Question

What is the volume occupied by 1 molecule of water, if its density is 1 g/cm$^3$?

• A. 9.09 $\times$ 10$^{-23}$ cm$^3$
• B. 2.98 $\times$ 10$^{-23}$ cm$^3$
• C. 4.03 $\times$ 10$^{-23}$ cm$^3$
• D. 5.50 $\times$ 10$^{-23}$ cm$^3$

Hint:

To find the volume occupied by one molecule, divide the molar volume of the substance by Avogadro's number.

Answer 1

The Correct Option is B

Step 1: Information Gathering.

• Water density: 1 g/cm³.
• Water molar mass (H₂O): 18.015 g/mol.
• Avogadro's number: 6.022 × 10²³ molecules/mol (number of molecules per mole).

Step 2: Volume of One Mole of Water Calculation.

Using the density, the volume of 1 gram of water is:

Volume = Mass / Density = 1 g / 1 g/cm³ = 1 cm³

Given that 1 mole of water weighs 18.015 grams, its volume is:

Volume of 1 mole = 18.015 g × (1 cm³/g) = 18.015 cm³

Step 3: Volume of One Water Molecule Calculation.

The volume of a single water molecule is calculated by dividing the volume of one mole by Avogadro's number:

Volume of 1 molecule = Volume of 1 mole / Avogadro's number = 18.015 cm³ / (6.022 × 10²³)

Volume of 1 molecule = 2.98 × 10^-23 cm³

Conclusion: The approximate volume of a single water molecule is 2.98 × 10^-23 cm³.