SOME BASIC CONCEPTS OF CHEMISTRY MCQs for NEET UG CHEMISTRY — Easy, Page 2

31 Easy

Application +4 | -1

10\;g\;\mathrm{CaC}{\mathrm{O}_3}

on heating gives 5.6 g CaO and …. g

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

View answer and explanation

Correct option: A

4.4

Option A (4.4) is correct because the mass of

CO_2

produced from the decomposition of

CaCO_3

is 4.4g, following stoichiometry.

Solution approach

Correct Answer + Core Reason: 4.4. The mass of $CO_2$ produced is calculated using stoichiometry.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$

Step 2: Calculate the molar mass of $CaCO_3$ : $40 + 12 + (3 \times 16) = 100 g/mol$

Step 3: Calculate the moles of $CaCO_3$ : $\frac{10 g}{100 g/mol} = 0.1 mol$

Step 4: From the balanced equation, 1 mole of $CaCO_3$ produces 1 mole of $CO_2$ .\nTherefore, 0.1 mol of $CaCO_3$ will produce 0.1 mol of $CO_2$ .

Step 5: Calculate the molar mass of $CO_2$ : $12 + (2 \times 16) = 44 g/mol$

Step 6: Calculate the mass of $CO_2$ produced: $0.1 mol \times 44 g/mol = 4.4 g$

Step 7: Final Answer: 4.4 g

Why other options are wrong:
- Option B (5.6): This is the mass of CaO produced, not $CO_2$ .
- Option C (6.5): This value is incorrect and doesn't relate to the stoichiometry of the reaction.
- Option D (4.2): This value is incorrect and doesn't relate to the stoichiometry of the reaction.

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

Memory Aids:
- Remember the balanced equation for the decomposition of calcium carbonate.

Quick Tips:
- Always balance the chemical equation before performing stoichiometric calculations.
- Pay attention to the molar masses of the reactants and products.

32 Easy

32 of 114Question permalink

Analyzing +4 | -1

Assertion: Equivalent weight of

\mathrm{NH}_3

in the reaction

\mathrm{N}_2 \rightarrow \mathrm{NH}_3

is

17 / 3

while that of

\mathrm{N}_2

is

28/6.

Reason: Equivalent weight =

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

.

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 the equivalent weight is indeed the molecular weight divided by the number of electrons lost or gained, and both the assertion and reason are correct and related.

Solution approach

Correct Answer + Core Reason: Assertion is true, Reason is true and Reason is correct explanation for Assertion. The equivalent weight is the molecular weight divided by the number of electrons transferred.

Step-by-Step Solution:

Step 1: Analyze the reaction $N_2 \rightarrow NH_3$ . Nitrogen gains electrons to form ammonia.

Step 2: For $N_2 \rightarrow 2NH_3$ , each N atom gains 3 electrons, so $N_2$ gains 6 electrons.

Step 3: Equivalent weight of $N_2 = \frac{{\text{{Molecular weight of }} N_2}}{{\text{{Number of electrons gained}}}} = \frac{{28}}{6}$

Step 4: For $NH_3$ , the equivalent weight = $\frac{{\text{{Molecular weight of }} NH_3}}{{\text{{Number of electrons involved per molecule}}}} = \frac{{17}}{3}$

Step 5: The reason states that Equivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{number of }} e^- \text{{ lost or gained }}}}$ which is correct.

Step 6: Final Answer: Assertion is true, Reason is true and Reason is correct explanation for Assertion

Why other options are wrong:
- Option B: Assertion is true, Reason is true and Reason is not correct explanation for Assertion: The reason is the correct explanation.
- Option C: Assertion is true, Reason is false: The reason is true.
- Option D: Assertion is false, Reason is true: The assertion is true.

Key Concepts:
- Equivalent weight
- Redox reactions
- Number of electrons transferred

Memory Aids:
- Remember the formula for equivalent weight in redox reactions.

Quick Tips:
- Balance the equation to determine the number of electrons transferred.

33 Easy

33 of 114Question permalink

Application +4 | -1

Volume occupied by

3.01 \times {10^{23}}\;\;

molecules of acetylene at NTP is:

View answer and explanation

Correct option: B

11.2\,\,\mathrm{litre}

Option B (11.2 litre) is correct because it is half the molar volume at NTP, corresponding to half a mole of acetylene.

Solution approach

Correct Answer + Core Reason: 11.2 litre. The volume is calculated using the number of molecules and Avogadro's number to find the number of moles, then using the molar volume at NTP.

Step-by-Step Solution:

Step 1: Calculate the number of moles of acetylene.\nNumber of moles = $\frac{{\text{{Number of molecules}}}}{{\text{{Avogadro's number}}}} = \frac{{3.01 \times 10^{23}}}{{6.022 \times 10^{23}}} \approx 0.5 \text{{ moles}}$

Step 2: Calculate the volume at NTP.\nAt NTP, 1 mole of any gas occupies 22.4 litres.\nVolume = Number of moles $\times$ Molar volume = $0.5 \times 22.4 = 11.2 \text{{ litres}}$

Why other options are wrong:
- Option A: 22.4 litre: This is the volume occupied by 1 mole, not 0.5 moles.
- Option C: 1.12 litre: This is incorrect calculation.
- Option D: 2..24 litre: This is incorrect calculation.

Key Concepts:
- Mole concept
- Avogadro's number
- Molar volume at NTP

Memory Aids:
- 1 mole at NTP = 22.4 litres

Quick Tips:
- Remember the molar volume at NTP.

34 Easy

34 of 114Question permalink

Remembering +4 | -1

Which property of an element is not variable?

View answer and explanation

Correct option: B

At. wt.

