Final answer:
The total number of moles of water molecules in the container at the triple point is approximately 3.3 x 10²¹ mol. We arrived at this conclusion by calculating the mole content of 1L of gas at STP conditions and adding it to the mol content in the liquid phase, where the gas phase's contribution is negligible. Option 1.
Step-by-step explanation:
The student's question involves calculating the total number of moles of water molecules in both liquid and gas phases at the triple point, where they coexist. Given that the triple point allows for the coexistence of solid, liquid, and gas states, and the conditions are assumed to resemble standard temperature and pressure (STP), we can apply our knowledge of molar volume and Avogadro's number to find the solution.
Firstly, the liquid phase contains 3.3 x 10²¹ mol of water molecules. For the gas phase, we know that at STP, 1 mole of an ideal gas occupies 22.4 liters. Hence, 1 L of gas will contain 1 / 22.4 moles of water molecules. Performing the calculation:
1 L gas * (1 mol / 22.4 L) = 0.04464 mol
However, the question provides the amount in molecules, so we convert it to moles:
1.6 x 10²° molecules * (1 mol / 6.022 x 10²³ molecules) = 2.66 x 10⁻´ mol
Now, we add the moles of water in the liquid phase to the gas phase to find the total moles:
3.3 x 10²¹ mol + 2.66 x 10⁻´ mol = 3.300266 x 10²¹ mol
This value is approximately 3.3 x 10²¹ mol, which matches the amount in the liquid phase, as the amount in the gas phase is negligible in comparison. Thus, the correct answer is 3.3 x 10²¹ mol, or answer choice 1 - 3.46 x 10²¹ mol, assuming the given value was rounded.