Question

In which case is the number of molecules of water maximum ?

• A. $10^{-3} $ mol of water
• B. 18 mL of water
• C. 0.00224 L of water vapours at 1 atm and 273 K
• D. 0.18 g of water

Answer 1

The Correct Option is B

 To determine in which case the number of water molecules is maximum, we will calculate the number of molecules of water in each option.

1. 10^-3 mol of water:
   • Number of moles = 10^-3 mol.
   • Number of molecules = Moles × Avogadro's number = \(10^{-3} \times 6.022 \times 10^{23}\).
   • Number of molecules = \(6.022 \times 10^{20}\).
2. 18 mL of water:
   • Density of water = 1 g/mL, so 18 mL of water = 18 g of water.
   • Molar mass of water = 18 g/mol.
   • Number of moles = 18 g / 18 g/mol = 1 mol.
   • Number of molecules = 1 mol × Avogadro's number = \(6.022 \times 10^{23}\) molecules.
3. 0.00224 L of water vapors at 1 atm and 273 K:
   • At 1 atm and 273 K, 1 mole of any gas occupies 22.4 L (ideal gas condition).
   • Number of moles = 0.00224 L / 22.4 L/mol = \(10^{-4}\) mol.
   • Number of molecules = \(10^{-4} \times 6.022 \times 10^{23}\).
   • Number of molecules = \(6.022 \times 10^{19}\).
4. 0.18 g of water:
   • Molar mass of water = 18 g/mol.
   • Number of moles = 0.18 g / 18 g/mol = 0.01 mol.
   • Number of molecules = 0.01 mol × Avogadro's number = \(0.01 \times 6.022 \times 10^{23}\).
   • Number of molecules = \(6.022 \times 10^{21}\).

Comparing the number of molecules for all the options, the option with the maximum number of molecules is 18 mL of water with \(6.022 \times 10^{23}\) molecules.

Thus, the correct answer is 18 mL of water.