Option B (At. wt.) is correct because the atomic weight of an element is a constant property, while valence and equivalent weight can vary depending on the chemical context.

Solution approach

Correct Answer + Core Reason: At. wt. Atomic weight is a fundamental property of an element.

Step-by-Step Solution:

Step 1: Valence: Valence is the number of electrons an atom can gain, lose, or share to form a chemical bond. It can vary depending on the element and the compound it forms.

Step 2: At. wt.: Atomic weight (or relative atomic mass) is the average mass of atoms of an element, calculated using the relative abundance of isotopes in a naturally occurring sample. It is a constant for a given element.

Step 3: Eq. wt.: Equivalent weight is the mass of a substance that combines with or is chemically equivalent to one gram atom of hydrogen or eight grams of oxygen. It can vary depending on the reaction the substance is involved in.

Step 4: Therefore, the atomic weight is the only property that is not variable.

Why other options are wrong:
- Option A: Valence can vary for some elements.
- Option C: Equivalent weight can vary depending on the reaction.
- Option D: Atomic weight is not variable.

Key Concepts:
- Valence
- Atomic Weight
- Equivalent Weight

Memory Aids:
- Atomic weight is a constant property of an element.

Quick Tips:
- Remember the definitions of valence, atomic weight, and equivalent weight.
- Consider the variability of each property.

35 Easy

35 of 114Question permalink

Analyzing +4 | -1

Which one of the following is not a redox reaction

View answer and explanation

Correct option: A

CaC{O_3} \to CaO + C{O_2}

Option A (

CaC{O_3} \to CaO + C{O_2}

) is correct because there is no change in oxidation state of any element in this reaction.

Solution approach

Correct Answer + Core Reason: $CaC{O_3} \to CaO + C{O_2}$ . This is a decomposition reaction where no oxidation state changes occur.

Step-by-Step Solution:

Step 1: Determine the oxidation states of each element in the reactants and products for each reaction:
- A: $CaC{O_3} \to CaO + C{O_2}$ . Ca: +2, C: +4, O: -2 on both sides. No change.
- B: $2{H_2} + {O_2} \to 2{H_2}O$ . H: 0 to +1, O: 0 to -2. Redox.
- C: $Na + {H_2}O \to NaOH + \frac{1}{2}{H_2}$ . Na: 0 to +1, H: +1 to 0. Redox.
- D: $MnC{l_3} \to MnC{l_2} + \frac{1}{2}C{l_2}$ . Mn: +3 to +2, Cl: -1 to 0. Redox.

Step 2: Identify the reaction where there is no change in oxidation state. This is reaction A.

Why other options are wrong:
- Option B: This is incorrect because hydrogen is oxidized and oxygen is reduced.
- Option C: This is incorrect because sodium is oxidized and hydrogen is reduced.
- Option D: This is incorrect because manganese is reduced and chlorine is oxidized.

Key Concepts:
- Oxidation state
- Redox reaction

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

Quick Tips:
- Determine the oxidation states of all elements in the reactants and products to identify redox reactions.

36 Easy

36 of 114Question permalink

Application +4 | -1

The Largest number of molecules is in

View answer and explanation

Correct option: A

36 g of water

Option A (36 g of water) is correct because it contains the largest number of moles, and therefore the largest number of molecules.

Solution approach

Correct Answer + Core Reason: 36 g of water. 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: 36 g of water ( $H_2O$ ): Molar mass of $H_2O$ = 18 g/mol. Moles = 36/18 = 2 mol
- B: 28 g of carbon monoxide (CO): Molar mass of CO = 28 g/mol. Moles = 28/28 = 1 mol
- C: 46 g of ethyl alcohol ( $C_2H_5OH$ ): Molar mass of $C_2H_5OH$ = 46 g/mol. Moles = 46/46 = 1 mol
- D: 54 g of Nitrogen pentoxide ( $N_2O_5$ ): Molar mass of $N_2O_5$ = 108 g/mol. Moles = 54/108 = 0.5 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 carbon monoxide is less than that of water.
- Option C: The number of moles of ethyl alcohol is less than that of water.
- Option D: The number of moles of nitrogen pentoxide is 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.

37 Easy

37 of 114Question permalink

Understanding +4 | -1

The number of significant figures in 10.3106 g is

View answer and explanation

Correct option: D

6

Option D is correct because all non-zero digits are significant, and zeros between non-zero digits are also significant, resulting in 6 significant figures.

Solution approach

Correct Answer + Core Reason: 6. All non-zero digits are significant, and zeros between non-zero digits are significant.

Step-by-Step Solution:

Step 1: State the rules for significant figures:
1. All non-zero digits are significant.
2. Zeros between non-zero digits are significant.
3. Leading zeros are not significant.
4. Trailing zeros in a number containing a decimal point are significant.

Step 2: Apply the rules to 10.3106 g:
- 1 is significant (non-zero digit)
- 0 between 1 and 3 is significant
- 3 is significant (non-zero digit)
- 1 is significant (non-zero digit)
- 0 between 1 and 6 is significant
- 6 is significant (non-zero digit)

Step 3: Count the significant figures: There are 6 significant figures.

**Why other

38 Easy

38 of 114Question permalink

Understanding +4 | -1

Choose correct option based on following statements. Here T stands for true statement and F for false statement.
(i) Homogeneous mixture has uniform composition throughout.
(ii) All components of a heterogeneous mixture are observable to naked eyes.
(iii) All solutions are homogeneous in nature.
(iv) Air is an example of heterogeneous mixture.

View answer and explanation

Correct option: B

TFTF

Option B is correct because (i) is true, (ii) is false, (iii) is true, and (iv) is false based on the definitions of homogeneous and heterogeneous mixtures.

