Mole Concept And Avogadro's Number
Calculate the number of molecules present in 9 g of water ((\text{H}_2\text{O})). ((N_A = 6.022 \times 10^{23}) mol(^{-1}), Molar mass of water = 18 g/mol)
Select the correct option:
Solution
\(3.011 \times 10^{23}\)
Avogadro's number specifies that one mole of any substance contains exactly (6.022 \times 10^{23}) entities. The number of molecules is found by first converting mass to moles, then multiplying by Avogadro's number. Moles of water (= 9/18 = 0.5) mol. Number of molecules (= 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}). Option (1.806 \times 10^{23}) is incorrect; it would correspond to (0.3) mol, not 0.5 mol. Option (6.022 \times 10^{23}) is the number for a full mole (18 g), not 9 g. Option (1.204 \times 10^{24}) corresponds to 2 mol (36 g of water), which is twice the given mass. This question directly applies the NCERT concept linking mass, moles, and number of particles via Avogadro's number. Plausibility check: 9 g is exactly half of 18 g (one mole), so the answer must be exactly half of (N_A), giving (3.011 \times 10^{23}).
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About This Question
- Subject
- chemistry
- Chapter
- some basic concepts in chemistry
- Topic
- mole concept and avogadro's number
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
\(3.011 \times 10^{23}\)
Avogadro's number specifies that one mole of any substance contains exactly (6.022 \times 10^{23}) entities. The number of molecules is found by first converting mass to moles, then multiplying by Avogadro's number. Moles of water (= 9/18 = 0.5) mol. Number of molecules (= 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}). Option (1.806 \times 10^{23}) is incorrect; it would correspond to (0.3) mol, not 0.5 mol. Option (6.022 \times 10^{23}) is the number for a full mole (18 g), not 9 g. Option (1.204 \times 10^{24}) corresponds to 2 mol (36 g of water), which is twice the given mass. This question directly applies the NCERT concept linking mass, moles, and number of particles via Avogadro's number. Plausibility check: 9 g is exactly half of 18 g (one mole), so the answer must be exactly half of (N_A), giving (3.011 \times 10^{23}).
This medium difficulty chemistry question is from the chapter some basic concepts in chemistry, covering the topic of mole concept and avogadro's number. It appeared in the 2025 exam.
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