91Ó°ÊÓ

To determine the volume of water when given a specific number of moles, we need to first find the mass of the water. In our example, we have 5.00 moles of water. The molar mass of water is 18.0 grams per mole, so the mass of 5.00 moles is calculated as follows: \ \[ \text{Mass} = 5.00 \, \text{mol} \times 18.0 \, \text{g/mol} = 90.0 \, \text{g} \].

Next, we use the density of water, which is approximately 1.00 \( \text{g/cm}^3 \), to find its volume. Density is mass per unit volume, so we rearrange the formula to solve for volume: \ \[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{90.0 \, \text{g}}{1.00 \, \text{g/cm}^3} = 90.0 \, \text{cm}^3 \].

This tells us that 5.00 moles of liquid water occupy a volume of 90.0 cubic centimeters.

Imagine water molecules spaced uniformly in a three-dimensional grid. Each molecule sits at the center of a cube, and the molecules are evenly distributed so each cube touches the adjacent ones. This idea helps us estimate the volume each molecule occupies.

First, we need to find out how many water molecules are present. We use Avogadro's number, which is \( 6.022 \times 10^{23} \) molecules per mole:\
\ \[ \text{Total Molecules} = 5.00 \, \text{mol} \times 6.022 \times 10^{23} = 3.011 \times 10^{24} \text{ molecules} \].

Given the water's total volume is 90.0 \( \text{cm}^3 \), the individual volume for each molecule's "cube" is: \ \[ \text{Volume per Molecule} = \frac{90.0 \, \text{cm}^3}{3.011 \times 10^{24}} = 2.988 \times 10^{-23} \, \text{cm}^3 \].

We can find the cube's side length \( a \) by solving \( a^3 = 2.988 \times 10^{-23} \, \text{cm}^3 \). The calculation results in the edge length of each molecule's cube.

Density is a measure of how much mass is contained in a given volume. For water, the standard density is about 1.00 \( \text{g/cm}^3 \) under typical conditions of temperature and pressure.

This property is useful because it allows us to easily convert between mass and volume, a common necessity in chemistry. For example, knowing a sample's mass and the fact that water typically has a density of 1.00 \( \text{g/cm}^3 \) makes it straightforward to compute the water's volume using the formula: \ \[ \text{Volume} = \frac{\text{mass}}{\text{density}} \].

This concept shows its significance when determining the space an amount of water occupies, as exercised in the problem above.

Avogadro's number, \( 6.022 \times 10^{23} \), represents the number of atoms, ions, or molecules in one mole of any substance. It is fundamental to connecting the micro-world of atoms and molecules to observable amounts of material.

In our example, to find out how many water molecules are in 5 moles, we multiply the number of moles by Avogadro's number: \ \[ \text{Total Molecules} = 5.00 \, \text{moles} \times 6.022 \times 10^{23} \text{ molecules/mol} = 3.011 \times 10^{24} \text{ molecules} \].

Understanding Avogadro's number helps in calculating how the microscopic world influences macroscopic properties such as volume and density.