Question

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

• A. 18 mL of water
• B. 0·00224 L of water vapours at 1 atm and 273 K
• C. 0·18 g of water
• D. 10^–3 mol of water

Answer 1

The Correct Option is A

To determine which case contains the maximum number of molecules of water, we need to calculate the number of molecules in each given option. We know that one mole of any substance contains Avogadro’s number of molecules, which is approximately \( 6.022 \times 10^{23} \) molecules.

1. 18 mL of water:
   • Density of water = 1 g/mL
   • Mass of 18 mL of water = 18 g
   • Molar mass of water (H₂O) = 18 g/mol

   Number of moles = \( \frac{18\, \text{g}}{18\, \text{g/mol}} = 1\, \text{mol} \)

   Number of molecules = \( 1\, \text{mol} \times 6.022 \times 10^{23}\, \text{molecules/mol} = 6.022 \times 10^{23}\, \text{molecules} \)

2. 0.00224 L of water vapours at 1 atm and 273 K:
   • We use the ideal gas equation \( PV = nRT \) to find moles. At standard temperature and pressure (STP), 1 mole of a gas occupies 22.4 L.
   • Volume of water vapour = 0.00224 L

   Number of moles = \( \frac{0.00224\, \text{L}}{22.4\, \text{L/mol}} = 0.0001\, \text{mol} \)

   Number of molecules = \( 0.0001\, \text{mol} \times 6.022 \times 10^{23}\, \text{molecules/mol} = 6.022 \times 10^{19}\, \text{molecules} \)

3. 0.18 g of water:
   • Molar mass of water = 18 g/mol

   Number of moles = \( \frac{0.18\, \text{g}}{18\, \text{g/mol}} = 0.01\, \text{mol} \)

   Number of molecules = \( 0.01\, \text{mol} \times 6.022 \times 10^{23}\, \text{molecules/mol} = 6.022 \times 10^{21}\, \text{molecules} \)

4. 10⁻³ mol of water:

   Number of molecules = \( 10^{-3}\, \text{mol} \times 6.022 \times 10^{23}\, \text{molecules/mol} = 6.022 \times 10^{20}\, \text{molecules} \)

Comparing the number of molecules calculated for each option, it is clear that 18 mL of water contains the maximum number of molecules.