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: Understand the given information.

• The density of water is given as 1 g/cm³.
• The molar mass of water (H₂O) is 18.015 g/mol.
• Avogadro's number is 6.022 × 10²³ molecules/mol, which represents the number of molecules in one mole of a substance.

Step 2: Calculate the volume of one mole of water.

Since the density is given as 1 g/cm³, the volume occupied by 1 gram of water is:

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

Now, we know that 1 mole of water weighs 18.015 grams, so the volume occupied by 1 mole of water will be:

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

Step 3: Calculate the volume of one molecule of water.

The volume occupied by a single molecule of water is the volume of 1 mole divided 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 volume occupied by one molecule of water is approximately 2.98 × 10^-23 cm³.