Question

Volume occupied by one molecule of water $(density\, = 1 g\, cm^{-3} )$ is

• A. $9.0 \times 10 ^{-23} cm^3$
• B. $6.023 \times 10 ^{-23} cm^3$
• C. $3.0 \times 10 ^{-23} cm^3$
• D. $5.5 \times 10 ^{-23} cm^3$

Answer 1

The Correct Option is C

To determine the volume occupied by one molecule of water, we can use the following approach:

1. We know from Avogadro's hypothesis that one mole of any substance contains 6.022 \times 10^{23} molecules.

2. The molar mass of water (H₂O) is approximately 18 g/mol.

3. The density of water is given as 1 \, g \, cm^{-3}. This implies that 1 mole of water, weighing 18 grams, occupies a volume of 18 cm³.

4. To find the volume occupied by one molecule of water, we calculate:

   \[ \text{Volume of one molecule of water} = \frac{\text{Volume of 1 mole of water}}{\text{Number of molecules in 1 mole}} = \frac{18 \, \text{cm}^3}{6.022 \times 10^{23}} \]

5. Performing the calculation gives:

   \[ \frac{18}{6.022 \times 10^{23}} \approx 2.99 \times 10^{-23} \, \text{cm}^3 \]

6. Rounding off to match the given options, we get 3.0 \times 10^{-23} \, cm^3.

Thus, the volume occupied by one molecule of water is 3.0 \times 10^{-23} \, cm^3, which matches the correct answer provided in the options.