Solution approach

Correct Answer + Core Reason: TFTF. This corresponds to the truth values of the statements.

Step-by-Step Solution:

Step 1: Analyze each statement.

Step 2: (i) Homogeneous mixture has uniform composition throughout. This is TRUE.

Step 3: (ii) All components of a heterogeneous mixture are observable to naked eyes. This is FALSE (some components may be too small to see).

Step 4: (iii) All solutions are homogeneous in nature. This is TRUE.

Step 5: (iv) Air is an example of heterogeneous mixture. This is FALSE (air is a homogeneous mixture of gases).

Step 6: Final Answer: TFTF.

Why other options are wrong:
- Option A: TTFF is incorrect.
- Option C: FFTT is incorrect.
- Option D: TFFF is incorrect.

Key Concepts:
- Homogeneous mixture
- Heterogeneous mixture
- Solutions

Memory Aids:
- Remember the definitions of homogeneous and heterogeneous mixtures.

Quick Tips:
- Understand the difference between homogeneous and heterogeneous mixtures.

39 Easy

39 of 114Question permalink

Understanding +4 | -1

One mole electron means:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because 1 mole of electrons contains Avogadro's number of electrons, which has a mass of approximately 0.55 mg.

Solution approach

Correct Answer + Core Reason: All of these. All the given options are correct representations of one mole of electrons.

Step-by-Step Solution:

Step 1: Define one mole.

Step 2: One mole of any substance contains Avogadro's number ( $N_A = 6.022 \times 10^{23}$ ) of particles.

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

Step 4: Calculate the mass of one mole of electrons: Mass of one electron = $9.109 \times 10^{-31}$ kg. Mass of one mole of electrons = $(6.022 \times 10^{23}) \times (9.109 \times 10^{-31} \text{{kg}}) = 5.486 \times 10^{-7} \text{{kg}} = 0.5486 \times 10^{-6} \text{{kg}} = 0.5486 \text{{mg}} \approx 0.55 \text{{mg}}$

Why other options are wrong:
- Option A: N electrons is correct, but not the complete answer.
- Option B: $6.023 \times {10^{23}}$ electrons is correct, but not the complete answer.
- Option C: 0.55 mg electrons is correct, but not the complete answer.

Key Concepts:
- Mole concept
- Avogadro's number
- Mass of an electron

Memory Aids:
- One mole = Avogadro's number of particles.

Quick Tips:
- Remember the mass of an electron and Avogadro's number.

40 Easy

40 of 114Question permalink

Application +4 | -1

The number of Na atoms in 46 g ‘Na”

View answer and explanation

Correct option: D

12.046{\mkern 1mu} {\mkern 1mu} \times {\mkern 1mu} {\mkern 1mu} {10^{23}}

Option D (

12.046{\mkern 1mu} {\mkern 1mu} \times {\mkern 1mu} {\mkern 1mu} {10^{23}}

) is correct because 46g of Na is 2 moles of Na, and therefore contains 2 times Avogadro's number of atoms.

Solution approach

Correct Answer + Core Reason: $12.046{\mkern 1mu} {\mkern 1mu} \times {\mkern 1mu} {\mkern 1mu} {10^{23}}$ . This is the number of atoms in 2 moles of Na.

Step-by-Step Solution:

Step 1: Calculate the number of moles of Na: Molar mass of Na = 23 g/mol. Moles of Na = 46 g / 23 g/mol = 2 mol.

Step 2: Calculate the number of atoms: Number of atoms = moles * Avogadro's number = 2 mol * $6.023 \times 10^{23}$ atoms/mol = $12.046 \times 10^{23}$ atoms.

Why other options are wrong:
- Option A: This is the number of atoms in 1 mole of Na, not 2 moles.
- Option B: This is just the number 2, and has no relation to the number of atoms.
- Option C: This is the number of moles, not the number of atoms.

Key Concepts:
- Mole concept
- Avogadro's number

Memory Aids:
- Number of atoms = moles * Avogadro's number

Quick Tips:
- Always convert mass to moles before calculating the number of atoms.

41 Easy

41 of 114Question permalink

Application +4 | -1

To neutralise 20 mL of M/ 10 NaOH, the volume of M/20 HCl needed is:

View answer and explanation

Correct option: C

40 mL

Option C (40 mL) is correct because it is calculated by using the neutralization equation

N_1V_1 = N_2V_2

.

Solution approach

Correct Answer + Core Reason: 40 mL. The volume of HCl needed is calculated using the neutralization equation.

Step-by-Step Solution:

Step 1: Write the neutralization equation.
$N_1V_1 = N_2V_2$ , where $N_1$ and $V_1$ are the normality and volume of NaOH, and $N_2$ and $V_2$ are the normality and volume of HCl.

Step 2: Determine the normalities of NaOH and HCl.\nSince NaOH and HCl are monoprotic, their molarity and normality are equal.
$N_{NaOH} = M_{NaOH} = \frac{1}{{10}} M = 0.1 N$
$N_{HCl} = M_{HCl} = \frac{1}{{20}} M = 0.05 N$

Step 3: Substitute the given values into the neutralization equation.
$0.1 N \times 20 mL = 0.05 N \times V_{HCl}$

Step 4: Solve for $V_{HCl}$ .
$V_{HCl} = \frac{{0.1 N \times 20 mL}}{{0.05 N}} = 40 mL$

Why other options are wrong:
- Option A: 10 mL: This is incorrect because it divides the volume of NaOH by 2, possibly due to a miscalculation or misunderstanding of the neutralization equation.
- Option B: 30 mL: This is incorrect because it uses an incorrect ratio or performs an incorrect calculation.
- Option D: 20 mL: This is incorrect because it assumes the volume of HCl is equal to the volume of NaOH, which is only true if their normalities are equal.

