Problem 5 - Chemistry and Chemical Reactivity

Question

What mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) must you add to \(125 \mathrm{g}\) of water to prepare \(0.200 \mathrm{m} \mathrm{Na}_{2} \mathrm{CO}_{3} ?\) What is the mole fraction of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in the resulting solution?

Step-by-Step Solution

Answer

Mass needed: 2.65 g; Mole fraction: 0.0036.

Step 1: Understanding Molality

Molality (\(m\)) is defined as the moles of solute per kilogram of solvent. The formula for molality is given by:\[m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\]

Step 2: Calculating Moles of Solute

We are given the molality of the solution as \(0.200 \, \mathit{m}\) and the mass of water as \(125 \, \text{g}\). First, convert the mass of water to kilograms:\[125 \, \text{g} = 0.125 \, \text{kg}\]Now, use the molality to find moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\):\[\text{moles of } \mathrm{Na}_{2} \mathrm{CO}_{3} = 0.200 \, \mathit{m} \times 0.125 \, \text{kg} = 0.025 \, \text{moles}\]

Step 3: Molar Mass of Na2CO3

The molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is calculated by summing the atomic masses:- Na: 22.99 g/mol (2 atoms)- C: 12.01 g/mol- O: 16.00 g/mol (3 atoms)\[Molar \, mass = 2 \times 22.99 + 12.01 + 3 \times 16.00 = 105.99 \, \text{g/mol}\]

Step 4: Calculating Mass of Na2CO3

Use the number of moles and the molar mass to find the mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\):\[\text{mass} = 0.025 \, \text{moles} \times 105.99 \, \text{g/mol} = 2.65 \, \text{g}\]

Step 5: Calculating Mole Fraction of Na2CO3

The mole fraction of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in the solution is calculated as the ratio of moles of solute to total moles in the solution. First, find the moles of water:\[\text{moles of water} = \frac{125 \, \text{g}}{18.015 \, \text{g/mol}} \approx 6.94 \, \text{moles}\]The mole fraction (\(X\)) is:\[X_{\mathrm{Na}_{2} \mathrm{CO}_{3}} = \frac{0.025}{0.025 + 6.94} \approx 0.0036\]

Step 6: Conclusion: Final Answers

The mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) needed to prepare the solution is \(2.65 \, \text{g}\), and the mole fraction of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in the solution is \(0.0036\).

Key Concepts

Mole Fraction

The concept of the mole fraction is an essential tool in chemistry to express the concentration of a component in a mixture. It is defined as the number of moles of a component (solute or solvent) divided by the total number of moles in the solution. This allows for a dimensionless number that provides insight into the concentration ratio between different components.

To calculate the mole fraction, you need two pieces of information: the number of moles of the solute and the number of moles of the solvent. For example, in the given problem, if we add 0.025 moles of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) to water, and the water contains approximately 6.94 moles, the total moles in the solution is the sum of these two values. Therefore, the mole fraction of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) is calculated using the formula:

• \( X = \frac{\text{moles of } \mathrm{Na}_{2} \mathrm{CO}_{3}}{\text{total moles}} \)

This results in a mole fraction of approximately 0.0036. This low value of the mole fraction indicates a dilute presence of the solute in the solution.

Molar Mass

Molar mass is a fundamental characteristic of chemical substances, telling us the mass of one mole of a particular substance, usually in grams per mole (g/mol). Understanding this concept is crucial for converting between the mass of a substance and the number of moles, which then helps in various calculations such as reactions and solutions.

To find the molar mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \), sum the atomic masses of all atoms present in the compound:

• Sodium (Na): 22.99 g/mol, with 2 atoms contributing \( 2 \times 22.99 \).
• Carbon (C): 12.01 g/mol.
• Oxygen (O): 16.00 g/mol, with 3 atoms contributing \( 3 \times 16.00 \).

The altogether molar mass is \( 105.99 \text{ g/mol} \). This number is used to convert moles of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) into grams, facilitating calculations such as determining the mass needed to achieve a specific molality in a solution.

Solution Preparation

Solution preparation is the process of mixing a solute with a solvent to create a homogenous mixture at the desired concentration. In chemistry, this often involves using concepts like molality, mole fraction, and molar mass to determine the correct proportions.

In our exercise, to prepare a solution of \( 0.200 \mathrm{m} \) molality of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \), you must first determine the mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) needed. Given the moles calculated from the molality formula \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \), it was calculated that 0.025 moles of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) are needed.

The molar mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \), already calculated as 105.99 g/mol, allows conversion to mass:

• Mass = 0.025 moles \( \times 105.99 \text{ g/mol} = 2.65 \text{ g} \)

This mass is then added to the solvent, ensuring that the process accounts for any temperature or pressure changes during mixing. This careful preparation results in the required molality, ensuring that experiments or applications relying on precise concentrations proceed accurately.