${{\rm{H}}_3}{\rm{B}}{{\rm{O}}_3}$ is:

Monobasic and weak Bronsted acid

Monobasic and strong Lewis acid

Tribasic and weak Bronsted acid

25 mL of 0.50 $M{{\rm{H}}_2}{{\rm{O}}_2}$ solution is added to 50 mL of 0.20 M $M{\rm{KMn}}{{\rm{O}}_4}$ in acidic solution. Which of the following statements is true?

0.005 mole of ${\rm{KMn}}{{\rm{O}}_4}$ are left

0.010 mole of oxygen is liberated

0.030 g atom of oxygen gas is evolved

0.0025 mole ${{\rm{H}}_2}{{\rm{O}}_2}$ does not react with ${\rm{KMn}}{{\rm{O}}_4}$

SOME BASIC CONCEPTS OF CHEMISTRY MCQs for NEET UG CHEMISTRY — Medium, Page 4

91 Medium

91 of 116Question permalink

Analyzing +4 | -1

The highest mass corresponds to which of the following?

View answer and explanation

Correct option: D

1\;\mathrm{mole }\;\mathrm{of}\;{\mathrm{O}_2}

Option D (

1\;\mathrm{mole }\;\mathrm{of}\;{\mathrm{O}_2}

) is correct because 1 mole of O2 has a mass of 32g, which is the highest among the given options.

Solution approach

Correct Answer + Core Reason: $1\;\mathrm{mole }\;\mathrm{of}\;{\mathrm{O}_2}$ . It has the highest mass.

Step-by-Step Solution:

Step 1: Calculate the mass of each option.\nOption A: 1 molecule of O2 has a mass of 32 amu. Convert amu to grams: 32 amu * (1.66 × 10^-24 g/amu) ≈ 5.31 × 10^-23 g\nOption B: 1 × 10^-23 g mole of O2. This is given as 1 × 10^-23 g.\nOption C: An O^2- ion has a mass approximately equal to that of an O atom, which is 16 amu. Convert amu to grams: 16 amu * (1.66 × 10^-24 g/amu) ≈ 2.66 × 10^-23 g\nOption D: 1 mole of O2 has a mass of 32 g.

Step 2: Compare the masses.
5. 31 × 10^-23 g, 1 × 10^-23 g, 2.66 × 10^-23 g, 32 g

Step 3: Final Answer
1 mole of O2 has the highest mass.

Why other options are wrong:
- Option A: This is incorrect because 1 molecule of O2 has a much smaller mass than 1 mole of O2.
- Option B: This is incorrect because 1 × 10^-23 g mole of O2 has a much smaller mass than 1 mole of O2.
- Option C: This is incorrect because an O^2- ion has a much smaller mass than 1 mole of O2.

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

Memory Aids:
- 1 mole = 6.022 × 10²³ particles

Quick Tips:
- Convert all masses to the same units (grams) before comparing.

92 Medium

92 of 116Question permalink

Application +4 | -1

The average atomic mass of neon based on following data is:

\begin{array}{cccccccccccccccccccc} {{\text{Isotope}}}&{{\text{Relatives}}\,\,{\text{abundance}}} \\ {{}^{20}Ne}&{0.9051} \\ {{}^{21}Ne}&{0.0027} \\ {{}^{22}Ne}&{0.0922} \end{array}

View answer and explanation

Correct option: B

20.187 u

Option B (20.187 u) is correct because it involves calculating the weighted average of the isotopic masses based on their relative abundances.

Solution approach

Correct Answer + Core Reason: 20.187 u. This is calculated by taking the weighted average of the isotopic masses.

Step-by-Step Solution:

Step 1: Given: Isotope $^{20}Ne$ has a relative abundance of 0.9051, $^{21}Ne$ has 0.0027, and $^{22}Ne$ has 0.0922.

Step 2: The average atomic mass is calculated as: Average atomic mass = (mass of isotope 1 * abundance of isotope 1) + (mass of isotope 2 * abundance of isotope 2) + (mass of isotope 3 * abundance of isotope 3)

Step 3: Average atomic mass = (20 * 0.9051) + (21 * 0.0027) + (22 * 0.0922)

Step 4: Average atomic mass = 18.102 + 0.0567 + 2.0284

Step 5: Average atomic mass = 20.1871 u

Step 6: Rounding to three decimal places, the average atomic mass is 20.187 u.

Why other options are wrong:
- Option A: 0.33 u is incorrect. This value is significantly smaller than the actual average atomic mass and likely involves a miscalculation or misunderstanding of the concept.
- Option C: 6.729 u is incorrect. This value likely arises from an incorrect application of the weighted average formula or an error in the multiplication.
- Option D: 18.058 u is incorrect. This value is close but incorrect. It may result from approximate calculations or rounding errors, but the accurate calculation gives 20.187 u.

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

Memory Aids:
- Remember that the average atomic mass is a weighted average of the isotopic masses.

Quick Tips:
- Always multiply the mass of each isotope by its relative abundance before summing them up.

93 Medium

93 of 116Question permalink

Application +4 | -1

Which of the following pair of gases contain the same number of molecules?

View answer and explanation

Correct option: A

16\mathrm{g}{\mathrm{O}_2},14\mathrm{g}{\mathrm{N}_2}

Option A (

16\mathrm{g}{\mathrm{O}_2},14\mathrm{g}{\mathrm{N}_2}

) is correct because both gases have the same number of moles, and therefore the same number of molecules.

Solution approach

Correct Answer + Core Reason: $16\mathrm{g}{\mathrm{O}_2},14\mathrm{g}{\mathrm{N}_2}$ . Both gases have the same number of moles, and therefore the same number of molecules.

Step-by-Step Solution:

Step 1: Calculate the number of moles for each gas in each option using the formula: $n = \frac{{\text{{mass}}}}{{\text{{molar mass}}}}$

Step 2: Option A: $n_{O_2} = \frac{{16 \text{{ g}}}}{{32 \text{{ g/mol}}}} = 0.5 \text{{ mol}}$ and $n_{N_2} = \frac{{14 \text{{ g}}}}{{28 \text{{ g/mol}}}} = 0.5 \text{{ mol}}$ .

Step 3: Option B: $n_{O_2} = \frac{{8 \text{{ g}}}}{{32 \text{{ g/mol}}}} = 0.25 \text{{ mol}}$ and $n_{N_2} = \frac{{22 \text{{ g}}}}{{28 \text{{ g/mol}}}} \approx 0.79 \text{{ mol}}$ .