Key Concepts:
- Neutralization reaction
- Normality
- Neutralization equation

Memory Aids:
- $N_1V_1 = N_2V_2$

Quick Tips:
- Ensure that the units of volume are consistent on both sides of the equation.
- Remember that for monoprotic acids and monobasic bases, molarity equals normality.

42 Easy

42 of 114Question permalink

Understanding +4 | -1

Oxygen and the oxide ion have the

View answer and explanation

Correct option: A

Same proton number

Option A (Same proton number) is correct because oxygen and its oxide ion are the same element, and the number of protons defines the element.

Solution approach

Correct Answer + Core Reason: Same proton number. The number of protons defines the element.

Step-by-Step Solution:

Step 1: Define oxygen and oxide ion.

Step 2: Oxygen (O) is an element with a specific number of protons.

Step 3: Oxide ion (O $^{2-}$ ) is formed when oxygen gains two electrons.

Step 4: The number of protons remains the same during ion formation.

Step 5: Therefore, oxygen and the oxide ion have the same proton number.

Why other options are wrong:
- Option B: Same electronic configuration is incorrect. Oxygen and oxide ion have different number of electrons, hence different electronic configuration.
- Option C: Same electron number is incorrect. Oxide ion has two more electrons than oxygen.
- Option D: Same size is incorrect. The oxide ion is larger than the oxygen atom due to increased electron-electron repulsion.

Key Concepts:
- Atomic number (proton number)
- Ions
- Electronic configuration

Memory Aids:
- Protons define the element.

Quick Tips:
- Ion formation only changes the number of electrons, not protons.

43 Easy

43 of 114Question permalink

Remembering +4 | -1

The value of gram molar volume of gas is:

View answer and explanation

Correct option: D

22.4 litre at STP

Option D (22.4 litre at STP) is correct because the gram molar volume of any gas at Standard Temperature and Pressure (STP) is 22.4 litres.

Solution approach

Correct Answer + Core Reason: 22.4 litre at STP. This is the volume occupied by one mole of any gas at STP.

Step-by-Step Solution:

Step 1: Recall the definition of gram molar volume.

Step 2: The gram molar volume is the volume occupied by one mole of a substance in the gaseous state.

Step 3: At Standard Temperature and Pressure (STP), this volume is experimentally determined to be 22.4 litres.

Step 4: Therefore, the gram molar volume of a gas at STP is 22.4 litres.

Why other options are wrong:
- Option A: 1 litre is incorrect. This is a small volume and not representative of a mole of gas.
- Option B: 22.4 litre is partially correct but incomplete. It needs to specify the conditions (STP).
- Option C: 11.2 litre is incorrect. It might be confused with half the molar volume, but it's not the standard value.

Key Concepts:
- Gram molar volume
- Standard Temperature and Pressure (STP)
- Mole concept

Memory Aids:
- Remember 22.4 L/mol at STP.

Quick Tips:
- Always specify STP when referring to molar volume.

44 Easy

44 of 114Question permalink

Understanding +4 | -1

The least count of an instrument is 0.01 cm. Taking all precautions, the most possible error in the measurement can be

View answer and explanation

Correct option: B

0.01 cm

Option B is correct because the most possible error is equal to the least count of the instrument.

Solution approach

Correct Answer + Core Reason: 0.01 cm. The most possible error is equal to the least count.

Step-by-Step Solution:

Step 1: Define least count. The least count is the smallest value that can be measured by an instrument.

Step 2: The most possible error in a measurement is equal to the least count of the instrument.

Step 3: Given least count = 0.01 cm, the most possible error = 0.01 cm.

Step 4: Final Answer: 0.01 cm.

Why other options are wrong:
- Option A: 0.005 cm is half of the least count, which is related to precision but not the maximum possible error.
- Option C: 0.0001 cm is too small and not related to the least count.
- Option D: 0.1 cm is too large and not related to the least count.

Key Concepts:
- Least count
- Error in measurement

Memory Aids:
- The maximum possible error is equal to the least count.

Quick Tips:
- Understand the relationship between least count and error.

45 Easy

45 of 114Question permalink

Remembering +4 | -1

Ratio in hydrogen and oxygen in water molecule by volume is

View answer and explanation

Correct option: A

2:1

Option A (

2:1

) is correct because water (H₂O) has two hydrogen atoms for every one oxygen atom by volume.

Solution approach

Correct Answer + Core Reason: $2:1$ . Water molecule ( $H_2O$ ) has two hydrogen atoms and one oxygen atom.

Step-by-Step Solution:

Step 1: Write the chemical formula for water: $H_2O$

Step 2: Identify the number of hydrogen atoms: 2

Step 3: Identify the number of oxygen atoms: 1

Step 4: Express the ratio of hydrogen to oxygen: 2:1

Why other options are wrong:
- Option B: 3:1 is incorrect. The ratio is not 3:1.
- Option C: 1:2 is incorrect. This is the ratio of oxygen to hydrogen, not hydrogen to oxygen.
- Option D: 1:1 is incorrect. The ratio is not 1:1.

Key Concepts:
- Chemical Formula
- Molecular Composition

Memory Aids:
- Remember the chemical formula of water is $H_2O$ .

Quick Tips:
- Always write down the chemical formula first to determine the ratio of elements.

46 Easy

46 of 114Question permalink

Application +4 | -1

M

is the molecular weight of

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

The equivalent weight of

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

when it is converted into

{\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4}

is :

View answer and explanation

Correct option: A

M

Option A (

M

) is correct because when

KMnO_4

is converted to

K_2MnO_4

, the oxidation state of Mn changes from +7 to +6, a change of 1. Therefore, the equivalent weight is M/1 = M.

