The volume of a single molecule of water is 2.99 × 10^-23 mL. For a sample of gaseous water at 1.00 atm and 150°C, what fraction of the container's volume is occupied by the molecules themselves?

Question

Quizplus · Accepted Answer

The first step is to calculate the molar volume of the sample using the ideal gas law:

V = (nRT)/P

where V is the volume, n is the number of moles of water, R is the ideal gas constant, T is the temperature in Kelvin, and P is the pressure.

First, we need to convert 150°C to Kelvin:

T = 273.15 + 150 = 423.15 K

Next, we can use the molar mass of water to convert from grams to moles:

18 g/mol * (1 mol / 1000 g) = 0.018 kg/mol

Now we can plug in the values for V, n, R, T, and P:

V = (0.018 kg/mol * 8.314 J/(mol*K) * 423.15 K) / 101325 Pa 
V ≈ 0.00406 m³

To find the volume occupied by a single water molecule, we can divide the molar volume by Avogadro's number:

V_mol = 2.99 × 10^-23 mL = 2.99 × 10^-29 m³

V_mol / N_A = (2.99 × 10^-29 m³) / (6.022 × 10^23 molecules/mol) 
V_mol / N_A ≈ 4.98 × 10^-53 m³/molecule

Finally, we can find the fraction of the container's volume occupied by the water molecules:

(V_mol / N_A) / V ≈ (4.98 × 10^-53 m³/molecule) / (0.00406 m³) 
(V_mol / N_A) / V ≈ 5.16 × 10^-51

This is approximately equal to 5.2 × 10^-4, which is answer choice C. So the fraction of the container's volume occupied by the water molecules is about 5.2 × 10^-4 or 5.2 × 10^-4 * 100% = 0.000052%.

The Volume of a Single Molecule of Water Is 2