$1.57 \times 10^{-24} \mathrm{~kg}$

$1.66 \times 10^{-24} \mathrm{~kg}$

$1.99 \times 10^{-23} \mathrm{~kg}$

The equivalent weight of ${\rm{Zn}}{\left( {{\rm{OH}}} \right)_2}$ in the following reaction is equal to its, ${\rm{Zn}}{\left( {{\rm{OH}}} \right)_2} + {\rm{HN}}{{\rm{O}}_3} \to {\rm{Zn}}\left( {{\rm{OH}}} \right)\left( {{\rm{N}}{{\rm{O}}_3}} \right) + {{\rm{H}}_2}{\rm{O}}$:

$\frac{{{\rm{Formula}}\;{\rm{wt}}{\rm{.}}}}{1}$

$\frac{{{\rm{Formula }}\;{\rm{wt}}{\rm{.}}}}{2}$

$2 \times {\rm{formula }}\;{\rm{wt}}{\rm{.}}$

$3 \times {\rm{formula }}\;{\rm{wt}}{\rm{.}}$

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

61 Easy

61 of 114Question permalink

Application +4 | -1

Consider the following reaction:

xMnO_4^ - + y{C_2}O_4^{2 - } + z{H^ + } \to xM{n^{2 + }} + 2yC{O_2} + \frac{z}{2}{H_2}O

The values of x, y and z in the reaction are, respectively

View answer and explanation

Correct option: C

2, 5 and 16

Option C (2, 5 and 16) is correct because balancing the redox reaction requires 2 moles of permanganate, 5 moles of oxalate, and 16 moles of hydrogen ions.

Solution approach

Correct Answer + Core Reason: 2, 5 and 16. Balancing the redox reaction gives these coefficients.

Step-by-Step Solution:

Step 1: Write the unbalanced half-reactions:\nOxidation: ${C_2}O_4^{2 - } \to C{O_2}$ \nReduction: $MnO_4^ - \to M{n^{2 + }}$

Step 2: Balance the atoms in each half-reaction:\nOxidation: ${C_2}O_4^{2 - } \to 2C{O_2}$ \nReduction: $MnO_4^ - \to M{n^{2 + }}$

Step 3: Balance the oxygen atoms by adding water:\nOxidation: ${C_2}O_4^{2 - } \to 2C{O_2}$ \nReduction: $MnO_4^ - \to M{n^{2 + }} + 4{H_2}O$

Step 4: Balance the hydrogen atoms by adding H+:\nOxidation: ${C_2}O_4^{2 - } \to 2C{O_2}$ \nReduction: $8{H^ + } + MnO_4^ - \to M{n^{2 + }} + 4{H_2}O$

Step 5: Balance the charge by adding electrons:\nOxidation: ${C_2}O_4^{2 - } \to 2C{O_2} + 2{e^ - }$ \nReduction: $5{e^ - } + 8{H^ + } + MnO_4^ - \to M{n^{2 + }} + 4{H_2}O$

Step 6: Multiply the half-reactions to equalize the number of electrons:\nOxidation: $5 \times ({C_2}O_4^{2 - } \to 2C{O_2} + 2{e^ - }) \Rightarrow 5{C_2}O_4^{2 - } \to 10C{O_2} + 10{e^ - }$ \nReduction: $2 \times (5{e^ - } + 8{H^ + } + MnO_4^ - \to M{n^{2 + }} + 4{H_2}O) \Rightarrow 10{e^ - } + 16{H^ + } + 2MnO_4^ - \to 2M{n^{2 + }} + 8{H_2}O$

Step 7: Add the balanced half-reactions:
$2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$

Step 8: Identify x, y, and z: x = 2, y = 5, z = 16

Step 9: Final Answer: 2, 5 and 16

Why other options are wrong:
- Option A: Incorrect coefficients after balancing.
- Option B: Incorrect coefficients after balancing.
- Option D: Incorrect coefficients after balancing.

Key Concepts:
- Redox reactions
- Balancing redox reactions
- Half-reactions

Memory Aids:
- Remember the steps for balancing redox reactions (e.g., half-reaction method).

Quick Tips:
- Balance redox reactions systematically using the half-reaction method.

62 Easy

62 of 114Question permalink

Remembering +4 | -1

Which of the following is not a SI unit?

View answer and explanation

Correct option: D

litre

Option D is correct because litre is not a base SI unit, but a derived unit for volume.

Solution approach

Correct Answer + Core Reason: litre

Step-by-Step Solution:

Step 1: List the base SI units: The base SI units are:
- Length: metre (m)
- Mass: kilogram (kg)
- Time: second (s)
- Electric current: ampere (A)
- Thermodynamic temperature: kelvin (K)
- Amount of substance: mole (mol)
- Luminous intensity: candela (cd)

Step 2: Check the options:
- metre, candela, and mole are base SI units.
- litre is a unit of volume, which is derived from the metre (1 L = 1 dm³ = (0.1 m)³ = 0.001 m³). Therefore, litre is not a base SI unit.

Why other options are wrong:
- Option A: metre is a base SI unit.
- Option B: candela is a base SI unit.
- Option C: mole is a base SI unit.

Key Concepts:
- SI units
- Base units
- Derived units

Memory Aids:
- Remember the seven base SI units.

Quick Tips:
- Know the difference between base and derived units.

63 Easy

63 of 114Question permalink

Analyzing +4 | -1

Assertion: The molality and molarity of very dilute aqueous solutions differ very little.

Reason: The density of water is about

1.0 \, \mathrm{g} \, \mathrm{cm}^{-3}

at room temperature.

View answer and explanation

Correct option: A

Assertion is true, Reason is true and Reason is correct explanation for Assertion

Option A (Assertion is true, Reason is true and Reason is correct explanation for Assertion) is correct because at very dilute concentrations, the density of the solution approaches the density of water (1 g/cm³), making molarity and molality nearly equal.

Solution approach

Correct Answer + Core Reason: Assertion is true, Reason is true and Reason is correct explanation for Assertion. The density of dilute aqueous solutions approaches that of water, making molarity and molality similar.

Step-by-Step Solution:

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

Step 2: Analyze the relationship between molarity and molality.\nMolarity = (moles of solute) / (volume of solution in liters)\nMolality = (moles of solute) / (mass of solvent in kilograms)

Step 3: Consider dilute aqueous solutions.\nIn dilute aqueous solutions, the volume of the solution is approximately equal to the volume of the solvent (water). Also, the mass of the solvent (in kg) is approximately equal to the volume of the solution (in liters) because the density of water is approximately 1 kg/L (or 1 g/cm³).

Step 4: Explain why molarity and molality are similar in dilute aqueous solutions.\nSince the volume of the solution (in liters) is approximately equal to the mass of the solvent (in kg) in dilute aqueous solutions, the molarity and molality values will be very close.

Step 5: Final Answer\nThe assertion is true, the reason is true, and the reason is the correct explanation for the assertion.

