Question:
Which of the following contains the maximum number of atoms?
Which of the following contains the maximum number of atoms?
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Convert each mass to moles, then multiply by the atoms per molecule and compare.
Updated On: Nov 12, 2026
- \(18\) grams of water
- \(24\) grams of dioxygen
- \(22\) grams of carbon dioxide
- \(16\) grams of methane
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Number of atoms \(=\) moles \(\times\) atoms per molecule \(\times N_A\). Compare the product moles \(\times\) atomicity.
Step 2: Option (A) and (B)
\(18\) g \(\text{H}_2\text{O}\): \(\dfrac{18}{18}\times3=3\).
\(24\) g \(\text{O}_2\): \(\dfrac{24}{32}\times2=1.5\).
Step 3: Option (C) and (D)
\(22\) g \(\text{CO}_2\): \(\dfrac{22}{44}\times3=1.5\).
\(16\) g \(\text{CH}_4\): \(\dfrac{16}{16}\times5=5\).
Step 4: Compare the options
Methane gives the largest value, 5 \(N_A\) atoms, so option (D).
Final Answer:
16 grams of methane contains the most atoms. \[ \boxed{\text{(D) }\text{16 grams of methane}} \]
Number of atoms \(=\) moles \(\times\) atoms per molecule \(\times N_A\). Compare the product moles \(\times\) atomicity.
Step 2: Option (A) and (B)
\(18\) g \(\text{H}_2\text{O}\): \(\dfrac{18}{18}\times3=3\).
\(24\) g \(\text{O}_2\): \(\dfrac{24}{32}\times2=1.5\).
Step 3: Option (C) and (D)
\(22\) g \(\text{CO}_2\): \(\dfrac{22}{44}\times3=1.5\).
\(16\) g \(\text{CH}_4\): \(\dfrac{16}{16}\times5=5\).
Step 4: Compare the options
Methane gives the largest value, 5 \(N_A\) atoms, so option (D).
Final Answer:
16 grams of methane contains the most atoms. \[ \boxed{\text{(D) }\text{16 grams of methane}} \]
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