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NEET CHEMISTRYSome Basic Concepts of ChemistryMedium

Question

The volume occupied by one water molecule (density = 1 g cm31 \text{ g cm}^{-3}) is:

A

9.0×1023 cm39.0 \times 10^{-23} \text{ cm}^3

B

6.023×1023 cm36.023 \times 10^{-23} \text{ cm}^3

C

3.0×1023 cm33.0 \times 10^{-23} \text{ cm}^3

D

5.5×1023 cm35.5 \times 10^{-23} \text{ cm}^3

Step-by-Step Solution

The molar mass of water (H2O\text{H}_2\text{O}) is approximately 18 g mol118 \text{ g mol}^{-1} . Given that the density of water is 1 g cm31 \text{ g cm}^{-3}, we can calculate the volume occupied by one mole of water: Volume=MassDensity\text{Volume} = \frac{\text{Mass}}{\text{Density}} Volume of 1 mole=18 g1 g cm3=18 cm3\text{Volume of } 1 \text{ mole} = \frac{18 \text{ g}}{1 \text{ g cm}^{-3}} = 18 \text{ cm}^3. One mole of water contains Avogadro's number of molecules (NA=6.022×1023N_A = 6.022 \times 10^{23}) . Therefore, the volume occupied by a single water molecule is: Volume of one molecule=Volume of one moleNA\text{Volume of one molecule} = \frac{\text{Volume of one mole}}{N_A} Volume of one molecule=18 cm36.022×10232.989×1023 cm33.0×1023 cm3\text{Volume of one molecule} = \frac{18 \text{ cm}^3}{6.022 \times 10^{23}} \approx 2.989 \times 10^{-23} \text{ cm}^3 \approx 3.0 \times 10^{-23} \text{ cm}^3.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Some Basic Concepts of Chemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYSome Basic Concepts of Chemistryvolumeoccupiedmoleculedensity

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