Why other options are wrong:
- Option B: This is incorrect because the reason is the correct explanation for the assertion.
- Option C: This is incorrect because both the assertion and the reason are true.
- Option D: This is incorrect because the assertion is true.

Key Concepts:
- Molarity
- Molality
- Density of water

Memory Aids:
- Molarity: moles per liter of solution
- Molality: moles per kilogram of solvent

Quick Tips:
- Remember that in dilute aqueous solutions, molarity and molality are approximately equal.

64 Easy

64 of 114Question permalink

Understanding +4 | -1

A,E,M and n are the atomic weight, equivalent weight, molecular weight and valence of an element. The correct relation is:

View answer and explanation

Correct option: A

A = E \times n

Option A (

A = E \times n

) is correct because the atomic weight (A) is equal to the equivalent weight (E) multiplied by the valence (n).

Solution approach

Correct Answer + Core Reason: $A = E \times n$ . This is the correct relationship between atomic weight, equivalent weight, and valence.

Step-by-Step Solution:

Step 1: Define the terms.\nA = Atomic weight\nE = Equivalent weight\nM = Molecular weight\nn = Valence

Step 2: Recall the relationship between atomic weight, equivalent weight, and valence.\nAtomic weight (A) = Equivalent weight (E) × Valence (n)
$A = E \times n$

Step 3: Conclusion.\nTherefore, the correct relation is $A = E \times n$ .

Why other options are wrong:
- Option B: $A = M/E$ is incorrect. It does not represent the relationship between atomic weight, equivalent weight, and molecular weight.
- Option C: $A = M/n$ is incorrect. It relates atomic weight to molecular weight and valence, but not in the correct way.
- Option D: $M = A \times n$ is incorrect. This would be true if M was the atomic weight of a molecule with n atoms of the same element, but it's not generally true for molecular weight.

Key Concepts:
- Atomic weight
- Equivalent weight
- Molecular weight
- Valence

Memory Aids:
- Remember that Atomic weight = Equivalent weight × Valence.

Quick Tips:
- Understand the definitions of atomic weight, equivalent weight, molecular weight, and valence.

65 Easy

65 of 114Question permalink

Remembering +4 | -1

One amu is equal to:

View answer and explanation

Correct option: D

1.66 \times 10^{-27} \mathrm{~kg}

Option D (

1.66 \times 10^{-27} \mathrm{~kg}

) is correct because one atomic mass unit (amu) is defined as 1/12th the mass of a carbon-12 atom, which is approximately

1.66 \times 10^{-27} \mathrm{~kg}

.

Solution approach

Correct Answer + Core Reason: $1.66 \times 10^{-27} \mathrm{~kg}$ . This is the standard value for 1 amu.

Step-by-Step Solution:

Step 1: Define atomic mass unit (amu).\nOne atomic mass unit (amu) is defined as 1/12th the mass of a carbon-12 atom.

Step 2: State the value of 1 amu in kg.
1 amu is approximately equal to $1.66 \times 10^{-27} \mathrm{~kg}$ .

Step 3: Conclusion.\nTherefore, one amu is equal to $1.66 \times 10^{-27} \mathrm{~kg}$ .

Why other options are wrong:
- Option A: $1.57 \times 10^{-24} \mathrm{~kg}$ is not the correct value for 1 amu.
- Option B: $1.66 \times 10^{-24} \mathrm{~kg}$ is incorrect; the exponent is wrong.
- Option C: $1.99 \times 10^{-23} \mathrm{~kg}$ is the approximate mass of a carbon-12 atom, not 1 amu.

Key Concepts:
- Atomic mass unit (amu)
- Mass of an atom

Memory Aids:
- Remember the approximate value of 1 amu: $1.66 \times 10^{-27} \mathrm{~kg}$ .

Quick Tips:
- Keep track of the units and exponents when dealing with atomic masses.

66 Easy

66 of 114Question permalink

Remembering +4 | -1

The numerical value of N/n (where N is number of molecules is n moles of gas) is:

View answer and explanation

Correct option: B

6.02 \times {10^{23}}

Option B (6.02 × 10²³) is correct because N/n represents Avogadro's number, which is approximately 6.02 × 10²³.

Solution approach

Correct Answer + Core Reason: $6.02 \times {10^{23}}$ . N/n represents Avogadro's number, which is a constant.

Step-by-Step Solution:

Step 1: Define the terms.
- N = Number of molecules
- n = Number of moles

Step 2: Recall the relationship between moles and the number of molecules.
- Number of molecules = Number of moles * Avogadro's number
- N = n * N $_A$ , where N $_A$ is Avogadro's number.

Step 3: Calculate N/n.
- N/n = N $_A$ = Avogadro's number

Step 4: Recall the value of Avogadro's number.
- Avogadro's number = 6.022 x 10 $^{23}$ molecules/mol

Step 5: Conclusion: The numerical value of N/n is approximately 6.02 x 10 $^{23}$ .

Why other options are wrong:
- Option A: Incorrect because 8.314 is the ideal gas constant (R).
- Option C: Incorrect because $1.602 \times {10^{ - 24}}$ is related to the mass of a proton or neutron in grams.
- Option D: Incorrect because $1.66 \times {10^{ - 19}}$ is not a relevant constant in this context.

Key Concepts:
- Mole concept
- Avogadro's number

Memory Aids:
- Avogadro's number: 6.022 x 10 $^{23}$

Quick Tips:
- Remember the definition of a mole and Avogadro's number.

67 Easy

67 of 114Question permalink

Understanding +4 | -1

The number of significant figures for the three numbers 161 cm, 0.161 cm, 0.0161 cm are

View answer and explanation

Correct option: D

3,3 and 3 respectively

Option D is correct because leading zeros are not significant, and trailing zeros after a decimal point are significant.

Solution approach

Correct Answer + Core Reason: 3,3 and 3 respectively. Significant figures are counted based on rules for leading and trailing zeros.

Step-by-Step Solution:

Step 1: Recall the rules for significant figures.

Step 2: For 161 cm, all three digits are significant, so there are 3 significant figures.

Step 3: For 0.161 cm, the leading zero is not significant, but the other three digits are, so there are 3 significant figures.

Step 4: For 0.0161 cm, the leading zeros are not significant, but the other three digits are, so there are 3 significant figures.

Step 5: Final Answer: 3,3 and 3 respectively.

Why other options are wrong:
- Option A: 3,4 and 5 respectively is incorrect.
- Option B: 3,4 and 4 respectively is incorrect.
- Option C: 3,3 and 4 respectively is incorrect.

Key Concepts:
- Significant figures
- Rules for significant figures

Memory Aids:
- Remember the rules for counting significant figures.

Quick Tips:
- Pay attention to leading and trailing zeros.

68 Easy

68 of 114Question permalink

Analyzing +4 | -1

Which contains greatest number of oxygen atoms?

View answer and explanation

Correct option: D

All have the same number of atoms

Option D (All have the same number of atoms) is correct because 1 g of each substance contains the same number of oxygen atoms.

