Question

At \(373 \mathrm{~K}\) and \(1 \mathrm{~atm}\), if the density of liquid water is \(1.0 \mathrm{~g} / \mathrm{ml}\) and that of water vapour is \(0.0006 \mathrm{~g} / \mathrm{ml}\), then the volume occupied by water molecules in 1 litre of steam at that temperature is (a) \(6 \mathrm{ml}\) (b) \(60 \mathrm{ml}\) (c) \(0.6 \mathrm{ml}\) (d) \(0.06 \mathrm{ml}\)

Short answer

The volume occupied by water molecules in 1 litre of steam at the given conditions is \(0.6 \mathrm{ml}\), making (c) the correct answer.

Step-by-step solution

Calculating the mass of 1 litre of steam

Since the density of water vapour at the given conditions is \(0.0006 \mathrm{~g/ml}\), calculate the mass of 1 litre (or 1000 ml) of steam by using the formula: mass = density \(\times\) volume. Here, the volume of the steam is 1000 ml. So, mass of steam = \(0.0006 \mathrm{~g/ml} \times 1000 \mathrm{~ml} = 0.6 \mathrm{~g}\).

Finding the volume occupied by the water molecules in steam

Now that we know the mass of the water molecules in 1 litre of steam is \(0.6 \mathrm{~g}\), use the density of liquid water to find the volume of this mass would occupy in liquid form, since the density of liquid water is \(1.0 \mathrm{~g/ml}\). Volume = \(\frac{mass}{density} = \frac{0.6 \mathrm{~g}}{1.0 \mathrm{~g/ml}} = 0.6 \mathrm{~ml}\).

Key concepts

Understanding Physical Chemistry

Physical chemistry is the branch of chemistry that deals with the principles and methodologies of physics to study the composition, structure, and properties of matter, as well as the changes these undergo during chemical reactions. It bridges the gap between the macroscopic world we can observe and the microscopic realm of atoms and molecules.

When tackling density problems in physical chemistry, we focus on how the mass of a substance relates to its volume. This concept helps us understand phase changes and predict the behavior of substances under different temperatures and pressures. In the example exercise, we considered water in its gaseous and liquid states at a specific temperature and pressure, providing a practical application of these abstract principles.

Comparing Gaseous versus Liquid States

The gaseous and liquid states of matter showcase distinct physical properties, chiefly due to differences in intermolecular forces and the amount of free space between particles.

In a liquid, such as water at room temperature, molecules are close together with only small movements, which accounts for its relatively high density compared to gas. In contrast, the gaseous state, like water vapor at 373 K, molecules are dispersed, moving rapidly and occupying more space, leading to a much lower density. The exercise demonstrates this stark contrast: liquid water has a density of 1.0 g/ml, whereas water vapor has a density of just 0.0006 g/ml. Understanding these differences is crucial for correctly solving problems related to conversion between the two states, as seen in volume calculations for water in its gaseous form.

Molar Mass Calculations and Their Importance

The molar mass of a substance is the mass of one mole of its particles, typically expressed in grams per mole ((\(g/mol\))). This concept is integral to converting between the mass of a substance and the number of moles.

In our example, though molar mass isn't directly used, understanding it is essential for more complex problems where you might need to calculate the number of moles of water vapor or liquid water to proceed with the solution. For water, the molar mass is approximately 18.01528 g/mol, which can be used alongside Avogadro's number ((\(6.022 \times 10^{23} \), molecules/mol)) to relate microscopic properties of individual molecules to their macroscopic bulk properties.