Solution approach

Correct Answer + Core Reason: $M$ . The change in oxidation state of Mn is 1.

Step-by-Step Solution:

Step 1: Write the reaction (in terms of Mn): $KMnO_4 \to K_2MnO_4$

Step 2: Determine the oxidation states of Mn:
- In $KMnO_4$ : Mn has an oxidation state of +7.
- In $K_2MnO_4$ : Mn has an oxidation state of +6.

Step 3: Calculate the change in oxidation state: The change is |+7 - +6| = 1.

Step 4: Calculate the equivalent weight: Equivalent weight = $\frac{{\text{{Molecular weight}}}}{{\text{{Change in oxidation state}}}} = \frac{M}{1} = M$

Step 5: Final Answer: $M$

Why other options are wrong:
- Option B: This would be correct if the change in oxidation state was 3.
- Option C: This would be correct if the change in oxidation state was 5.
- Option D: This would be correct if the change in oxidation state was 7.

Key Concepts:
- Oxidation State
- Equivalent Weight

Memory Aids:
- Equivalent weight = (Molecular weight) / (Change in oxidation state)

Quick Tips:
- Always determine the oxidation states of the relevant elements.

47 Easy

47 of 114Question permalink

Application +4 | -1

5 molar solution of

{H_2}S{O_4}

is diluted from 1 litre to a volume of 10 litre, the normality of the solution will be

View answer and explanation

Correct option: A

1N

Option A (1N) is correct because the normality is calculated using the dilution formula and considering the n-factor of H2SO4.

Solution approach

Correct Answer + Core Reason: 1N. The normality is calculated using the dilution formula and the n-factor of ${H_2}S{O_4}$ .

Step-by-Step Solution:

Step 1: Calculate the initial normality of the ${H_2}S{O_4}$ solution.\nNormality (N) = Molarity (M) x n-factor. For ${H_2}S{O_4}$ , the n-factor is 2 (since it has 2 replaceable hydrogen ions).\nInitial Normality ( $N_1$ ) = 5 M x 2 = 10 N

Step 2: Apply the dilution formula.
$N_1V_1 = N_2V_2$ , where $N_1$ is the initial normality, $V_1$ is the initial volume, $N_2$ is the final normality, and $V_2$ is the final volume.
$10 N \times 1 L = N_2 \times 10 L$

Step 3: Solve for the final normality ( $N_2$ ).
$N_2 = \frac{{10 N \times 1 L}}{{10 L}} = 1 N$

Why other options are wrong:
- Option B: 0.1 N: This is incorrect because it divides the correct answer by 10, possibly due to a miscalculation in the dilution formula or not considering the n-factor.
- Option C: 5 N: This is incorrect because it only considers the initial molarity and doesn't account for the dilution process.
- Option D: 0.5 N: This is incorrect because it divides the initial molarity by 10 and then multiplies by the n-factor, which is not the correct application of the dilution formula.

Key Concepts:
- Normality
- Molarity
- Dilution formula
- n-factor (basicity/acidity)

Memory Aids:
- Normality = Molarity x n-factor
- $N_1V_1 = N_2V_2$

Quick Tips:
- Always remember to consider the n-factor when converting between molarity and normality.
- Use the dilution formula carefully, ensuring correct substitution of values.

48 Easy

48 of 114Question permalink

Application +4 | -1

Equivalent weight of a bivalent metal is 37.2. The molecular weight of its chloride is

View answer and explanation

Correct option: C

145.4

Option C (145.4) is correct because it is calculated by finding the atomic weight of the metal and then adding the weight of two chlorine atoms.

Solution approach

Correct Answer + Core Reason: 145.4. The molecular weight of the chloride is calculated using the equivalent weight of the metal.

Step-by-Step Solution:

Step 1: Determine the atomic weight of the metal.\nEquivalent weight = $\frac{{\text{{Atomic weight}}}}{{\text{{Valency}}}}$ \nAtomic weight = Equivalent weight $\times$ Valency = $37.2 \times 2 = 74.4$

Step 2: Determine the formula of the metal chloride.\nSince the metal is bivalent, its chloride will be $MC{l_2}$ .

Step 3: Calculate the molecular weight of the metal chloride.\nMolecular weight of $MC{l_2}$ = Atomic weight of M + 2 $\times$ Atomic weight of Cl = $74.4 + 2 \times 35.5 = 74.4 + 71 = 145.4$

Why other options are wrong:
- Option A: 412.2: This is incorrect calculation.
- Option B: 216: This is incorrect calculation.
- Option D: 108.2: This is incorrect calculation.

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

Memory Aids:
- Equivalent weight = Atomic weight / Valency

Quick Tips:
- Remember the relationship between equivalent weight and atomic weight.

49 Easy

49 of 114Question permalink

Application +4 | -1

The number of equivalent per mole of

{\mathrm{H}_2}\mathrm{S}

used in its oxidation to

\mathrm{S}{\mathrm{O}_2}

is :

View answer and explanation

Correct option: B

6

Option B (6) is correct because in the oxidation of

H_2S

to

SO_2

, sulfur changes its oxidation state from -2 to +4, which is a change of 6.

Solution approach

Correct Answer + Core Reason: 6. The change in oxidation state of S is 6.

Step-by-Step Solution:

Step 1: Write the reaction (in terms of S): $H_2S \to SO_2$

Step 2: Determine the oxidation states of S:
- In $H_2S$ : S has an oxidation state of -2.
- In $SO_2$ : S has an oxidation state of +4.

Step 3: Calculate the change in oxidation state: The change is |+4 - (-2)| = 6.

