Problem 29 - University Physics with Modern Physics

Question

Consider 5.00 mol of liquid water. (a) What volume is occupied by this amount of water? The molar mass of water is 18.0 \(\mathrm{g} / \mathrm{mol}\) (b) Imagine the molecules to be, on average, uniformly spaced, with each molecule at the center of a small cube. What is the length of an edge of each small cube if adjacent cubes touch but don't overlap? (c) How does this distance compare with the diameter of a molecule?

Step-by-Step Solution

Answer

(a) 90.0 cm³, (b) edge length of cube in nm, (c) compare to 0.275 nm.

Step 1: Calculate the Mass of Water

To find the mass of 5.00 mol of water, use the molar mass given: \[\text{Mass} = \text{Moles} \times \text{Molar Mass} = 5.00 \, \text{mol} \times 18.0 \, \text{g/mol} = 90.0 \, \text{g}\]

Step 2: Calculate the Volume of Water

The density of water is generally around 1.00 \( \text{g/cm}^3 \). Using the mass calculated: \[\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{90.0 \, \text{g}}{1.00 \, \text{g/cm}^3} = 90.0 \, \text{cm}^3\]

Step 3: Calculate Number of Molecules

Since 1 mole contains Avogadro's number of molecules (\(6.022 \times 10^{23}\)), calculate the total number of molecules in 5.00 mol:\[\text{Total Molecules} = 5.00 \times 6.022 \times 10^{23} \]

Step 4: Calculate Volume per Molecule

From Step 2, the total volume of water is \(90.0 \, \text{cm}^3\), and Step 3 gives the total number of molecules. The volume per molecule can be calculated as:\[\text{Volume per Molecule} = \frac{90.0 \, \text{cm}^3}{5.00 \times 6.022 \times 10^{23}}\]

Step 5: Find Edge Length of Cube

Assuming each molecule occupies a cube with the derived volume, the side length of such a cube \(a\) can be calculated using the formula for the volume of a cube \(a^3 = \text{Volume per Molecule}\):\[a = \sqrt[3]{\text{Volume per Molecule}}\]

Step 6: Compare to Molecular Diameter

Water molecules have an approximate diameter of 0.275 nm. Convert \(a\) from \(\text{cm}\) to \(\text{nm}\) (where 1 cm = \(10^7\) nm) and compare it with the molecular diameter.

Key Concepts

Volume Calculation

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.

Molecular Spacing

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 of Water

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

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.