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 $\degree$C, what fraction of the container's volume is occupied by the molecules themselves?

Question

Quizplus · Accepted Answer

First, we need to calculate the volume occupied by one molecule of water at the given conditions:

V = 2.99 x 10^(-23) L/molecule (conversion from mL to L)

Next, we need to calculate the total volume occupied by all the molecules in the container:

V_total = (n/V) x k x T/P

where n is the number of molecules, k is the Boltzmann constant, T is the temperature in Kelvin, and P is the pressure.

To simplify the calculation, we can assume a container with a volume of 1 L, so n/V is simply the concentration of water in the gas phase at the given conditions. This can be calculated using the ideal gas law:

n/V = (P/RT) x (1/1000)

where R is the gas constant and T is the temperature in Kelvin.

Substituting the values given in the problem, we get:

n/V = (1.00 atm / 0.0821 L atm/mol K / (273 + 150) K) x (1/1000) = 0.0417 mol/L

Now we can calculate the total volume occupied by all the molecules:

V_total = (0.0417 mol/L / 2.99 x 10^(-23) L/molecule) x (1.38 x 10^(-23) J/K / molecule) x (423 K) / (1.00 atm) = 0.0189 L

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

fraction = V_total / 1 L = 0.0189

Therefore, the answer is C, which is closest to 5.2 x 10^(-15) (the value obtained by dividing V by the volume of the container in mL).

The Volume of a Single Molecule of Water Is 2 °\degree°

The Volume of a Single Molecule of Water Is 2 $\degree$