Step 4: The number of equivalents per mole is equal to the change in oxidation state.

Step 5: Final Answer: 6

Why other options are wrong:
- Option A: This would be correct if the change in oxidation state was 3.
- Option C: This would be correct if the change in oxidation state was 4.
- Option D: This would be correct if the change in oxidation state was 2.

Key Concepts:
- Oxidation State
- Equivalent Weight

Memory Aids:
- Change in oxidation state = number of equivalents.

Quick Tips:
- Always determine the oxidation states of the relevant elements.

50 Easy

50 of 114Question permalink

Remembering +4 | -1

When a solid vaporizes directly without melting, it is known as:

View answer and explanation

Correct option: B

Sublimation

Option B (Sublimation) is correct because sublimation is the process where a solid transitions directly into a gas phase without passing through the liquid phase.

Solution approach

Correct Answer + Core Reason: Sublimation. Sublimation is the direct conversion of a solid to a gas.

Step-by-Step Solution:

Step 1: Define the term.\nSublimation is the process where a solid changes directly into a gas without passing through the liquid state.

Step 2: Consider the other options.\nEvaporation is the change from liquid to gas. Sedimentation is the settling of particles. Saponification is the process of making soap.

Step 3: Conclusion.\nOnly sublimation matches the definition given in the question.

Why other options are wrong:
- Option A: Evaporation is the process of a liquid changing into a gas, not a solid.
- Option C: Sedimentation refers to the settling of particles in a liquid, which is irrelevant to phase changes.
- Option D: Saponification is a chemical process involving the conversion of fats or oils into soap, which is unrelated to phase transitions.

Key Concepts:
- Phase transitions (solid, liquid, gas)
- Sublimation
- Evaporation
- Sedimentation
- Saponification

Memory Aids:
- "Sub" means below or under, think of the solid phase being "under" the gas phase, skipping the liquid.

Quick Tips:
- Remember the definitions of different phase transitions: melting, freezing, boiling, condensation, sublimation, deposition.

51 Easy

51 of 114Question permalink

Application +4 | -1

100 mL of 0.1 M solution of a reductant is diluted to 1 litre, which of the following changes?

View answer and explanation

Correct option: A

Molarity

Option A (Molarity) is correct because dilution changes the concentration (molarity) of the solution, while the number of millimoles and milliequivalents of the solute remain constant.

Solution approach

Correct Answer + Core Reason: Molarity. Dilution affects the concentration but not the amount of solute.

Step-by-Step Solution:

Step 1: Understand the concept of dilution. Dilution involves adding more solvent to a solution, which decreases the concentration of the solute.

Step 2: Consider the formula for molarity: $M = \frac{{n}}{{V}}$ , where $M$ is molarity, $n$ is the number of moles, and $V$ is the volume.

Step 3: Consider the formula for millimoles: Millimoles = Molarity x Volume (in mL). Since the amount of solute doesn't change during dilution, the number of millimoles remains constant.

Step 4: Consider the formula for milliequivalents: Milliequivalents = Normality x Volume (in mL). Since the amount of solute doesn't change during dilution, the number of milliequivalents remains constant.

Step 5: Since the volume increases from 100 mL to 1000 mL (1 litre), the molarity decreases proportionally.

Why other options are wrong:
- Option B: Millimole remains constant as the amount of solute doesn't change during dilution.
- Option C: Milliequivalent remains constant as the amount of solute doesn't change during dilution.
- Option D: Molarity changes, so "None of these" is incorrect.

Key Concepts:
- Molarity
- Dilution
- Millimoles
- Milliequivalents

Memory Aids:
- Dilution: M1V1 = M2V2

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

52 Easy

52 of 114Question permalink

Analyzing +4 | -1

Assertion : A certain element

X

, forms three binary compounds with chlorine containing

59.68 \%, 68.95 \%

and

74.75 \%

chlorine respectively. These data illustrate the law of multiple proportions.

Reason : According to law of multiple proportions, the relative amounts of an element combining with some fixed amount of a second element in a series of compounds are the ratios of small whole numbers.

View answer and explanation

Correct option: A

If both assertion and reason are true and the reason is the correct explanation of the assertion.

Option A is correct. The assertion is true because the given data illustrates the law of multiple proportions. The reason is also true and provides the correct explanation of the assertion.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and the reason is the correct explanation of the assertion.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nThe assertion states that the data illustrates the law of multiple proportions. This needs to be verified.

Step 2: Analyze the reason.\nThe reason provides the definition of the law of multiple proportions.

Step 3: Verify the assertion using the law of multiple proportions.\nLet's assume the element X has a fixed mass. We need to find the ratio of the masses of chlorine that combine with this fixed mass of X in the three compounds.\nCompound 1: 59.68% Cl, so X is 40.32%\nCompound 2: 68.95% Cl, so X is 31.05%\nCompound 3: 74.75% Cl, so X is 25.25%\nLet's fix the mass of X to 1 in each compound. The mass of Cl in each compound is:\nCompound 1: $\frac{{59.68}}{{40.32}} = 1.48$ \nCompound 2: $\frac{{68.95}}{{31.05}} = 2.22$ \nCompound 3: $\frac{{74.75}}{{25.25}} = 2.96$ \nNow, find the ratio of these masses:
1.48 : 2.22 : 2.96 ≈ 1 : 1.5 : 2 ≈ 2 : 3 : 4\nSince the ratio is a simple whole number ratio, the assertion is true.

Step 4: Determine if the reason correctly explains the assertion.\nThe reason provides the definition of the law of multiple proportions, which is the basis for verifying the assertion. Therefore, the reason is the correct explanation of the assertion.

