -1

Choose the correct option based on following statements. Here ‘T’ stands for true and ‘F’ stands for false statement.
(i) Molecular mass of cane sugar

(C_{12}H_{22}O_{11})

is 182 amu.
(ii) 1 mole of cane sugar contains

6.022 \times 10^{23}

molecules of cane sugar.
(iii) 34.20 g of cane sugar contains

6.022 \times 10^{21}

molecules of cane sugar.

View answer and explanation

Correct option: C

FTF

Option C (FTF) is correct because statement (i) is false (molecular mass is 342 amu), statement (ii) is true, and statement (iii) is false (34.20 g contains

6.022 \times 10^{22}

molecules).

Solution approach

Correct Answer + Core Reason: FTF is the correct option because the first and third statements are false, while the second statement is true.

Step-by-Step Solution:

Step 1: Evaluate statement (i): Molecular mass of cane sugar $(C_{12}H_{22}O_{11})$ is 182 amu.\nMolecular mass = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 amu. Therefore, statement (i) is FALSE.

Step 2: Evaluate statement (ii): 1 mole of cane sugar contains $6.022 \times 10^{23}$ molecules of cane sugar.\nThis is the definition of a mole. Therefore, statement (ii) is TRUE.

Step 3: Evaluate statement (iii): 34.20 g of cane sugar contains $6.022 \times 10^{21}$ molecules of cane sugar.\nMolar mass of cane sugar = 342 g/mol.\nMoles of cane sugar in 34.20 g = $\frac{{34.20}}{{342}} = 0.1$ moles.\nNumber of molecules = 0.1 x $6.022 \times 10^{23} = 6.022 \times 10^{22}$ molecules. Therefore, statement (iii) is FALSE.

Step 4: Conclusion: The statements are False, True, False (FTF).

Why other options are wrong:
- Option A: TTF is incorrect because statement (i) is false and statement (iii) is false.
- Option B: TFT is incorrect because statement (i) is false and statement (iii) is false.
- Option D: FTT is incorrect because statement (iii) is false.

Key Concepts:
- Molecular Mass
- Mole Concept
- Avogadro's Number

Memory Aids:
- 1 mole = $6.022 \times 10^{23}$ particles

Quick Tips:
- Calculate the molecular mass carefully.
- Remember the definition of a mole.

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