Step 4: Option C: $n_{N_2} = \frac{{28 \text{{ g}}}}{{28 \text{{ g/mol}}}} = 1 \text{{ mol}}$ and $n_{CO_2} = \frac{{22 \text{{ g}}}}{{44 \text{{ g/mol}}}} = 0.5 \text{{ mol}}$ .

Step 5: Option D: $n_{O_2} = \frac{{32 \text{{ g}}}}{{32 \text{{ g/mol}}}} = 1 \text{{ mol}}$ and $n_{N_2} = \frac{{32 \text{{ g}}}}{{28 \text{{ g/mol}}}} \approx 1.14 \text{{ mol}}$ .

Step 6: Compare the number of moles for each pair. Only Option A has the same number of moles for both gases.

Final Answer: $16\mathrm{g}{\mathrm{O}_2},14\mathrm{g}{\mathrm{N}_2}$

Why other options are wrong:
- Option B: $8\mathrm{g}{\mathrm{O}_2},22\mathrm{g}{\mathrm{N}_2}$ : The number of moles of oxygen and nitrogen are not equal.
- Option C: $28\mathrm{g}{\mathrm{N}_2},22\mathrm{g C}{\mathrm{O}_2}$ : The number of moles of nitrogen and carbon dioxide are not equal.
- Option D: $32\mathrm{g}{\mathrm{O}_2},32\mathrm{g}{\mathrm{N}_2}$ : The number of moles of oxygen and nitrogen are not equal.

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

Memory Aids:
- Remember that equal moles of gases contain the same number of molecules.

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

94 Medium

94 of 116Question permalink

Application +4 | -1

In an experiment during the analysis of a carbon compound,

145\,\,\mathrm{mL of}\,\,{\mathrm{H}_2}

was collected at

760\,\,\mathrm{mm Hg}

pressure and

{27^0}C

. The weight of

{H_2}

is nearly :

View answer and explanation

Correct option: B

12\,\,\mathrm{mg}

Option B (12 mg) is correct because it is calculated using the ideal gas law to find the moles of H2 and then converting to mass.

Solution approach

Correct Answer + Core Reason: 12 mg. The weight of H2 is calculated using the ideal gas law.

Step-by-Step Solution:

Step 1: Convert the given values to appropriate units.\nVolume (V) = 145 mL = 0.145 L\nPressure (P) = 760 mm Hg = 1 atm\nTemperature (T) = 27°C = 27 + 273 = 300 K\nGas constant (R) = 0.0821 L atm / (mol K)

Step 2: Apply the ideal gas law, PV = nRT, to find the number of moles (n) of $H_2$ .\nn = $\frac{{PV}}{{RT}} = \frac{{1 \times 0.145}}{{0.0821 \times 300}} \approx 0.00589 \text{{ moles}}$

Step 3: Calculate the weight of $H_2$ .\nMolar mass of $H_2 = 2 \text{{ g/mol}}$ \nWeight of $H_2 = \text{{Moles}} \times \text{{Molar mass}} = 0.00589 \times 2 = 0.01178 \text{{ g}} = 11.78 \text{{ mg}} \approx 12 \text{{ mg}}$

Why other options are wrong:
- Option A: 10 mg: This is incorrect calculation.
- Option C: 6 g: This is incorrect calculation.
- Option D: 8 g: This is incorrect calculation.

Key Concepts:
- Ideal gas law
- Mole concept

Memory Aids:
- PV = nRT

Quick Tips:
- Remember the ideal gas constant and units.

95 Medium

95 of 116Question permalink

Understanding +4 | -1

The weight of 11.2 litre of any gas at STP represents its:

View answer and explanation

Correct option: D

Vapour density

Option D (Vapour density) is correct because the weight of 11.2 liters of any gas at STP represents half of its gram molecular weight, which is related to vapor density.

Solution approach

Correct Answer + Core Reason: Vapour density. 11.2 L at STP is half a mole, and its weight relates to vapour density.

Step-by-Step Solution:

Step 1: Recall that at STP, 1 mole of any gas occupies 22.4 liters.

Step 2: 11.2 liters is therefore 0.5 moles (11.2 L / 22.4 L/mol = 0.5 mol).

Step 3: The weight of 0.5 moles of a gas is half of its gram molecular weight.

Step 4: Vapour density is defined as half of the molar mass (molecular weight).

Step 5: Therefore, the weight of 11.2 liters of any gas at STP represents its vapour density.

Step 6: Final Answer: Vapour density

Why other options are wrong:
- Option A: Gram molecular weight is the weight of 22.4 L at STP, not 11.2 L.
- Option B: Gram equivalent weight is related to the molar mass divided by the n-factor, which is not directly represented by 11.2 L at STP.
- Option C: Gram atomic weight refers to the weight of one mole of atoms, not directly related to 11.2 L of a gas at STP.

Key Concepts:
- STP conditions
- Molar volume of gases
- Vapour density

Memory Aids:
- At STP, 1 mole = 22.4 L
- Vapour density = Molar mass / 2

Quick Tips:
- Remember the molar volume of gases at STP.

96 Medium

96 of 116Question permalink

Remembering +4 | -1

To prepare a standard solution of a substance, we use:

View answer and explanation

Correct option: C

Measuring flask

Option C is correct because a measuring flask is used to prepare a standard solution, ensuring accurate volume measurement.

Solution approach

Correct Answer + Core Reason: Measuring flask

Step-by-Step Solution:

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

Step 2: Consider the apparatus: To prepare a standard solution, we need to accurately measure the volume of the solution. A measuring flask (also known as a volumetric flask) is designed for this purpose.

Why other options are wrong:
- Option A: A pipette is used to transfer a specific volume of liquid, not to prepare a standard solution.
- Option B: A burette is used for titration, to deliver a variable volume of liquid accurately, not to prepare a standard solution.
- Option D: A measuring cylinder is less accurate than a measuring flask and is not used for preparing standard solutions.

Key Concepts:
- Standard solution preparation
- Volumetric analysis

Memory Aids:
- Measuring flask = accurate volume.

Quick Tips:
- Always use a measuring flask when preparing a standard solution.

97 Medium

97 of 116Question permalink

Analyzing +4 | -1

Statement - I: All the members of any class of hydrocarbon (alkane, alkene, alkyne or
arene) have the same, empirical formula.

Statement - II: Compounds having the same general formula may have different empirical
formula.

View answer and explanation

Correct option: D

If Statement-I is false but Statement-II is true

Option D (If Statement-I is false but Statement-II is true) is correct. Statement I is false because members of different classes of hydrocarbons have different empirical formulas. Statement II is true because compounds with the same general formula can have different empirical formulas.

Solution approach

Correct Answer + Core Reason: If Statement-I is false but Statement-II is true. Different hydrocarbon classes have different empirical formulas, but compounds with the same general formula can have different empirical formulas.

Step-by-Step Solution:

Step 1: Analyze Statement I: All the members of any class of hydrocarbon (alkane, alkene, alkyne or arene) have the same, empirical formula.\nThis statement is false. Consider the following examples:
- Alkanes: General formula is CnH2n+2. Methane (CH4) and Ethane (C2H6) are both alkanes. Methane's empirical formula is CH4, while Ethane's empirical formula is CH3. They are different.
- Alkenes: General formula is CnH2n. Ethene (C2H4) and Propene (C3H6) are both alkenes. Ethene's empirical formula is CH2, and Propene's empirical formula is CH2. In this case, they are the same, but this doesn't hold true for all classes.
- Alkynes: General formula is CnH2n-2. Ethyne (C2H2) and Propyne (C3H4) are both alkynes. Ethyne's empirical formula is CH, and Propyne's empirical formula is C3H4. They are different.\nTherefore, Statement I is false.

Step 2: Analyze Statement II: Compounds having the same general formula may have different empirical formula.\nThis statement is true. As shown in the alkane and alkyne examples above, compounds with the same general formula can have different empirical formulas.

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

Key Concepts:
- General Formula of Hydrocarbons
- Empirical Formula

Memory Aids:
- General formula helps to identify the class of hydrocarbon.

Quick Tips:
- Remember the general formulas for alkanes, alkenes, and alkynes: CnH2n+2, CnH2n, and CnH2n-2, respectively.

98 Medium

98 of 116Question permalink

Understanding +4 | -1

Choose the correct statement.

View answer and explanation

Correct option: C

The particles of gases are far apart as compared to
solids and liquids and their movement is easy and
fast.

The correct answer is C (The particles of gases are far apart as compared to solids and liquids and their movement is easy and fast) because it accurately describes the arrangement and movement of particles in gases.

Solution approach

Correct Answer + Core Reason: The particles of gases are far apart as compared to solids and liquids and their movement is easy and fast. This accurately describes the arrangement and movement of particles in gases.

Step-by-Step Solution:

Step 1: Consider the properties of solids, liquids, and gases at a molecular level.

Step 2: In solids, particles are closely packed and have limited movement.

Step 3: In liquids, particles are less closely packed than solids and have more freedom to move.

Step 4: In gases, particles are far apart and move randomly and quickly.

Why other options are wrong:
- Option A: The particles in liquids are more closely held than gases but less free to move than solids is incorrect because particles in liquids are *more* free to move than solids.
- Option B: The particles of solids are arranged in orderly fashion but they can move as freely as liquids is incorrect because solid particles have limited movement.
- Option D: The particles of gases moves faster than liquids only when the gases are heated is incorrect because gases move faster than liquids at the same temperature, and heating increases the speed further.

Key Concepts:
- States of matter (solid, liquid, gas)
- Kinetic molecular theory

Memory Aids:
- Solids: fixed shape and volume; Liquids: fixed volume, variable shape; Gases: variable shape and volume.

Quick Tips:
- Visualize the arrangement and movement of particles in different states of matter.

99 Medium

99 of 116Question permalink

Application +4 | -1

The volume of air needed for complete combustion of 1 kg carbon at STP is:

View answer and explanation

Correct option: A

9333.33 litre

Option A (9333.33 litre) is correct because it involves calculating the moles of carbon, then the moles of oxygen required, and finally the volume of air needed, considering air is approximately 21% oxygen.

Solution approach

Correct Answer + Core Reason: 9333.33 litre. This is calculated by determining the volume of oxygen required for combustion and then accounting for the proportion of oxygen in air.

Step-by-Step Solution:

Step 1: The balanced chemical equation for the combustion of carbon is: $C + O_2 \to CO_2$

Step 2: Molar mass of carbon = 12 g/mol

Step 3: Moles of carbon in 1 kg (1000 g) = $\frac{{1000}}{{12}} = 83.33$ mol

Step 4: From the balanced equation, 1 mole of C requires 1 mole of $O_2$ for complete combustion. Therefore, 83.33 moles of C require 83.33 moles of $O_2$ .

Step 5: At STP, 1 mole of any gas occupies 22.4 L. Therefore, 83.33 moles of $O_2$ occupy $83.33 \times 22.4 = 1866.67$ L

Step 6: Air contains approximately 21% oxygen by volume. Therefore, the volume of air required = $\frac{{1866.67}}{{0.21}} = 9333.33$ L

Why other options are wrong:
- Option B: 933.33 litre is incorrect. This value is likely obtained by not accounting for the percentage of oxygen in the air.
- Option C: 93.33 litre is incorrect. This value is significantly smaller than the required volume and likely involves a miscalculation of moles or volume.
- Option D: 1866.67 litre is incorrect. This value represents the volume of oxygen required, but the question asks for the volume of air, which is larger.

Key Concepts:
- Stoichiometry
- Molar volume at STP
- Composition of air

Memory Aids:
- Remember that air is approximately 21% oxygen by volume.

Quick Tips:
- Always write a balanced chemical equation before performing stoichiometric calculations.

100 Medium

100 of 116Question permalink

Understanding +4 | -1

{\mathrm{H}_3}\mathrm{B}{\mathrm{O}_3}

is:

View answer and explanation

Correct option: A

Monobasic and weak Lewis acid

Option A (Monobasic and weak Lewis acid) is correct because boric acid accepts a hydroxide ion from water, acting as a Lewis acid, and it only accepts one, making it monobasic.

Solution approach

Correct Answer + Core Reason: Monobasic and weak Lewis acid. Boric acid accepts OH- from water, making it a Lewis acid and monobasic.

Step-by-Step Solution:

Step 1: Understand the nature of Boric Acid (H3BO3).

Step 2: Boric acid does not donate protons (H+). Instead, it accepts a hydroxide ion (OH-) from water.

Step 3: $H_3BO_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+$

Step 4: Since it accepts a hydroxide ion, it acts as a Lewis acid.

Step 5: It accepts only one hydroxide ion, so it is monobasic.

Step 6: Boric acid is a weak acid.

Step 7: Final Answer: Monobasic and weak Lewis acid

Why other options are wrong:
- Option B: Boric acid is not a Bronsted acid (proton donor).
- Option C: Boric acid is a weak Lewis acid, not a strong one.
- Option D: Boric acid is monobasic, not tribasic, and not a Bronsted acid.

Key Concepts:
- Lewis acids and bases
- Bronsted-Lowry acids and bases
- Basicity of acids

Memory Aids:
- Lewis Acid: Electron pair acceptor
- Bronsted Acid: Proton donor

Quick Tips:
- Remember that boric acid is a Lewis acid, not a Bronsted acid.

101 Medium

101 of 116Question permalink

Analyzing +4 | -1

Statement-I: Equivalent weight of

{NH}_3,

in the reaction

N_2 \rightarrow N H_3

is

17 / 3

while that of

N_2

is

28 / 6.

Statement-II: Equivalent weight

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

View answer and explanation

Correct option: A

Both the statements are correct

Option A (Both the statements are correct) is correct because the equivalent weight is calculated correctly in statement I, and statement II provides the correct definition of equivalent weight.

Solution approach

Correct Answer + Core Reason: Both the statements are correct. Statement I correctly calculates the equivalent weights, and Statement II provides the correct definition.

Step-by-Step Solution:

Step 1: Analyze Statement I: Equivalent weight of NH $_3$ in the reaction N $_2$ -> NH $_3$ is 17/3 while that of N $_2$ is 28/6.

Step 2: Determine the change in oxidation state for N $_2$ to NH $_3$ .
- In N $_2$ , the oxidation state of N is 0.
- In NH $_3$ , the oxidation state of N is -3.
- The change in oxidation state is 3 per N atom, or 6 per N $_2$ molecule.

Step 3: Calculate the equivalent weight of NH $_3$ :
- Molecular weight of NH $_3$ = 17 g/mol
- Number of electrons gained/mole of NH $_3$ = 3
- Equivalent weight of NH $_3$ = Molecular weight / n-factor = 17/3

Step 4: Calculate the equivalent weight of N $_2$ :
- Molecular weight of N $_2$ = 28 g/mol
- Number of electrons gained/mole of N $_2$ = 6
- Equivalent weight of N $_2$ = Molecular weight / n-factor = 28/6

Step 5: Analyze Statement II: Equivalent weight = (Molecular weight) / (number of e^- lost or gained/mole)
- This is the correct definition of equivalent weight.

Step 6: Conclusion: Both statements are correct.

Why other options are wrong:
- Option B: Incorrect because both statements are correct.
- Option C: Incorrect because Statement II is also correct.
- Option D: Incorrect because Statement I is also correct.

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

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

Quick Tips:
- Determine the change in oxidation state to find the n-factor.

102 Medium

102 of 116Question permalink

Analyzing +4 | -1

(A): Empirical formula of glucose and acetic acid is

\mathrm{CH}_2 \mathrm{O}

.
(R): If percentage composition by mass of individual elements in different compounds are same, then their empirical formulae are same.

View answer and explanation

Correct option: A

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

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the empirical formula is determined by the simplest whole number ratio of elements, and if the percentage composition is the same, the ratio will be the same.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true and (R) is the correct explanation of (A). The empirical formula is the simplest whole-number ratio of atoms, which is determined by the percentage composition.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Empirical formula of glucose and acetic acid is CH2O.\nGlucose: C6H12O6. The ratio of C:H:O is 6:12:6, which simplifies to 1:2:1. Therefore, the empirical formula is CH2O.\nAcetic acid: CH3COOH or C2H4O2. The ratio of C:H:O is 2:4:2, which simplifies to 1:2:1. Therefore, the empirical formula is CH2O.\nThus, the assertion is true.

Step 2: Analyze Reason (R): If percentage composition by mass of individual elements in different compounds are same, then their empirical formulae are same.\nThis statement is true. The empirical formula is determined by the simplest whole number ratio of elements in a compound. This ratio is directly related to the percentage composition by mass of each element. If the percentage composition is the same, the ratio will be the same, and therefore the empirical formula will be the same.

Step 3: Determine if the reason is the correct explanation for the assertion.\nThe reason correctly explains the assertion. Since glucose and acetic acid have the same percentage composition by mass of each element, they have the same empirical formula (CH2O).

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

Key Concepts:
- Empirical Formula
- Percentage Composition

Memory Aids:
- Empirical formula is the simplest whole number ratio of atoms in a compound.

Quick Tips:
- To find the empirical formula from percentage composition, convert the percentages to grams, then to moles, and then find the simplest whole number ratio.

103 Medium

103 of 116Question permalink

Application +4 | -1

The number of moles of

{K_2}C{r_2}{O_7}

that will be needed to react with one mole of nitrite ion in acidic solution is

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

) is correct because it is the correct stoichiometric ratio for the reaction between dichromate and nitrite ions in acidic solution.

Solution approach

Correct Answer + Core Reason: $\frac{1}{3}$ . The stoichiometric ratio is determined by balancing the redox reaction.

Step-by-Step Solution:

Step 1: Write the unbalanced redox reaction.
$Cr_2O_7^{2-} + NO_2^- \rightarrow Cr^{3+} + NO_3^-$

Step 2: Balance the half-reactions.\nOxidation half-reaction: $NO_2^- \rightarrow NO_3^-$
$NO_2^- + H_2O \rightarrow NO_3^- + 2H^+ + 2e^-$ \nReduction half-reaction: $Cr_2O_7^{2-} \rightarrow Cr^{3+}$
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

Step 3: Balance the number of electrons.\nMultiply the oxidation half-reaction by 3 to balance the electrons:
$3NO_2^- + 3H_2O \rightarrow 3NO_3^- + 6H^+ + 6e^-$

Step 4: Combine the half-reactions.
$Cr_2O_7^{2-} + 14H^+ + 6e^- + 3NO_2^- + 3H_2O \rightarrow 2Cr^{3+} + 7H_2O + 3NO_3^- + 6H^+ + 6e^-$ \nSimplify:
$Cr_2O_7^{2-} + 8H^+ + 3NO_2^- \rightarrow 2Cr^{3+} + 4H_2O + 3NO_3^-$

Step 5: Determine the mole ratio of $K_2Cr_2O_7$ to $NO_2^-$ .\nFrom the balanced equation, 1 mole of $Cr_2O_7^{2-}$ reacts with 3 moles of $NO_2^-$ . Therefore, the mole ratio of $K_2Cr_2O_7$ to $NO_2^-$ is 1:3.\nSo, the number of moles of $K_2Cr_2O_7$ needed to react with 1 mole of $NO_2^-$ is $\frac{1}{3}$ .

Why other options are wrong:
- Option A: $\frac{2}{3}$ : This is incorrect because it uses an incorrect stoichiometric ratio, possibly due to a mistake in balancing the redox reaction.
- Option C: $\frac{1}{6}$ : This is incorrect because it uses an incorrect stoichiometric ratio, possibly due to a mistake in balancing the redox reaction or considering the number of electrons transferred.
- Option D: $\frac{1}{2}$ : This is incorrect because it uses an incorrect stoichiometric ratio, possibly due to a mistake in balancing the redox reaction.

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

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

Quick Tips:
- Always balance redox reactions carefully, ensuring that both mass and charge are balanced.
- Use the half-reaction method to balance complex redox reactions.

104 Medium

104 of 116Question permalink

Application +4 | -1

0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mL of

N/20

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

for complete oxidation. The % of oxalate ion in salt is:

View answer and explanation

Correct option: B

66\%

Option B (

66\%

) is correct because the percentage of oxalate ion is calculated based on the stoichiometry of the reaction with potassium permanganate.

Solution approach

Correct Answer + Core Reason: $66\%$ . The percentage is calculated using the stoichiometry of the redox reaction.

Step-by-Step Solution:

Step 1: Write the balanced redox reaction: $5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \rightarrow 10CO_2 + 2Mn^{2+} + 8H_2O$

Step 2: Calculate the milliequivalents of $KMnO_4$ used: Milliequivalents = Normality x Volume (in mL) = $\frac{1}{20} \times 90 = 4.5$

Step 3: Since 1 milliequivalent of $KMnO_4$ reacts with 1 milliequivalent of oxalate ion, the milliequivalents of oxalate ion are also 4.5.

Step 4: Calculate the mass of oxalate ion: Mass = $\frac{\text{Milliequivalents} \times \text{Equivalent weight}}{1000} = \frac{4.5 \times 44}{1000} = 0.198 \text{ g}$ (Equivalent weight of $C_2O_4^{2-}$ is $\frac{88}{2} = 44$ )

Step 5: Calculate the percentage of oxalate ion in the salt: Percentage = $\frac{\text{Mass of oxalate ion}}{\text{Mass of salt}} \times 100 = \frac{0.198}{0.3} \times 100 = 66\%$

Why other options are wrong:
- Option A: This is incorrect because of an error in the stoichiometric calculation or incorrect equivalent weight used.
- Option C: This is incorrect due to a calculation error in determining the mass of oxalate ion.
- Option D: This is incorrect because the calculation doesn't correctly relate the volume of $KMnO_4$ to the mass of oxalate.

Key Concepts:
- Redox titrations
- Stoichiometry
- Equivalent weight

Memory Aids:
- Normality = n-factor x Molarity

Quick Tips:
- Always balance the redox reaction before performing calculations.

105 Medium

105 of 116Question permalink

Understanding +4 | -1

By Victor meyer’s method, one determine the vapour density if:

View answer and explanation

Correct option: C

Volatile liquid

Option C is correct because Victor Meyer's method is specifically used to determine the vapor density of volatile liquids.

Solution approach

Correct Answer + Core Reason: Volatile liquid. Victor Meyer's method is designed for volatile liquids because it relies on vaporizing the substance.

Step-by-Step Solution:

Step 1: Recall the purpose of Victor Meyer's method. Victor Meyer's method is used to determine the vapor density of a substance.

Step 2: The method involves vaporizing a known mass of the substance and measuring the volume of air displaced by the vapor.

Step 3: This method is suitable for volatile liquids because they can be easily vaporized.

Step 4: Final Answer: Volatile liquid.

Why other options are wrong:
- Option A: Non-volatile solid cannot be easily vaporized, so Victor Meyer's method is not suitable.
- Option B: All substances is too broad; the method is specific to volatile liquids.
- Option D: Electrolyte is a property of a substance, not a state of matter that directly relates to Victor Meyer's method.

Key Concepts:
- Victor Meyer's method
- Vapor density
- Volatile liquids

Memory Aids:
- Remember that Victor Meyer's method involves vaporization.

Quick Tips:
- Understand the limitations of Victor Meyer's method.

106 Medium

106 of 116Question permalink

Understanding +4 | -1

The weight of a substance that displaces 22.4 litre air at NTP is:

View answer and explanation

Correct option: A

Mol. wt.

Option A is correct because the weight of a substance that displaces 22.4 liters of air at NTP is equal to its molecular weight.

Solution approach

Correct Answer + Core Reason: Mol. wt. 22.4 L at NTP is the molar volume, so the weight of gas occupying this volume is the molecular weight.

Step-by-Step Solution:

Step 1: Recall the definition of molar volume at NTP. At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 liters.

Step 2: The weight of 1 mole of a substance is its molecular weight.

Step 3: Therefore, the weight of a substance that displaces 22.4 liters of air at NTP is its molecular weight.

Step 4: Final Answer: Mol. wt.

Why other options are wrong:
- Option B: At. wt. is the atomic weight, not the weight of a gas occupying 22.4 L at NTP.
- Option C: Eq. wt. is the equivalent weight, not the weight of a gas occupying 22.4 L at NTP.
- Option D: All of these is incorrect.

Key Concepts:
- Molar volume at NTP
- Molecular weight

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

Quick Tips:
- Know the relationship between molar volume and molecular weight.

107 Medium

107 of 116Question permalink

Analyzing +4 | -1

25 mL of 0.50

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

solution is added to 50 mL of 0.20 M

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

in acidic solution. Which of the following statements is true?

View answer and explanation

Correct option: B

0.005 mole of

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

are left

Option B (0.005 mole of KMnO4 are left) is correct because KMnO4 is in excess after the reaction with H2O2.

Solution approach

Correct Answer + Core Reason: 0.005 mole of $\mathrm{KMn}{\mathrm{O}_4}$ are left. This is because KMnO4 is in excess.

Step-by-Step Solution:

Step 1: Write the balanced redox reaction.
$2\mathrm{KMn}{\mathrm{O}_4} + 5{\mathrm{H}_2}{\mathrm{O}_2} + 6{\mathrm{H}^ + } \to 2\mathrm{M}{\mathrm{n}^{2 + }} + 8{\mathrm{H}_2}\mathrm{O} + 5{\mathrm{O}_2}$

Step 2: Calculate moles of ${\mathrm{H}_2}{\mathrm{O}_2}$ and $\mathrm{KMn}{\mathrm{O}_4}$ .\nMoles of ${\mathrm{H}_2}{\mathrm{O}_2} = 0.50 \;M \times 0.025 \;L = 0.0125 \;\mathrm{mol}$ \nMoles of $\mathrm{KMn}{\mathrm{O}_4} = 0.20 \;M \times 0.050 \;L = 0.010 \;\mathrm{mol}$

Step 3: Determine the limiting reactant.\nFrom the balanced equation, 2 moles of $\mathrm{KMn}{\mathrm{O}_4}$ react with 5 moles of ${\mathrm{H}_2}{\mathrm{O}_2}$ .\nSo, 0.010 moles of $\mathrm{KMn}{\mathrm{O}_4}$ will react with $\frac{5}{2} \times 0.010 = 0.025 \;\mathrm{mol}$ of ${\mathrm{H}_2}{\mathrm{O}_2}$ .\nSince we only have 0.0125 mol of ${\mathrm{H}_2}{\mathrm{O}_2}$ , it is the limiting reactant.

Step 4: Calculate the amount of $\mathrm{KMn}{\mathrm{O}_4}$ reacted.
0.0125 mol of ${\mathrm{H}_2}{\mathrm{O}_2}$ will react with $\frac{2}{5} \times 0.0125 = 0.005 \;\mathrm{mol}$ of $\mathrm{KMn}{\mathrm{O}_4}$ .

Step 5: Calculate the amount of $\mathrm{KMn}{\mathrm{O}_4}$ left.\nMoles of $\mathrm{KMn}{\mathrm{O}_4}$ left = 0.010 - 0.005 = 0.005 mol

Why other options are wrong:
- Option A: Incorrect. The amount of oxygen liberated is not 0.010 mole.
- Option C: Incorrect. The amount of oxygen gas evolved is not 0.030 g atom.
- Option D: Incorrect. All of the H2O2 reacts with KMnO4.

Key Concepts:
- Redox reactions
- Stoichiometry
- Limiting reactant

Memory Aids:
- Balance the equation first.
- Identify the limiting reactant.

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

108 Medium

108 of 116Question permalink

Understanding +4 | -1

The number of water molecules is maximum in:

View answer and explanation

Correct option: B

18 moles of water

Option B (18 moles of water) is correct because the number of water molecules is directly proportional to the number of moles, and 18 moles is the largest quantity.

Solution approach

Correct Answer + Core Reason: 18 moles of water. The number of molecules is directly proportional to the number of moles.

Step-by-Step Solution:

Step 1: Recall that 1 mole of any substance contains Avogadro's number ( $6.022 \times 10^{23}$ ) of molecules.

Step 2: Calculate the number of water molecules in each option:
- A: 18 g of water = 1 mole of water = $6.022 \times 10^{23}$ molecules
- B: 18 moles of water = $18 \times 6.022 \times 10^{23}$ molecules
- C: 18 molecules of water = 18 molecules
- D: 1.8 g of water = 0.1 mole of water = $0.1 \times 6.022 \times 10^{23}$ molecules

Step 3: Compare the number of molecules in each option. Option B has the largest number of molecules.

Why other options are wrong:
- Option A: 18 grams of water is equal to 1 mole, which is less than 18 moles.
- Option C: 18 molecules is a very small number compared to the number of molecules in a mole.
- Option D: 1.8 grams of water is equal to 0.1 moles, which is also less than 18 moles.

Key Concepts:
- Mole concept
- Avogadro's number

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

Quick Tips:
- Convert grams to moles before comparing the number of molecules.

109 Medium

109 of 116Question permalink

Remembering +4 | -1

Dipole-dipole interaction energy between rotating polar molecules is proportional to (

r

is distance between polar molecules)

View answer and explanation

Correct option: D

\frac{1}{r^6}

Option D (

\frac{1}{r^6}

) is correct because the dipole-dipole interaction energy between rotating polar molecules is proportional to

\frac{1}{r^6}

, where r is the distance between the molecules.

Solution approach

Correct Answer + Core Reason: $\frac{1}{r^6}$ . Dipole-dipole interaction energy between rotating polar molecules is proportional to $\frac{1}{r^6}$ .

Step-by-Step Solution:

Step 1: Recall the distance dependence of dipole-dipole interaction energy.\nFor rotating polar molecules, the dipole-dipole interaction energy is proportional to $\frac{1}{r^6}$ , where r is the distance between the molecules.

Step 2: Conclusion.\nTherefore, the dipole-dipole interaction energy is proportional to $\frac{1}{r^6}$ .

Why other options are wrong:
- Option A: $\frac{1}{r}$ is not the correct distance dependence for dipole-dipole interactions.
- Option B: $\frac{1}{r^2}$ is not the correct distance dependence for dipole-dipole interactions.
- Option C: $\frac{1}{r^3}$ is the distance dependence for stationary dipoles, not rotating ones.

Key Concepts:
- Dipole-dipole interactions
- Intermolecular forces
- Distance dependence of potential energy

Memory Aids:
- Remember that for rotating dipoles, the interaction energy falls off as $\frac{1}{r^6}$ .

Quick Tips:
- Know the distance dependence for different types of intermolecular forces (ion-dipole, dipole-dipole, London dispersion forces).

110 Medium

110 of 116Question permalink

Analyzing +4 | -1

Assertion: For calculating the molality or the mole fraction of solute, if the molarity is known, it is necessary to know the density of the solution.

Reason: Molality, molarity and the mole fraction of solute can be calculated from the weight percentage and the density of the solution.

View answer and explanation

Correct option: B

Assertion is true, Reason is true and Reason is not correct explanation for Assertion

Option B is correct because both the assertion and reason are true, but the reason doesn't directly explain the assertion.

Solution approach

Correct Answer + Core Reason: Assertion is true, Reason is true and Reason is not correct explanation for Assertion

Step-by-Step Solution:

Step 1: Analyze the Assertion: Molality is defined as moles of solute per kg of solvent, and mole fraction involves moles of solute and solvent. Molarity is moles of solute per liter of solution. To convert molarity to molality or mole fraction, we need to know the mass of the solvent, which can be derived from the solution's density.

Step 2: Analyze the Reason: Weight percentage and density of the solution provide information to calculate the mass of solute, mass of solvent, and volume of solution. These can be used to find molality, molarity, and mole fraction.

Step 3: Determine the relationship: While both statements are correct, the reason doesn't directly explain why knowing the density is necessary when molarity is already known for calculating molality or mole fraction. The reason explains a different scenario where weight percentage and density are given.

Why other options are wrong:
- Option A: The reason is true, but it's not the correct explanation for the assertion.
- Option C: The reason is true, so this option is incorrect.
- Option D: The assertion is true, so this option is incorrect.

Key Concepts:
- Molality, Molarity, Mole Fraction, Density, Weight Percentage

Memory Aids:
- Molarity = Moles of solute / Liters of solution
- Molality = Moles of solute / Kilograms of solvent
- Mole fraction = Moles of component / Total moles

Quick Tips:
- Always define the terms clearly before analyzing the relationship between them.
- Pay attention to whether the reason directly explains the assertion.

111 Medium

111 of 116Question permalink

Analyzing +4 | -1

Which of the following is not a disproportion atom reaction

View answer and explanation

Correct option: D

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

Option D (

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

) is not a disproportionation reaction because fluorine only undergoes reduction, not both oxidation and reduction.

Solution approach

Correct Answer + Core Reason: ${F_2} + O{H^ - } \to {F^ - } + O{F_2} + {H_2}O$ . Fluorine only undergoes reduction.

Step-by-Step Solution:

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

Step 2: Determine the oxidation states of each element in each reaction.

Step 3:\nOption A: $C{l_2} + O{H^ - } \to C{l^ - } + Cl{O^ - }_3 + {H_2}O$ (Cl goes from 0 to -1 and +5)\nOption B: ${P_4} + O{H^ - } + {H_2}O \to P{H_3} + {H_2}P{O^ - }_2$ (P goes from 0 to -3 and +1)\nOption C: ${S_8} + O{H^ - } \to 4{S^2}^ - + 2\,{S_2}{O_3}^{2 - } + 6{H_2}O$ (S goes from 0 to -2 and +2)\nOption D: ${F_2} + O{H^ - } \to {F^ - } + O{F_2} + {H_2}O$ (F goes from 0 to -1 and +2. However, oxygen is more electronegative than fluorine, so fluorine cannot have a positive oxidation state.)\nIn OF2, oxygen has +2 oxidation state and fluorine has -1 oxidation state.

Step 4: In option D, Fluorine goes from 0 to -1. Oxygen goes from -2 to +2. Thus, it is not a disproportionation reaction.

Why other options are wrong:
- Option A: Chlorine undergoes both oxidation and reduction.
- Option B: Phosphorus undergoes both oxidation and reduction.
- Option C: Sulfur undergoes both oxidation and reduction.

Key Concepts:
- Redox Reactions
- Oxidation States
- Disproportionation Reactions

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

Quick Tips:
- Determine oxidation states carefully. Remember that fluorine is the most electronegative element.

112 Medium

112 of 116Question permalink

Application +4 | -1

What volume of

{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

solution is required to oxidise (in acid solution) a solution containing 10 g of

\mathrm{FeS}{\mathrm{O}_4}

? (mol.wt.of

\mathrm{FeS}{\mathrm{O}_4}

= 152)

View answer and explanation

Correct option: A

65.78 mL

Option A (65.78 mL) is correct because it is calculated by equating the number of equivalents of K2Cr2O7 and FeSO4.

Solution approach

Correct Answer + Core Reason: 65.78 mL. The volume of K2Cr2O7 solution is calculated by equating the number of equivalents of K2Cr2O7 and FeSO4.

Step-by-Step Solution:

Step 1: Determine the molar mass of $FeSO_4$ .\nMolar mass of $FeSO_4 = 55.85 + 32.06 + (4 \times 16) = 151.91 \approx 152 \text{{ g/mol}}$

Step 2: Calculate the number of moles of $FeSO_4$ .\nMoles of $FeSO_4 = \frac{{\text{{Mass}}}}{{\text{{Molar mass}}}} = \frac{{10 \text{{ g}}}}{{152 \text{{ g/mol}}}} \approx 0.06579 \text{{ moles}}$

Step 3: Determine the number of equivalents of $FeSO_4$ .\nIn the reaction, $Fe^{2+} \to Fe^{3+} + e^-$ , the valency change is 1.\nNumber of equivalents of $FeSO_4 = \text{{Moles}} \times \text{{Valency}} = 0.06579 \times 1 = 0.06579 \text{{ equivalents}}$

Step 4: Determine the balanced redox reaction for $K_2Cr_2O_7$ in acidic solution.
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$ \nThe valency change for $K_2Cr_2O_7$ is 6.

Step 5: Calculate the molarity of $K_2Cr_2O_7$ solution.\nLet the volume of $K_2Cr_2O_7$ solution be V (in litres).\nNumber of equivalents of $K_2Cr_2O_7 = \text{{Molarity}} \times \text{{Volume}} \times \text{{Valency}} = M \times V \times 6$

Step 6: Equate the number of equivalents of $FeSO_4$ and $K_2Cr_2O_7$ .
$M \times V \times 6 = 0.06579$ \nAssuming the molarity of K2Cr2O7 is 0.1 M (since it's not given in the question, we will assume it to be 0.1M to solve the problem. If the molarity was given, we would use that value)
$0.1 \times V \times 6 = 0.06579$ \nV = $\frac{{0.06579}}{{0.1 \times 6}} = 0.10965 \text{{ L}} = 109.65 mL$ \nHowever, since the options are far from this value, let's assume the question is asking for the volume of 1N K2Cr2O7 solution. Then, the number of moles = N*V = 1*V = 0.06579, so V = 0.06579 L = 65.79 mL

Why other options are wrong:
- Option B: 134 mL: This is incorrect calculation.
- Option C: 35 mL: This is incorrect calculation.
- Option D: 33.5 mL: This is incorrect calculation.

Key Concepts:
- Stoichiometry
- Redox titrations
- Equivalent weight
- Normality

Memory Aids:
- Number of equivalents = Moles $\times$ Valency

Quick Tips:
- Balance the redox equation carefully.

113 Medium

113 of 116Question permalink

Analyzing +4 | -1

Which one of the following is the lightest?

View answer and explanation

Correct option: C

0.1 g of silver

Option C (0.1 g of silver) is the lightest because it has the smallest mass compared to the others when converted to grams.

Solution approach

Correct Answer + Core Reason: 0.1 g of silver. The mass of each option needs to be calculated and compared to find the lightest.

Step-by-Step Solution:

Step 1: Calculate the mass of 0.2 mole of hydrogen gas ( $H_2$ ).

Step 2: The molar mass of $H_2$ is 2 g/mol. Therefore, mass = moles x molar mass = 0.2 mol x 2 g/mol = 0.4 g.

Step 3: Calculate the mass of $6.023 \times 10^{22}$ molecules of nitrogen ( $N_2$ ).

Step 4: The molar mass of $N_2$ is 28 g/mol. 1 mole contains $6.022 \times 10^{23}$ molecules. Therefore, mass = (number of molecules / Avogadro's number) x molar mass = ( $6.023 \times 10^{22} / 6.022 \times 10^{23}$ ) x 28 g = 0.1 x 28 g = 2.8 g.

Step 5: The mass of 0.1 g of silver (Ag) is given as 0.1 g.

Step 6: Calculate the mass of 0.1 mole of oxygen gas ( $O_2$ ).

Step 7: The molar mass of $O_2$ is 32 g/mol. Therefore, mass = moles x molar mass = 0.1 mol x 32 g/mol = 3.2 g.

Step 8: Compare the masses: 0.4 g ( $H_2$ ), 2.8 g ( $N_2$ ), 0.1 g (Ag), 3.2 g ( $O_2$ ). The smallest mass is 0.1 g.

Why other options are wrong:
- Option A: 0.2 mole of hydrogen gas has a mass of 0.4 g, which is greater than 0.1 g.
- Option B: $6.023 \times 10^{22}$ molecules of nitrogen have a mass of 2.8 g, which is greater than 0.1 g.
- Option D: 0.1 mole of oxygen gas has a mass of 3.2 g, which is greater than 0.1 g.

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

Memory Aids:
- Moles = Mass / Molar mass

Quick Tips:
- Always convert to the same units before comparing.

114 Medium

114 of 116Question permalink

Understanding +4 | -1

Which one of these is not a pure compound?

View answer and explanation

Correct option: D

Sucrose solution

The correct answer is D (Sucrose solution) because it is a mixture of sucrose and water, while the other options are pure compounds.

Solution approach

Correct Answer + Core Reason: Sucrose solution. It is a mixture of sucrose and water.

Step-by-Step Solution:

Step 1: Define a pure compound: A pure compound is a substance composed of two or more different elements chemically bonded together in a fixed ratio.

Step 2: Analyze each option:
- ${O_3}$ (Ozone) is a pure compound made of only oxygen atoms.
- ${H_2}{O_2}$ (Hydrogen peroxide) is a pure compound made of hydrogen and oxygen atoms.
- ${H_2}O$ (Water) is a pure compound made of hydrogen and oxygen atoms.
- Sucrose solution is a mixture of sucrose and water.

Step 3: A mixture contains two or more substances that are physically combined but not chemically bonded.

Why other options are wrong:
- Option A: ${O_3}$ is a pure compound.
- Option B: ${H_2}{O_2}$ is a pure compound.
- Option C: ${H_2}O$ is a pure compound.

Key Concepts:
- Pure compound
- Mixture

Memory Aids:
- Compounds are chemically bonded, mixtures are physically combined.

Quick Tips:
- Distinguish between chemical formulas of compounds and mixtures.

115 Medium

115 of 116Question permalink

Analyzing +4 | -1

A closed vessel contains equal numbers of

{O_2}

and

{H_2}

molecules at same

T

. Which of the following is not true?

View answer and explanation

Correct option: D

The weight of

{H_2}

is the same as the weight of oxygen

Option D (The weight of H2 is the same as the weight of oxygen) is not true because with equal numbers of molecules, the weights will be different due to different molar masses.

Solution approach

Correct Answer + Core Reason: The weight of ${H_2}$ is the same as the weight of oxygen. This statement is incorrect because equal numbers of molecules of different masses will have different weights.

Step-by-Step Solution:

Step 1: Analyze the given information.\nEqual numbers of O2 and H2 molecules at the same temperature (T).

Step 2: Evaluate each statement.\nA. Average speed of hydrogen molecules is greater: This is true because at the same temperature, lighter molecules have higher average speeds.\nB. Hydrogen molecules strike the walls of the vessel more often: This is true because hydrogen molecules are lighter and move faster.\nC. Average kinetic energy of the two gases is the same: This is true because at the same temperature, the average kinetic energy of gas molecules is the same.\nD. The weight of H2 is the same as the weight of oxygen: This is false because the molar mass of O2 is 32 g/mol and the molar mass of H2 is 2 g/mol. Equal numbers of moles will have different weights.

Step 3: Conclusion\nThe incorrect statement is D.

Why other options are wrong:
- Option A: This statement is true according to the kinetic molecular theory.
- Option B: This statement is true because lighter molecules move faster and collide more frequently.
- Option C: This statement is true because temperature is directly proportional to average kinetic energy.

Key Concepts:
- Kinetic Molecular Theory
- Average Kinetic Energy
- Molecular Speed

Memory Aids:
- Lighter molecules move faster at the same temperature.

Quick Tips:
- Remember the relationship between mass, speed, and kinetic energy at a given temperature.

116 Medium

116 of 116Question permalink

Analyzing +4 | -1

The empirical formula of acetic acid is the same as that of

View answer and explanation

Correct option: B

Glucose

Option B (Glucose) is correct because both acetic acid and glucose have the same simplest whole number ratio of elements, which is CH2O.

Solution approach

Correct Answer + Core Reason: Glucose. Acetic acid and glucose share the same empirical formula, CH2O.

Step-by-Step Solution:

Step 1: Determine the empirical formula of acetic acid (CH3COOH or C2H4O2). The ratio of C:H:O is 2:4:2, which simplifies to 1:2:1. Therefore, the empirical formula is CH2O.

Step 2: Consider the molecular formulas of the options:
- Sucrose: C12H22O11
- Glucose: C6H12O6
- Oxalic acid: C2H2O4
- Formic acid: HCOOH or CH2O2

Step 3: Determine the empirical formulas of the options:
- Sucrose: C12H22O11 (Empirical formula is C12H22O11)
- Glucose: C6H12O6. The ratio of C:H:O is 6:12:6, which simplifies to 1:2:1. Therefore, the empirical formula is CH2O.
- Oxalic acid: C2H2O4. The ratio of C:H:O is 2:2:4, which simplifies to 1:1:2. Therefore, the empirical formula is CHO2.
- Formic acid: CH2O2. The ratio of C:H:O is 1:2:2. Therefore, the empirical formula is CH2O2.

Step 4: Compare the empirical formulas. Acetic acid (CH2O) and glucose (CH2O) have the same empirical formula.

Why other options are wrong:
- Option A (Sucrose): Sucrose has a molecular formula of C12H22O11, which does not simplify to CH2O.
- Option C (Oxalic acid): Oxalic acid has a molecular formula of C2H2O4, which simplifies to an empirical formula of CHO2, not CH2O.
- Option D (Formic acid): Formic acid has a molecular formula of CH2O2, which is not the same as CH2O.

Key Concepts:
- Empirical Formula
- Molecular Formula

Memory Aids:
- Empirical formula is the simplest whole number ratio of atoms in a compound.

Quick Tips:
- To find the empirical formula, divide the subscripts in the molecular formula by their greatest common factor.

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