Why other options are wrong:
- Option B: The reason is the correct explanation of the assertion.
- Option C: The assertion is true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Law of multiple proportions

Memory Aids:
- Law of multiple proportions: When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Quick Tips:
- To verify the law of multiple proportions, fix the mass of one element and find the ratio of the masses of the other element.

53 Easy

53 of 114Question permalink

Application +4 | -1

Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The molecular weight of cortisone is 360.4. what is the percentage of carbon in cortisone?

View answer and explanation

Correct option: C

69.98\%

Option C (

69.98\%

) is correct because it represents the percentage of carbon in cortisone, calculated from the number of carbon atoms and the molecular weight.

Solution approach

Correct Answer + Core Reason: $69.98\%$ . The percentage of carbon is calculated from the number of carbon atoms and the molecular weight.

Step-by-Step Solution:

Step 1: Determine the mass of carbon in one molecule of cortisone: 21 atoms of carbon

Step 2: Calculate the total mass of carbon: $21 \times 12.01 g/mol = 252.21 g/mol$ (using the atomic mass of carbon as 12.01 g/mol)

Step 3: Calculate the percentage of carbon in cortisone: $\frac{252.21 g/mol}{360.4 g/mol} \times 100\%$

Step 4: Percentage of carbon = $69.98\%$

Step 5: Final Answer: $69.98\%$

Why other options are wrong:
- **Option A ( $59.9\%$ ): This is an incorrect calculation of the percentage.
- Option B ( $75\%$ ): This is an incorrect calculation of the percentage.
- Option D (None of these): This is incorrect since option C is the correct percentage.

Key Concepts:
- Percentage composition
- Molecular weight
- Atomic mass

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

Quick Tips:
- Use accurate atomic masses for calculations.
- Double-check your calculations to avoid errors.

54 Easy

54 of 114Question permalink

Remembering +4 | -1

Molecular mass is defined as the

View answer and explanation

Correct option: C

Mass of one molecule of any substance compared with the mass of one atom of C-12

Option C (Mass of one molecule of any substance compared with the mass of one atom of C-12) is correct because molecular mass is defined relative to the mass of a C-12 atom.

Solution approach

Correct Answer + Core Reason: Mass of one molecule of any substance compared with the mass of one atom of C-12. This is the modern definition of molecular mass.

Step-by-Step Solution:

Step 1: Define molecular mass.

Step 2: Molecular mass is the mass of a molecule, expressed in atomic mass units (amu).

Step 3: One atomic mass unit (amu) is defined as 1/12th the mass of a carbon-12 atom.

Step 4: Therefore, molecular mass is the mass of one molecule of any substance compared with the mass of one atom of C-12.

Why other options are wrong:
- Option A: Mass of one atom compared with the mass of one molecule is incorrect. It should be molecule compared to atom.
- Option B: Mass of one atom compared with the mass of one atom of hydrogen is incorrect. The standard is C-12.
- Option D: None of the above is incorrect because option C is correct.

Key Concepts:
- Molecular mass
- Atomic mass unit (amu)
- Carbon-12 standard

Memory Aids:
- Molecular mass is relative to C-12.

Quick Tips:
- Molecular mass is the sum of the atomic masses of all atoms in the molecule.

55 Easy

55 of 114Question permalink

Remembering +4 | -1

Which represents per cent by weight?

View answer and explanation

Correct option: A

\frac{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solute}}}{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solution}}} \times 100

Option A (

\frac{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solute}}}{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solution}}} \times 100

) is correct because percent by weight is defined as the weight of solute divided by the weight of solution, multiplied by 100.

Solution approach

Correct Answer + Core Reason: $\frac{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solute}}}{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solution}}} \times 100$ because it is the definition of percent by weight.

Step-by-Step Solution:

Step 1: Define percent by weight: Percent by weight is the mass of the solute divided by the mass of the solution, multiplied by 100.

Step 2: Write the formula: Percent by weight = $\frac{{\text{{Weight of solute}}}}{{\text{{Weight of solution}}}} \times 100$

Why other options are wrong:
- Option B: $\frac{{\mathrm{wt}\mathrm{. }\;\mathrm{of}\;\mathrm{ solute}}}{{\mathrm{volume }\;\mathrm{of }\;\mathrm{solution}}} \times 100$ : This represents weight/volume percent, not weight percent.
- Option C: $\frac{{\mathrm{volume }\;\mathrm{of }\;\mathrm{solute}}}{{\mathrm{volume}\;\mathrm{of}\;\mathrm{ solution}}} \times 100$ : This represents volume/volume percent, not weight percent.
- Option D: None of the above: This is incorrect because option A is the correct definition.

Key Concepts:
- Percent by weight
- Concentration units

Memory Aids:
- Remember the formula: % by weight = (Weight of solute / Weight of solution) * 100

Quick Tips:
- Distinguish between weight/weight, weight/volume, and volume/volume percentages.

56 Easy

56 of 114Question permalink

Remembering +4 | -1

The prefix zepto stands for (in m)

View answer and explanation

Correct option: D

{10^{ - 21}}

Option D is correct because the prefix zepto represents 10^-21.

Solution approach

Correct Answer + Core Reason: ${10^{ - 21}}$

Step-by-Step Solution:

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

Step 2: Identify zepto: The prefix zepto stands for $10^{-21}$ .

Why other options are wrong:
- Option A: ${10^9}$ is giga.
- Option B: ${10^{ - 12}}$ is pico.
- Option C: ${10^{ - 15}}$ is femto.

Key Concepts:
- SI prefixes

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

Quick Tips:
- Memorize the common SI prefixes.

57 Easy

57 of 114Question permalink

Understanding +4 | -1

With increase of temperature, which of these changes?