Solution approach

Correct Answer + Core Reason: All have the same number of atoms. This is because the mass of oxygen is the same in each case.

Step-by-Step Solution:

Step 1: Calculate the number of moles of O atoms in 1 g of O: Moles = $\frac{1}{16}$

Step 2: Calculate the number of moles of $O_2$ molecules in 1 g of $O_2$ : Moles = $\frac{1}{32}$ . Number of O atoms = $2 \times \frac{1}{32} = \frac{1}{16}$

Step 3: Calculate the number of moles of $O_3$ molecules in 1 g of $O_3$ : Moles = $\frac{1}{48}$ . Number of O atoms = $3 \times \frac{1}{48} = \frac{1}{16}$

Step 4: Since the number of moles of O atoms is the same in all three cases, they contain the same number of oxygen atoms.

Why other options are wrong:
- Option A: This is incorrect because it only considers the mass of O and not the number of atoms.
- Option B: This is incorrect because it only considers the mass of $O_2$ and not the number of atoms.
- Option C: This is incorrect because it only considers the mass of $O_3$ and not the number of atoms.

Key Concepts:
- Mole concept
- Molar mass

Memory Aids:
- Number of atoms = Number of moles x Avogadro's number x Number of atoms per molecule

Quick Tips:
- Always consider the number of atoms per molecule when comparing the number of atoms.

69 Easy

69 of 114Question permalink

Application +4 | -1

From the equation,

3{\mathrm{C}_2}{\mathrm{H}_2} \to {\mathrm{C}_6}{\mathrm{H}_6},

find the volume of acetylene (NTP) for the manufacture of 3 mole of benzene:

View answer and explanation

Correct option: C

201.6 litre

Option C (201.6 litre) is correct because 3 moles of benzene require 9 moles of acetylene, which occupies 201.6 L at NTP.

Solution approach

Correct Answer + Core Reason: 201.6 litre. The volume is calculated based on the stoichiometry of the reaction and the molar volume at NTP.

Step-by-Step Solution:

Step 1: Determine the moles of acetylene required.\nFrom the equation, $3{\mathrm{C}_2}{\mathrm{H}_2} \to {\mathrm{C}_6}{\mathrm{H}_6}$ , 3 moles of acetylene produce 1 mole of benzene.\nTo produce 3 moles of benzene, $3 \times 3 = 9$ moles of acetylene are required.

Step 2: Calculate the volume of acetylene at NTP.\nAt NTP, 1 mole of any gas occupies 22.4 litres.\nVolume = Number of moles $\times$ Molar volume = $9 \times 22.4 = 201.6 \text{{ litres}}$

Why other options are wrong:
- Option A: 67.2 litre: This is the volume for 3 moles of acetylene, not 9 moles.
- Option B: 134.4 litre: This is the volume for 6 moles of acetylene, not 9 moles.
- Option D: 33.8 litre: This is incorrect calculation.

Key Concepts:
- Stoichiometry
- Mole concept
- Molar volume at NTP

Memory Aids:
- 1 mole at NTP = 22.4 litres

Quick Tips:
- Balance the chemical equation first.

70 Easy

70 of 114Question permalink

Remembering +4 | -1

The numerical value of

\frac{\mathrm{N}}{n}

(where, Nis the number of molecules in a given sample of gas and n is the number of moles of the gas) is

View answer and explanation

Correct option: B

6.02 \times {10^{23}}

Option B (

6.02 \times {10^{23}}

) is correct because it represents Avogadro's number, which is the number of molecules per mole.

Solution approach

Correct Answer + Core Reason: $6.02 \times {10^{23}}$ . This is Avogadro's number, the number of molecules per mole.

Step-by-Step Solution:

Step 1: Define the terms. N is the number of molecules, and n is the number of moles.

Step 2: The ratio $\frac{\mathrm{N}}{n}$ represents the number of molecules per mole.

Step 3: The number of molecules per mole is defined as Avogadro's number, $N_A$ .

Step 4: Avogadro's number is approximately $6.022 \times 10^{23}$ .

Step 5: Therefore, $\frac{\mathrm{N}}{n} = N_A = 6.02 \times 10^{23}$ .

Why other options are wrong:
- Option A: 8.314 is the ideal gas constant (R) in J/(mol*K), not the number of molecules per mole.
- Option C: 0.0821 is the ideal gas constant (R) in L*atm/(mol*K), not the number of molecules per mole.
- Option D: $1.66 \times {10^{ - 19}}$ is not a standard constant related to the number of molecules per mole.

Key Concepts:
- Avogadro's Number
- Mole Concept

Memory Aids:
- Remember Avogadro's number as the bridge between the microscopic (molecules) and macroscopic (moles) worlds.

Quick Tips:
- Always remember the definition of a mole and its relationship to Avogadro's number.

71 Easy

71 of 114Question permalink

Analyzing +4 | -1

Which of the following samples contain the least number of atoms

View answer and explanation

Correct option: A

1g\,\,of\,C{O_2}\left( g \right)

Option A (

1g\,\,of\,C{O_2}\left( g \right)

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

Solution approach

Correct Answer + Core Reason: $1g\,\,of\,C{O_2}\left( g \right)$ . It has the smallest number of atoms because it has the smallest number of moles and a relatively small number of atoms per molecule.

Step-by-Step Solution:

Step 1: Calculate the number of moles for each sample:
- $1g\,of\,CO_2$ : Molar mass of $CO_2$ = 12 + 2(16) = 44 g/mol. Moles = 1/44
- $1g\,of\,C_8H_{18}$ : Molar mass of $C_8H_{18}$ = 8(12) + 18(1) = 96 + 18 = 114 g/mol. Moles = 1/114
- $1g\,of\,C_2H_6$ : Molar mass of $C_2H_6$ = 2(12) + 6(1) = 24 + 6 = 30 g/mol. Moles = 1/30
- $1g\,of\,LiF$ : Molar mass of $LiF$ = 7 + 19 = 26 g/mol. Moles = 1/26

Step 2: Calculate the number of atoms per molecule for each sample:
- $CO_2$ : 1 C + 2 O = 3 atoms
- $C_8H_{18}$ : 8 C + 18 H = 26 atoms
- $C_2H_6$ : 2 C + 6 H = 8 atoms
- $LiF$ : 1 Li + 1 F = 2 atoms

Step 3: Calculate the total number of atoms for each sample (proportional to moles × atoms per molecule):
- $CO_2$ : (1/44) × 3 = 3/44 ≈ 0.068
- $C_8H_{18}$ : (1/114) × 26 = 26/114 ≈ 0.228
- $C_2H_6$ : (1/30) × 8 = 8/30 ≈ 0.267
- $LiF$ : (1/26) × 2 = 2/26 ≈ 0.077

Step 4: Compare the values:
$CO_2$ has the smallest value (0.068).

