Problem 74 - Chemistry and Chemical Reactivity

Question

Water at \(25^{\circ} \mathrm{C}\) has a density of \(0.997 \mathrm{g} / \mathrm{cm}^{3}\) Calculate the molality and molarity of pure water at this temperature.

Step-by-Step Solution

Answer

Molarity is 55.36 M and molality is 55.54 m.

Step 1: Define Density

The density of water at \( 25^{\circ} \text{C} \) is given as \( 0.997 \text{ g/cm}^3 \). This will help us convert between mass and volume.

Step 2: Convert Mass to Moles

We need the molar mass of water (\( \text{H}_2\text{O} \)) which is \( 18.015 \text{ g/mol} \). Given that 1 liter (1000 mL) of water weighs \( 0.997 \times 1000 = 997 \text{ g} \), we find:\[\text{moles of water} = \frac{997 \text{ g}}{18.015 \text{ g/mol}} \approx 55.36 \text{ moles}\]

Step 3: Calculate Molarity

Molarity is defined as the number of moles of solute (here, water itself) per liter of solution. The volume of the solution is 1 liter (since we have 1000 mL). Thus, the molarity is:\[\text{molarity} = \frac{55.36 \text{ moles}}{1 \text{ L}} = 55.36 \text{ M}\]

Step 4: Calculate Molality

Molality is defined as the number of moles of solute per kilogram of solvent (here, water itself). We have 997 grams of water, which is 0.997 kilograms. Using the moles calculated previously:\[\text{molality} = \frac{55.36 \text{ moles}}{0.997 \text{ kg}} \approx 55.54 \text{ m}\]

Key Concepts

Understanding the Density of Water

The density of water is a fundamental concept in solution chemistry. It is often used to convert between mass and volume measurements, which are crucial in chemical calculations. At room temperature, specifically at \(25^{\circ} \text{C}\), water has a density of \(0.997\, \text{g/cm}^3\). This means that each cubic centimeter (or milliliter) of water weighs 0.997 grams.

Density can vary slightly with temperature, so it’s essential to use the correct value for calculations involving solutions at different temperatures. This property helps us understand the behavior of water under varying conditions and is particularly useful when calculating other properties like molarity and molality.

Calculating Molarity

Molarity is a way of expressing the concentration of a solution. It is defined as the number of moles of solute (here, water itself) per liter of solution. To calculate molarity, you need two key pieces of information: the moles of your solute and the volume of the solution in liters.

In the context of pure water:

• The molar mass of water (\(\text{H}_2\text{O}\)) is \(18.015 \, \text{g/mol}\).
• To find out how many moles of water are in 1 liter, you first calculate the mass of 1 liter using the given density: \(997 \, \text{g}\).
• Then, convert grams to moles: \(\frac{997 \, \text{g}}{18.015 \, \text{g/mol}} \approx 55.36 \, \text{moles}\).

Thus, the molarity of pure water at \(25^{\circ} \text{C}\) is approximately \(55.36 \, \text{M}\). This result tells us the concentration in terms of the volume of the solution.

Understanding and Calculating Molality

Molality is another measure of concentration but is different from molarity. Instead of focusing on the volume of the solution, molality concentrates on the mass of the solvent. It is defined as the number of moles of solute per kilogram of solvent.

For calculating the molality of pure water:

• We already know the number of moles of water from the molarity calculation: \(55.36 \, \text{moles}\).
• Since the solvent is also water, we consider its mass: remember that 1 liter of water weighs \(997 \, \text{g}\), or \(0.997 \, \text{kg}\).
• Calculate molality using the formula: \(\frac{55.36 \, \text{moles}}{0.997 \, \text{kg}} \approx 55.54 \, \text{m}\).

Molality is especially useful in situations where volume can change with temperature or pressure, as it relies on mass, which remains constant. Understanding both molarity and molality allows chemists to describe solution concentrations precisely under varying conditions.