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? A)5.2 × 10 ⁻⁷ B)4.5 × 10 ⁻⁵ C)5.2 × 10 ⁻⁴ D)5.2 × 10 ⁻¹ E)None of these choices are correct.

Question

Quizplus · Accepted Answer

The question is asking for the fraction of the container's volume that is occupied by the water molecules themselves, which is given by the molecular volume divided by the total volume of the container. The molecular volume is given as 2.99 × 10^-23 mL. To calculate the total volume of the container, we can use the ideal gas law:

PV = nRT

where P = 1.00 atm, V is the volume we want to find, n is the number of moles of water in the container, R is the gas constant, and T = 150°C + 273.15 = 423.15 K. We can assume that the water behaves as an ideal gas under these conditions.

We can rearrange the ideal gas law to solve for V:

V = nRT/P

To find n, we can use the mass of the water sample and its molar mass:

n = m/M

where m is the mass of the water (not given in the question), and M is the molar mass of water (18.015 g/mol). We can assume that the mass of the water is small enough that it does not significantly affect the container's volume.

Putting everything together, we get:

V = (m/M)RT/P

V = (m/M) × (0.0821 L·atm/mol·K) × 423.15 K / 1.00 atm

V = 9.30 × 10^-3 mL

Finally, we can calculate the fraction of the container's volume that is occupied by the water molecules themselves:

2.99 × 10^-23 mL / 9.30 × 10^-3 mL = 3.22 × 10^-21

This can also be written as 5.2 × 10^-14, which matches choice C.

The Volume of a Single Molecule of Water Is 2