Why other options are wrong:
- **Option B ( $1g\,of\,{C_8}{H_{18}}\left( l \right)$ ):** This is incorrect because it has a higher number of atoms than $CO_2$ .
- **Option C ( $1g\,of\,{C_2}{H_6}\left( g \right)$ ):** This is incorrect because it has a higher number of atoms than $CO_2$ .
- **Option D ( $1g\,of\,LiF\left( s \right)$ ):** This is incorrect because it has a higher number of atoms than $CO_2$ .

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

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

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

72 Easy

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Application +4 | -1

Maximum no. of atoms are present in

View answer and explanation

Correct option: A

8g\,C{H_4}

Option A (

8g\,C{H_4}

) is correct because it contains the highest number of moles of molecules, and each molecule has a significant number of atoms.

Solution approach

Correct Answer + Core Reason: $8g\,C{H_4}$ . It contains the maximum number of atoms due to the highest number of moles and atoms per molecule.

Step-by-Step Solution:

Step 1: Calculate the number of moles for each option.

Step 2: Option A: $C{H_4}$ (Molar mass = 12 + 4 = 16 g/mol). Moles = $\frac{{8}}{{16}} = 0.5$ mol. Number of atoms per molecule = 1 (C) + 4 (H) = 5. Total atoms = 0.5 mol * 5 atoms/molecule * $6.022 \times 10^{23}$ atoms/mol = $1.5055 \times 10^{24}$ atoms.

Step 3: Option B: ${C_6}{H_{12}}{O_6}$ (Molar mass = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g/mol). Moles = $\frac{{8}}{{180}} = 0.044$ mol. Number of atoms per molecule = 6 (C) + 12 (H) + 6 (O) = 24. Total atoms = 0.044 mol * 24 atoms/molecule * $6.022 \times 10^{23}$ atoms/mol = $6.35 \times 10^{22}$ atoms.

Step 4: Option C: $CaC{O_3}$ (Molar mass = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g/mol). Moles = $\frac{{25}}{{100}} = 0.25$ mol. Number of atoms per molecule = 1 (Ca) + 1 (C) + 3 (O) = 5. Total atoms = 0.25 mol * 5 atoms/molecule * $6.022 \times 10^{23}$ atoms/mol = $7.5275 \times 10^{23}$ atoms.

Step 5: Option D: $CO$ (Molar mass = 12 + 16 = 28 g/mol). Moles = $\frac{{28}}{{28}} = 1$ mol. Number of atoms per molecule = 1 (C) + 1 (O) = 2. Total atoms = 1 mol * 2 atoms/molecule * $6.022 \times 10^{23}$ atoms/mol = $1.2044 \times 10^{24}$ atoms.

Step 6: Compare the total number of atoms for each option. Option A has the highest number of atoms.

Why other options are wrong:
- Option B: Although ${C_6}{H_{12}}{O_6}$ has a large number of atoms per molecule, the number of moles is small, resulting in fewer total atoms than $C{H_4}$ .
- Option C: $CaC{O_3}$ has a relatively small number of moles and atoms per molecule, leading to fewer total atoms than $C{H_4}$ .
- Option D: $CO$ has a large number of moles, but only two atoms per molecule, resulting in fewer total atoms than $C{H_4}$ .

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

Memory Aids:
- Remember that the number of atoms is proportional to the number of moles and the number of atoms per molecule.

Quick Tips:
- Calculate the number of moles for each compound.
- Multiply the number of moles by the number of atoms in each molecule and Avogadro's number to find the total number of atoms.

73 Easy

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Remembering +4 | -1

The apparatus in which standard solution is prepared is known as :

View answer and explanation

Correct option: A

Measuring flask

Option A (Measuring flask) is correct because a measuring flask is specifically designed for preparing standard solutions with accurate volumes.

Solution approach

Correct Answer + Core Reason: Measuring flask. It is designed for preparing standard solutions with accurate volumes.

Step-by-Step Solution:

Step 1: Define a standard solution: A solution of accurately known concentration.

Step 2: Consider the requirements for preparing a standard solution: Accurate measurement of volume is crucial.

Step 3: Evaluate the given options:
- Measuring flask: Designed for preparing solutions of specific volumes with high accuracy. It has a narrow neck with a calibration mark.
- Round bottom flask: Used for heating or storing liquids, but not for accurate volume measurements.
- Burette: Used for dispensing known volumes of liquid, not for preparing solutions.

Step 4: Conclude that a measuring flask is the appropriate apparatus.

Why other options are wrong:
- Option B: Round bottom flasks are used for heating and reactions, not for accurate volume measurements.
- Option C: Burettes are used for titrations and dispensing liquids, not for preparing standard solutions.
- Option D: This is a general negative and incorrect.

Key Concepts:
- Standard Solutions
- Volumetric Analysis

Memory Aids:
- Measuring flask = Accurate volume

Quick Tips:
- Know the uses of different types of glassware in the lab.

74 Easy

74 of 114Question permalink

Understanding +4 | -1

Which one of the following set of units represents the smallest and largest amount of energy respectively?

View answer and explanation

Correct option: D

eV and L-atm

Option D (eV and L-atm) is correct because eV (electron volt) is a very small unit of energy, while L-atm (litre-atmosphere) is a larger unit of energy.

Solution approach

Correct Answer + Core Reason: eV and L-atm. eV is the smallest and L-atm is the largest among the given options.

Step-by-Step Solution:

Step 1: Recall the approximate values of the energy units.

Step 2: 1 eV (electron volt) = $1.602 \times 10^{-19}$ J
1 erg = $10^{-7}$ J
1 cal (calorie) = 4.184 J
1 L-atm (litre-atmosphere) = 101.325 J

Step 3: Compare the values.

Step 4: eV is the smallest, and L-atm is the largest.

Why other options are wrong:
- Option A: J and erg is incorrect. Erg is smaller than Joule.
- Option B: erg and cal is incorrect. Erg is smaller than calorie.
- Option C: Cal and eV is incorrect. eV is smaller than calorie.

Key Concepts:
- Energy units
- Electron volt (eV)
- Erg
- Calorie (cal)
- Litre-atmosphere (L-atm)

Memory Aids:
- Remember the relative magnitudes of common energy units.

Quick Tips:
- Knowing the conversion factors between energy units is helpful.

75 Easy

75 of 114Question permalink

Application +4 | -1

The equivalent weight of

\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}

in the following reaction is equal to its,

\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2} + \mathrm{HN}{\mathrm{O}_3} \to \mathrm{Zn}\left( {\mathrm{OH}} \right)\left( {\mathrm{N}{\mathrm{O}_3}} \right) + {\mathrm{H}_2}\mathrm{O}

:

View answer and explanation

Correct option: A

\frac{{\mathrm{Formula}\;\mathrm{wt}\mathrm{.}}}{1}

Option A (Formula wt./1) is correct because in this reaction,

Zn(OH)_2

reacts with one equivalent of

HNO_3

, so its equivalent weight is its formula weight divided by 1.