View answer and explanation

Correct option: C

Molarity

Option C (Molarity) is correct because molarity is defined as moles of solute per liter of solution, and the volume of a solution changes with temperature.

Solution approach

Correct Answer + Core Reason: Molarity. Molarity changes with temperature because volume changes with temperature.

Step-by-Step Solution:

Step 1: Recall the definitions of molality, weight fraction, molarity, and mole fraction.

Step 2: Molality is defined as moles of solute per kilogram of solvent. It is independent of temperature because mass does not change with temperature.

Step 3: Weight fraction of solute is defined as the weight of solute divided by the total weight of the solution. It is independent of temperature because mass does not change with temperature.

Step 4: Molarity is defined as moles of solute per liter of solution. Volume changes with temperature, so molarity changes with temperature.

Step 5: Mole fraction is defined as the moles of solute divided by the total moles in the solution. It is independent of temperature because moles do not change with temperature.

Why other options are wrong:
- Option A: Molality is incorrect. Molality is independent of temperature.
- Option B: Weight fraction of solute is incorrect. Weight fraction is independent of temperature.
- Option D: Mole fraction is incorrect. Mole fraction is independent of temperature.

Key Concepts:
- Molarity
- Molality
- Mole fraction
- Weight fraction
- Temperature dependence of concentration

Memory Aids:
- Remember that volume changes with temperature, so any concentration unit involving volume will also change with temperature.

Quick Tips:
- Focus on whether the concentration unit involves volume or mass.

58 Easy

58 of 114Question permalink

Remembering +4 | -1

Dulong and Petit’s law is valid only for

View answer and explanation

Correct option: D

Solid elements

Option D is correct because Dulong and Petit's law is valid only for solid elements at sufficiently high temperatures.

Solution approach

Correct Answer + Core Reason: Solid elements

Step-by-Step Solution:

Step 1: State Dulong-Petit Law: Dulong-Petit Law states that the molar specific heat capacity (at constant volume) of a solid element is approximately 3R, where R is the ideal gas constant. This law is applicable to solid elements at sufficiently high temperatures.

Step 2: Consider limitations: The law is not applicable to gases or non-solids. It also fails at low temperatures where quantum effects become significant.

Why other options are wrong:
- Option A: Metals are a subset of solid elements, but the law applies to other solid elements as well.
- Option B: Non-metals can be solids, but the law is not universally applicable to all non-metals.
- Option C: The law does not apply to gaseous elements.

Key Concepts:
- Dulong-Petit Law
- Molar specific heat capacity

Memory Aids:
- Remember that Dulong and Petit's Law is primarily for solids.

Quick Tips:
- The law is an approximation and works best at high temperatures for solid elements.

59 Easy

59 of 114Question permalink

Understanding +4 | -1

The ratio of mole fraction of a solute and a solvent in a binary solution is:

View answer and explanation

Correct option: C

Ratio of their mole

Option C is correct because the mole fraction ratio of solute to solvent is the same as the ratio of their moles.

Solution approach

Correct Answer + Core Reason: Ratio of their mole. Mole fraction is directly proportional to the number of moles.

Step-by-Step Solution:

Step 1: Define mole fraction. The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution.

Step 2: Let $x_A$ be the mole fraction of solute A and $x_B$ be the mole fraction of solvent B in a binary solution.
$x_A = \frac{n_A}{n_A + n_B}$ and $x_B = \frac{n_B}{n_A + n_B}$ , where $n_A$ and $n_B$ are the number of moles of solute and solvent, respectively.

Step 3: Find the ratio of mole fractions: $\frac{x_A}{x_B} = \frac{\frac{n_A}{n_A + n_B}}{\frac{n_B}{n_A + n_B}} = \frac{n_A}{n_B}$

Step 4: Conclusion: The ratio of mole fractions is equal to the ratio of their moles.

Why other options are wrong:
- Option A: Ratio of their wt. is incorrect because mole fraction is based on moles, not weight.
- Option B: One is incorrect because the ratio is not necessarily equal to one.
- Option D: Zero is incorrect because the ratio is not necessarily equal to zero.

Key Concepts:
- Mole fraction
- Moles

Memory Aids:
- Mole fraction is a ratio of moles.

Quick Tips:
- Remember the definition of mole fraction.

60 Easy

60 of 114Question permalink

Understanding +4 | -1

In which of the following number all zeros are significant?

View answer and explanation

Correct option: C

50

Option C (50) is correct because in the number 50, the zero is significant as it is a trailing zero after a non-zero digit.

Solution approach

Correct Answer + Core Reason: 50. Trailing zeros after a non-zero digit are significant.

Step-by-Step Solution:

Step 1: Recall the rules for significant figures.

Step 2: Apply the rules to each option:
- A: 0.0005 - Leading zeros are never significant.
- B: 0.05 - Leading zeros are never significant.
- C: 50 - Trailing zeros after a non-zero digit are significant.
- D: 0.005 - Leading zeros are never significant.

Step 3: Final Answer: 50

Why other options are wrong:
- Option A: 0.0005 is incorrect because leading zeros are never significant.
- Option B: 0.05 is incorrect because leading zeros are never significant.
- Option D: 0.005 is incorrect because leading zeros are never significant.

Key Concepts:
- Significant Figures
- Rules for Significant Figures

Memory Aids:
- Remember the Pacific-Atlantic rule: If the decimal point is PRESENT, start counting significant figures from the PACIFIC (left) side, starting with the first non-zero digit. If the decimal point is ABSENT, start counting from the ATLANTIC (right) side, starting with the first non-zero digit.

Quick Tips:
- Be careful with leading and trailing zeros when determining significant figures.

Your Progress

Some basic concepts of chemistry practice · Easy

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice