Calculate the % empty space in 1 mole of water at 25 degree C (d=0.9971g/ml). The volume of a water molecule can be taken as the sum of the volumes of the 2 hydrogen molecules and the oxygen atom. Use the single bonded covalent radii of 37pm for hydrogen and 66pm for oxygen.

1 answer

First, we need to find the total volume of 1000 ml (1L) of water and the total volume of the water molecules in 1 mole of water. Then we can find the % empty space.

1. Find the mass of 1 mole of water.
Molar mass of water (H2O) = 2 * molar mass of H + molar mass of O
Molar mass of water = 2 * 1 g/mol (Hydrogen) + 16 g/mol (Oxygen) = 18 g/mol

2. Find the mass of 1 L (1000 ml) of water at 25 °C.
Density at 25°C = 0.9971 g/ml
Mass = Volume * Density
Mass = 1000 ml * 0.9971 g/ml = 997.1 g

3. Calculate the number of moles in 1L of water.
Number of moles = mass / molar mass = 997.1 g / 18 g/mol = 55.39 moles

4. Calculate the volume of one water molecule.
Volume of 1 H atom = 4/3 * pi * r^3 = 4/3 * pi * (37 pm)^3
Volume of 1 O atom = 4/3 * pi * r^3 = 4/3 * pi * (66 pm)^3
We assume that the H2O molecule is formed by overlapping volumes of 2 H atoms and 1 O atom.
Volume of 1 water molecule = (1 * Volume of O atom) + (2 * Volume of H atom)

5. Calculate the total volume of 1 mole of water molecules.
Total volume of water molecules = (number of moles) * (volume of 1 water molecule)

6. Calculate the volume of water containing 1 mole of water at 25 °C (assuming 1 mole occupies 1000 ml).
Total volume of 1 L water = 1000 ml

7. Calculate the % empty space.
% empty space = ((total volume of 1 L water - total volume of water molecules) / total volume of 1 L water) * 100

Let's calculate these values:

Volume of 1 H atom = 4/3 * pi * (37*10^-12 m)^3 = 2.11 * 10^-28 m^3
Volume of 1 O atom = 4/3 * pi * (66*10^-12 m)^3 = 1.20 * 10^-27 m^3
Volume of 1 water molecule = (1 * 1.20*10^-27) + (2 * 2.11*10^-28) = 1.54 * 10^-27 m^3
Total volume of water molecules = 55.39 moles * 1.54 * 10^-27 m^3/mole = 8.52 * 10^-26 m^3 (divide by 1000 to convert to liters)
Total volume of 1 L water = 1000 ml / 1000 = 1 L = 1 m^3
% empty space = ((1 - 8.52 * 10^-26) / 1) * 100 = 99.99999991 %

The % empty space in 1 mole of water at 25 °C is approximately 99.99999991 %.