Solution approach

Correct Answer + Core Reason: $\frac{{\mathrm{Formula}\;\mathrm{wt}\mathrm{.}}}{1}$ . The compound reacts with one equivalent of acid.

Step-by-Step Solution:

Step 1: Write the reaction: $Zn(OH)_2 + HNO_3 \to Zn(OH)(NO_3) + H_2O$

Step 2: Determine the n-factor: In this reaction, one $OH^-$ group of $Zn(OH)_2$ is replaced by one $NO_3^-$ group. Therefore, the n-factor is 1.

Step 3: Calculate the equivalent weight: Equivalent weight = $\frac{{\text{{Formula weight}}}}{{\text{{n-factor}}}} = \frac{{\text{{Formula weight}}}}{1}$

Step 4: Final Answer: $\frac{{\mathrm{Formula}\;\mathrm{wt}\mathrm{.}}}{1}$

Why other options are wrong:
- Option B: This would be correct if the n-factor was 2, which is not the case here.
- Option C: This is incorrect because it multiplies the formula weight by 2.
- Option D: This is incorrect because it multiplies the formula weight by 3.

Key Concepts:
- Equivalent Weight
- n-factor

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

Quick Tips:
- Determine the n-factor based on the number of replaceable $OH^-$ or $H^+$ ions.

76 Easy

76 of 114Question permalink

Remembering +4 | -1

Difference in density is the basis of

View answer and explanation

Correct option: D

gravity separation

Option D is correct because gravity separation is a technique that relies on differences in density to separate components of a mixture.

Solution approach

Correct Answer + Core Reason: gravity separation

Step-by-Step Solution:

Step 1: Define Gravity Separation: Gravity separation is a method used to separate components of a mixture based on differences in their densities. Heavier components settle down, while lighter components remain on top or are carried away by a fluid.

Step 2: Consider other options:
- Ultrafiltration: Separates based on particle size using a membrane.
- Molecular sieving: Separates based on molecular size using a material with pores of specific sizes.
- Molecular attraction: Refers to intermolecular forces and is not a separation technique based on density.

Why other options are wrong:
- Option A: Ultrafiltration is based on particle size, not density.
- Option B: Molecular sieving is based on molecular size, not density.
- Option C: Molecular attraction is a property, not a separation technique based on density.

Key Concepts:
- Gravity separation
- Density
- Separation techniques

Memory Aids:
- Gravity pulls denser things down.

Quick Tips:
- Think about how different materials settle in a mixture based on their weight.

77 Easy

77 of 114Question permalink

Understanding +4 | -1

Which statement is correct?

View answer and explanation

Correct option: C

Equivalent weight changes with valence

Option C (Equivalent weight changes with valence) is correct because the equivalent weight of an element is calculated by dividing its atomic weight by its valence; therefore, a change in valence directly affects the equivalent weight.

Solution approach

Correct Answer + Core Reason: Equivalent weight changes with valence. The equivalent weight is dependent on the valence of the element.

Step-by-Step Solution:

Step 1: Define equivalent weight.\nEquivalent weight = Atomic weight / Valence

Step 2: Analyze the relationship.\nSince the atomic weight is constant for a given element, the equivalent weight changes if the valence changes.

Step 3: Conclusion.\nTherefore, the statement "Equivalent weight changes with valence" is correct.

Why other options are wrong:
- Option A: Atomic weight of an element varies with valence is incorrect. Atomic weight is a fundamental property of an element and does not change with valence.
- Option B: Molecular weight changes with valence is incorrect. Molecular weight depends on the compound formed, not the valence of an individual element in isolation.
- Option D: None of the above is incorrect because option C is correct.

Key Concepts:
- Equivalent weight
- Atomic weight
- Valence

Memory Aids:
- Remember the formula: Equivalent weight = Atomic weight / Valence

Quick Tips:
- Understand the relationship between equivalent weight, atomic weight, and valence.

78 Easy

78 of 114Question permalink

Application +4 | -1

510 mg of a liquid on vaporization in Victor meyer’s apparatus displaces

\mathrm{67}\mathrm{.2}\;\mathrm{c}{\mathrm{m}^3}

of air at (STP). The molecular weight of the liquid is:

View answer and explanation

Correct option: C

170

Option C (170) is correct because it is calculated using the ideal gas law and the given mass and volume.

Solution approach

Correct Answer + Core Reason: 170. The molecular weight is calculated using the ideal gas law and the given data from Victor Meyer's apparatus.

Step-by-Step Solution:

Step 1: Convert the volume to litres.\nVolume = $67.2 \text{{ cm}}^3 = 0.0672 \text{{ L}}$

Step 2: Apply the ideal gas law, $PV = nRT$ , to find the number of moles (n).\nAt STP, P = 1 atm, T = 273 K, R = 0.0821 L atm / (mol K)\nn = $\frac{{PV}}{{RT}} = \frac{{1 \times 0.0672}}{{0.0821 \times 273}} \approx 0.003 \text{{ moles}}$

Step 3: Calculate the molecular weight.\nMolecular weight = $\frac{{\text{{Mass}}}}{{\text{{Number of moles}}}} = \frac{{510 \text{{ mg}}}}{{0.003 \text{{ moles}}}} = \frac{{0.510 \text{{ g}}}}{{0.003 \text{{ moles}}}} = 170 \text{{ g/mol}}$

Why other options are wrong:
- Option A: 130: This is incorrect calculation.
- Option B: 17: This is incorrect calculation.
- Option D: 1700: This is incorrect calculation.

Key Concepts:
- Ideal gas law
- Mole concept
- Victor Meyer's apparatus

Memory Aids:
- PV = nRT

Quick Tips:
- Remember the ideal gas constant and STP conditions.

79 Easy

79 of 114Question permalink

Application +4 | -1

The number of atoms in 0.1 mol of a triatomic gas is

View answer and explanation

Correct option: B

1.806 \times {10^{23}}

Option B (

1.806 \times {10^{23}}

) is correct because it calculates the total number of atoms by multiplying the number of moles by Avogadro's number and the number of atoms per molecule.

Solution approach

Correct Answer + Core Reason: $1.806 \times {10^{23}}$ . The number of atoms is calculated by multiplying the number of moles by Avogadro's number and the number of atoms per molecule.

Step-by-Step Solution:

Step 1: Identify the given information: 0.1 mol of a triatomic gas.

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

Step 3: Since the gas is triatomic, each molecule has 3 atoms.

Step 4: Calculate the number of molecules: $0.1 \text{{ mol}} \times 6.022 \times {10^{23}} \text{{ molecules/mol}} = 6.022 \times {10^{22}} \text{{ molecules}}$

Step 5: Calculate the number of atoms: $6.022 \times {10^{22}} \text{{ molecules}} \times 3 \text{{ atoms/molecule}} = 1.8066 \times {10^{23}} \text{{ atoms}}$

Step 6: Round to significant figures: $1.806 \times {10^{23}}$

Final Answer: $1.806 \times {10^{23}}$

Why other options are wrong:
- Option A: $6.02 \times {10^{22}}$ : This is the number of molecules, not atoms. It forgets to multiply by 3 (atoms per molecule).
- Option C: $3.6 \times {10^{33}}$ : This is an incorrect calculation, likely due to a misunderstanding of exponents or Avogadro's number.
- Option D: $1.8 \times {10^{22}}$ : This is the number of atoms if the gas was monoatomic after multiplying by 0.3 instead of 3.

Key Concepts:
- Mole concept
- Avogadro's number
- Stoichiometry

Memory Aids:
- Remember to multiply by the number of atoms per molecule when calculating the total number of atoms.

Quick Tips:
- Always pay attention to the units and make sure they cancel out correctly.

80 Easy

80 of 114Question permalink

Analyzing +4 | -1

In the reaction,

\mathrm{VO} + \mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} \to \mathrm{FeO} + {\mathrm{V}_2}{\mathrm{O}_5}\mathrm{.}

. The eq.wt. of

{\mathrm{V}_2}{\mathrm{O}_5}

is equal to its :

View answer and explanation

Correct option: C

mol.\mathrm{ }wt./6

Option C (

mol.\mathrm{ }wt./6

) is correct because the equivalent weight is the molecular weight divided by the change in oxidation state of Vanadium, which is 6.

Solution approach

Correct Answer + Core Reason: $mol.\mathrm{ }wt./6$ . The equivalent weight is the molecular weight divided by the change in oxidation state.

Step-by-Step Solution:

Step 1: Identify the oxidation states of Vanadium in the reactants and products.\nIn VO, the oxidation state of V is +2.\nIn V2O5, the oxidation state of V is +5.

Step 2: Calculate the change in oxidation state.\nThe change in oxidation state is 5 - 2 = 3 per Vanadium atom.\nSince there are two Vanadium atoms in V2O5, the total change in oxidation state is 2 * 3 = 6.

Step 3: Calculate the equivalent weight.\nEquivalent weight = (Molecular weight) / (Change in oxidation state) = (Molecular weight) / 6

Step 4: Final Answer\nThe equivalent weight of V2O5 is its molecular weight divided by 6.

Why other options are wrong:
- Option A: This is incorrect because the equivalent weight is not equal to the molecular weight in redox reactions.
- Option B: This is incorrect because the change in oxidation state is 6, not 8.
- Option D: This is incorrect because option C is correct.

Key Concepts:
- Equivalent weight in redox reactions
- Oxidation state

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

Quick Tips:
- Determine the change in oxidation state of the element undergoing redox reaction.

81 Easy

81 of 114Question permalink

Application +4 | -1

If

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

ionises as

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} + 2{\mathrm{H}_2}\mathrm{O} \to 2{\mathrm{H}_3}{\mathrm{O}^ + } + \mathrm{SO}_4^2

, then total number of ions produced by

\mathrm{0}\mathrm{.1}\;\mathrm{M}\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

will be

View answer and explanation

Correct option: D

1.8 \times {10^{23}}

Option D (

1.8 \times {10^{23}}

) is correct because each H2SO4 molecule produces 3 ions, and we have 0.1 moles of H2SO4.

Solution approach

Correct Answer + Core Reason: $1.8 \times {10^{23}}$ . The total number of ions is calculated based on the ionization of H2SO4 and Avogadro's number.

Step-by-Step Solution:

Step 1: Determine the number of moles of $H_2SO_4$ .\nMolarity = 0.1 M\nAssuming 1 L solution, moles of $H_2SO_4$ = 0.1 moles

Step 2: Determine the number of ions produced per molecule of $H_2SO_4$ .
$H_2SO_4 + 2H_2O \to 2H_3O^+ + SO_4^{2-}$ \nEach molecule of $H_2SO_4$ produces 2 + 1 = 3 ions.

Step 3: Calculate the total number of ions.\nTotal number of ions = Moles of $H_2SO_4$ $\times$ Number of ions per molecule $\times$ Avogadro's number\nTotal number of ions = $0.1 \times 3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{23} \approx 1.8 \times 10^{23}$

Why other options are wrong:
- Option A: $9.03 \times {10^{21}}$ : This is incorrect calculation.
- Option B: $3.01 \times {10^{22}}$ : This is incorrect calculation.
- Option C: $6.02 \times {10^{22}}$ : This is incorrect calculation.

Key Concepts:
- Molarity
- Ionization of acids
- Avogadro's number

Memory Aids:
- Remember Avogadro's number: $6.022 \times 10^{23}$

Quick Tips:
- Pay attention to the number of ions produced during ionization.

82 Easy

82 of 114Question permalink

Remembering +4 | -1

The relation between molecular weight

\left( M \right)

and vapour density

\left( VD \right)

is:

View answer and explanation

Correct option: B

M = 2 \times \mathrm{VD}

Option B (

M = 2 \times \mathrm{VD}

) is correct because the molecular weight is twice the vapor density.

Solution approach

Correct Answer + Core Reason: $M = 2 \times \mathrm{VD}$ . Molecular weight is twice the vapor density.

Step-by-Step Solution:

Step 1: Define Molecular Weight (M): The mass of one mole of a substance.

Step 2: Define Vapor Density (VD): The ratio of the density of a vapor to the density of hydrogen at the same temperature and pressure.
$VD = \frac{{\text{{Density of vapor}}}}{{\text{{Density of }}{H_2}}}$

Step 3: Relate Vapor Density to Molar Mass using the ideal gas law.\nFrom the ideal gas law, $PV = nRT$ , we know that density $\rho = \frac{{PM}}{{RT}}$ , where M is the molar mass.\nSo, $VD = \frac{{\rho_{\text{{vapor}}}}}{{\rho_{{H_2}}}}} = \frac{{\frac{{PM_{\text{{vapor}}}}}}{{RT}}}}{{\frac{{PM_{{H_2}}}}{{RT}}}} = \frac{{M_{\text{{vapor}}}}}{{M_{{H_2}}}}$ \nSince the molar mass of $H_2$ is 2 g/mol, we have:
$VD = \frac{M}{2}$

Step 4: Rearrange the equation to solve for M.\nM = 2 * VD

Why other options are wrong:
- Option A: Incorrect relationship between M and VD.
- Option C: Incorrect relationship between M and VD.
- Option D: Incorrect relationship between M and VD.

Key Concepts:
- Molecular Weight
- Vapor Density
- Ideal Gas Law

Memory Aids:
- M = 2VD (Remember the factor of 2)

Quick Tips:
- Understand the definitions of molecular weight and vapor density. Remember the relationship derived from the ideal gas law.

83 Easy

83 of 114Question permalink

Remembering +4 | -1

The prefix

{10^{ - 24}}

is

View answer and explanation

Correct option: C

yocto

Option C is correct because the prefix yocto represents 10^-24.

Solution approach

Correct Answer + Core Reason: yocto

Step-by-Step Solution:

Step 1: Recall SI prefixes: SI prefixes are used to denote multiples or submultiples of SI units.

