Question

Match List-I with List-II.

• A. A-IV, B-III, C-I, D-II
• B. A-III, B-II, C-IV, D-I
• C. A-III, B-IV, C-II, D-I
• D. A-III, B-IV, C-I, D-II

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To find the number of atoms, first calculate the moles of the substance using $n = \text{mass} / \text{molar mass}$. Then, multiply the moles by $N_A$ and the atomicity of the molecule. Note: $1\text{ mg} = 10^{-3}\text{ g}$.
Step 2: Key Formula or Approach:
Number of atoms $= \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}} \times N_A \times \text{Atomicity}$.
Step 3: Detailed Explanation:
A. Water ($H_2O$): Molar mass = $18\text{ g/mol}$, Atomicity = $3$. \[ \text{Atoms} = \frac{1.8 \times 10^{-3}}{18} \times 3 \times N_A = 10^{-4} \times 3 \times N_A = 3 \times 10^{-4} \times N_A \text{ (III)} \] B. Sulphuric acid ($H_2SO_4$): Molar mass = $98\text{ g/mol}$, Atomicity = $7$. \[ \text{Atoms} = \frac{9.8 \times 10^{-3}}{98} \times 7 \times N_A = 10^{-4} \times 7 \times N_A = 7 \times 10^{-4} \times N_A \text{ (IV)} \] C. Carbon ($C$): Molar mass = $12\text{ g/mol}$, Atomicity = $1$. \[ \text{Atoms} = \frac{1.8 \times 10^{-3}}{12} \times 1 \times N_A = 1.5 \times 10^{-4} \times N_A \text{ (I - Note: Adjusted list indexing)} \] D. Salt ($NaCl$): Molar mass = $58.5\text{ g/mol}$, Atomicity = $2$. \[ \text{Atoms} = \frac{5.85 \times 10^{-3}}{58.5} \times 2 \times N_A = 2 \times 10^{-4} \times N_A \text{ (II)} \]
Step 4: Final Answer:
The correct matching is A-III, B-IV, C-I, D-II.