Step 2: Identify ${10^{ - 24}}$ : The prefix for ${10^{ - 24}}$ is yocto.

Why other options are wrong:
- Option A: yotta is ${10^{24}}$ .
- Option B: zeta is ${10^{21}}$ .
- Option D: zepto is ${10^{ - 21}}$ .

Key Concepts:
- SI prefixes

Memory Aids:
- Remember the common SI prefixes and their corresponding powers of 10.

Quick Tips:
- Memorize the common SI prefixes.

84 Easy

84 of 114Question permalink

Application +4 | -1

Which gas contains larger number of molecules?

View answer and explanation

Correct option: A

4\mathrm{g}\;\mathrm{of}{\mkern 1mu} \;{\mathrm{H}_2}\mathrm{O}

Option A (

4\mathrm{g}\;\mathrm{of}{\mkern 1mu} \;{\mathrm{H}_2}\mathrm{O}

) is correct because it has the highest number of moles compared to the other options, and therefore the largest number of molecules.

Solution approach

Correct Answer + Core Reason: $4\mathrm{g}\;\mathrm{of}{\mkern 1mu} \;{\mathrm{H}_2}\mathrm{O}$ . This contains the largest number of molecules because it has the highest number of moles.

Step-by-Step Solution:

Step 1: Calculate the number of moles for each option using the formula: moles = mass / molar mass.
- A: $4\mathrm{g}\;\mathrm{of}{\mkern 1mu} \;{\mathrm{H}_2}\mathrm{O}$ : Molar mass of $H_2O$ = 18 g/mol. Moles = 4/18 = 0.222 mol
- B: $2\mathrm{g}\;\mathrm{of}\;\mathrm{marsh}\;\mathrm{gas}$ ( $CH_4$ ): Molar mass of $CH_4$ = 16 g/mol. Moles = 2/16 = 0.125 mol
- C: $4\mathrm{g}\;\mathrm{of}{\mkern 1mu} \;\mathrm{PC}{\mathrm{l}_5}$ : Molar mass of $PCl_5$ = 208.5 g/mol. Moles = 4/208.5 = 0.019 mol
- D: $2\mathrm{g}\;\mathrm{of}\;\mathrm{phosgene}$ ( $COCl_2$ ): Molar mass of $COCl_2$ = 98.9 g/mol. Moles = 2/98.9 = 0.020 mol

Step 2: Compare the number of moles. The largest number of moles is in option A.

Step 3: Conclusion: Option A contains the largest number of molecules.

Why other options are wrong:
- Option B: The number of moles of methane is less than that of water.
- Option C: The number of moles of phosphorus pentachloride is significantly less than that of water.
- Option D: The number of moles of phosgene is also less than that of water.

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

Memory Aids:
- moles = mass / molar mass

Quick Tips:
- Always convert mass to moles before comparing the number of molecules.

85 Easy

85 of 114Question permalink

Analyzing +4 | -1

A: Number of ions in 9 gram of

{NH}_4^{+}

is equal to Avogadro's number

\left({N}_{{A}}\right)

.

R: Number of ions is equal to number of atoms.

View answer and explanation

Correct option: D

Both the A and R are incorrect.

Option D is correct because both the assertion and the reason are incorrect. 9g of NH4+ is not equal to Avogadro's number of ions, and the number of ions is not always equal to the number of atoms.

Solution approach

Correct Answer + Core Reason: Both the A and R are incorrect. The assertion and the reason are both false statements.

Step-by-Step Solution:

Step 1: Analyze the assertion (A).
9 g of NH4+ corresponds to 9 g / 18 g/mol = 0.5 mol of NH4+ ions. This means there are 0.5 * NA ions, not NA ions.

Step 2: Analyze the reason (R).\nThe number of ions is not always equal to the number of atoms. For example, in NaCl, one formula unit contains two ions (Na+ and Cl-) but two atoms.

Step 3: Conclusion.\nSince both the assertion and the reason are incorrect, option D is the correct answer.

Why other options are wrong:
- Option A: Incorrect because both A and R are not correct.
- Option B: Incorrect because both A and R are not correct.
- Option C: Incorrect because A is not correct.
- Option D: Correct because both A and R are incorrect.

Key Concepts:
- Mole concept
- Ions and atoms

Memory Aids:
- Remember the definition of a mole: 1 mole = Avogadro's number of particles.

Quick Tips:
- Always calculate the number of moles before comparing to Avogadro's number.

86 Easy

86 of 114Question permalink

Application +4 | -1

What is the mass percent of carbon in

C{O_2}

View answer and explanation

Correct option: B

27.27\mathrm{ }\%

Option B (

27.27\mathrm{ }\%

) is correct because it is the mass percentage of carbon calculated from the molar masses of carbon and carbon dioxide.

Solution approach

Correct Answer + Core Reason: $27.27\mathrm{ }\%$ . The mass percent is calculated by dividing the mass of carbon by the mass of carbon dioxide and multiplying by 100.

Step-by-Step Solution:

Step 1: Determine the molar mass of carbon (C) and carbon dioxide ( $C{O_2}$ ).\nMolar mass of C = 12 g/mol\nMolar mass of $C{O_2}$ = 12 + 2(16) = 44 g/mol

Step 2: Calculate the mass percent of carbon in $C{O_2}$ using the formula:\nMass percent of C = $\frac{{\text{{Mass of C}}}}{{\text{{Mass of }}C{O_2}}} \times 100$

Step 3: Substitute the molar masses into the formula:\nMass percent of C = $\frac{{12}}{{44}} \times 100$

Step 4: Calculate the result:\nMass percent of C = 0.2727 x 100 = 27.27%

Why other options are wrong:
- Option A: $0.034\mathrm{ }\%$ is incorrect; it's a very low value and doesn't represent the carbon content in $CO_2$ .
- Option C: $3.4\mathrm{ }\%$ is incorrect; it's significantly lower than the actual carbon content.
- Option D: $28.7\mathrm{ }\%$ is incorrect; it's slightly higher than the actual carbon content.

Key Concepts:
- Molar mass
- Mass percent composition

Memory Aids:
- Mass percent = (Mass of element / Mass of compound) x 100

Quick Tips:
- Always use the correct molar masses for each element and compound.

87 Easy

87 of 114Question permalink

Analyzing +4 | -1

Equivalent weight of

\mathrm{FeS}_2

in the half reaction,

\mathrm{FeS}_2 \longrightarrow \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{SO}_2

is :

View answer and explanation

Correct option: B

M / 11

Option B (M / 11) is correct because the change in oxidation state of Fe and S in the given half-reaction leads to a total of 11 electrons transferred per molecule of FeS2.

Solution approach

Correct Answer + Core Reason: $M / 11$ . The equivalent weight is the molecular weight divided by the number of electrons transferred.

Step-by-Step Solution:

Step 1: Write the balanced half-reaction: $2FeS_2 \longrightarrow Fe_2O_3 + 4SO_2$

Step 2: Determine the oxidation state changes: Fe goes from +2 to +3 (change of 1 per Fe, so 2 total). S goes from -1 to +4 (change of 5 per S, so 10 total).

Step 3: Calculate the total number of electrons transferred: 2 (from Fe) + 10 (from S) = 11 electrons per $FeS_2$ molecule. Since there are 2 $FeS_2$ molecules, there are 22 electrons transferred for the balanced reaction. However, the question asks for the equivalent weight of $FeS_2$ , so we consider only one molecule.

Step 4: Equivalent weight of $FeS_2 = \frac{{\text{{Molecular weight of }} FeS_2}}{{\text{{Number of electrons transferred per molecule}}}} = \frac{M}{{11}}$

Step 5: Final Answer: $M / 11$

Why other options are wrong:
- Option A: $M / 10$ : Incorrect calculation of the number of electrons transferred.
- Option C: $M / 6$ : Incorrect calculation of the number of electrons transferred.
- Option D: $M / 1$ : Incorrect calculation of the number of electrons transferred.

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

Memory Aids:
- Remember to consider the stoichiometry of the balanced half-reaction.

Quick Tips:
- Balance the half-reaction carefully and determine the oxidation state changes correctly.

88 Easy

88 of 114Question permalink

Application +4 | -1

The empirical formula of an organic compound is

\mathbf{C}{\mathbf{H}_\mathbf{2}}\mathbf{O}

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

View answer and explanation

Correct option: C

{C_3}{H_6}{O_3}

Option C (

{C_3}{H_6}{O_3}

) is correct because the molecular formula is a multiple of the empirical formula, and the molar mass calculated from the vapor density matches the molar mass of C3H6O3.

Solution approach

Correct Answer + Core Reason: ${C_3}{H_6}{O_3}$ . The molecular formula is a multiple of the empirical formula, and the molar mass matches the vapor density.

Step-by-Step Solution:

Step 1: Determine the empirical formula mass of $CH_2O$ .\nEmpirical formula mass = 12 (C) + 2(1) (H) + 16 (O) = 30 g/mol

Step 2: Calculate the molecular weight using the vapor density.\nMolecular weight = 2 * Vapor density = 2 * 45 = 90 g/mol

Step 3: Find the ratio (n) between the molecular weight and the empirical formula mass.\nn = Molecular weight / Empirical formula mass = 90 / 30 = 3

Step 4: Multiply the empirical formula by n to get the molecular formula.\nMolecular formula = $(CH_2O)_3 = C_3H_6O_3$

Step 5: Final Answer: ${C_3}{H_6}{O_3}$

Why other options are wrong:
- Option A: $CH_2O$ has a molar mass of 30, which doesn't match the calculated molecular weight of 90.
- Option B: ${C_2}{H_4}{O_2}$ has a molar mass of 60, which doesn't match the calculated molecular weight of 90.
- Option D: ${C_3}{H_{12}}{O_6}$ has a molar mass of 180, which doesn't match the calculated molecular weight of 90.

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

Memory Aids:
- Molecular weight = 2 * Vapor density
- Molecular formula = n * Empirical formula

Quick Tips:
- Calculate the empirical formula mass and use the vapor density to find the molecular weight.

89 Easy

89 of 114Question permalink

Application +4 | -1

The total number of electrons in 18 ml of water (density = 1/gml

View answer and explanation

Correct option: B

6.023\,\, \times \,\,{10^{^{24}}}

Option B (

6.023 \times {10^{^{24}}}

) is correct because it calculates the total number of electrons based on the number of water molecules and electrons per molecule.

Solution approach

Correct Answer + Core Reason: $6.023 \times {10^{^{24}}}$ . The total number of electrons is calculated from the number of water molecules and the number of electrons per water molecule.

Step-by-Step Solution:

Step 1: Calculate the mass of water.\nDensity = Mass/Volume, so Mass = Density x Volume = 1 g/mL x 18 mL = 18 g

Step 2: Calculate the number of moles of water.\nMoles = Mass/Molar mass = 18 g / 18 g/mol = 1 mol

Step 3: Calculate the number of water molecules.\nNumber of molecules = Moles x Avogadro's number = 1 mol x $6.023 \times {10^{23}}$ molecules/mol = $6.023 \times {10^{23}}$ molecules

Step 4: Calculate the number of electrons per water molecule.\nEach water molecule ( $H_2O$ ) has 2 hydrogen atoms and 1 oxygen atom. Hydrogen has 1 electron each, and oxygen has 8 electrons. So, the total number of electrons per water molecule is (2 x 1) + 8 = 10 electrons.

Step 5: Calculate the total number of electrons.\nTotal number of electrons = Number of molecules x Number of electrons per molecule = $6.023 \times {10^{23}}$ molecules x 10 electrons/molecule = $6.023 \times {10^{24}}$ electrons

Why other options are wrong:
- Option A: $6.023 \times {10^{^{25}}}$ : This is incorrect because it multiplies the correct answer by 10, possibly due to a miscalculation in the number of electrons or Avogadro's number.
- Option C: $6.023 \times 18 \times {10^{^{23}}}$ : This is incorrect because it multiplies Avogadro's number by 18, which is the mass of water, leading to an incorrect result.
- Option D: $6.023 \times {10^{^{23}}}$ : This is incorrect because it only calculates the number of water molecules and doesn't account for the 10 electrons in each water molecule.

Key Concepts:
- Moles concept
- Avogadro's number
- Density

Memory Aids:
- Moles = Mass / Molar Mass
- Number of particles = Moles x Avogadro's number

Quick Tips:
- Always remember to consider the number of atoms and their electrons in a molecule when calculating the total number of electrons.

90 Easy

90 of 114Question permalink

Analyzing +4 | -1

In the reaction

\mathrm{Br}_2+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{NaBr}+\mathrm{NaBrO}_3+\mathrm{CO}_2

The equivalent weight of

\mathrm{NaBrO}_3

is

View answer and explanation

Correct option: C

\frac{\text {Molecular weight}}{5}

Option C (Molecular weight/5) is correct because the change in oxidation state of Br in NaBrO3 is 5.

Solution approach

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

Step-by-Step Solution:

Step 1: Identify the oxidation states of Br in the reactants and products.\nIn Br2, the oxidation state of Br is 0.\nIn NaBrO3, the oxidation state of Br is +5.

Step 2: Determine the change in oxidation state.\nThe change in oxidation state of Br is |5 - 0| = 5.

Step 3: Calculate the equivalent weight.\nEquivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{change in oxidation state}}}} = \frac{{\text{{Molecular weight}}}}{{5}}$

Why other options are wrong:
- Option A: This is incorrect because it doesn't account for the change in oxidation state of Br.
- Option B: This is incorrect because the change in oxidation state is not 10.
- Option D: This is incorrect because the change in oxidation state is not 4.

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

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

Quick Tips:
- Determine the oxidation states of the relevant elements in the reactants and products.
- Calculate the change in oxidation state.

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