Problem 81 - Chemistry Matter and Change

Question

How many grams of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) must be dissolved into 155 \(\mathrm{g}\) of water to create a solution with a molality of 8.20 mollkg?

Step-by-Step Solution

Answer

134.68 grams of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) are needed.

Step 1: Understand Molality Concept

Molality ( extit{m}) is defined as the number of moles of solute per kilogram of solvent. The formula is given by \( extit{m} = \frac{n}{ ext{kg of solvent}} \), where \( n \) is the number of moles of solute.

Step 2: Convert Solvent Mass to Kilograms

The mass of water provided is 155 grams. To convert this to kilograms, divide by 1000: \( 0.155 \text{ kg} = \frac{155}{1000} \).

Step 3: Calculate Moles Required

Using the formula for molality: \( 8.20 = \frac{n}{0.155} \). Solve for \( n \) to find the moles of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) required.\[ n = 8.20 \times 0.155 = 1.271 \text{ moles} \].

Step 4: Calculate Molar Mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \)

The molar mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) is calculated as follows: Sodium (Na) is 22.99 g/mol, Carbon (C) is 12.01 g/mol, and Oxygen (O) is 16.00 g/mol. \[ (2 \times 22.99) + 12.01 + (3 \times 16.00) = 105.99 \text{ g/mol} \].

Step 5: Calculate Grams of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \)

The number of grams needed is calculated using the molar mass: Grams of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) = 1.271 moles \( \times 105.99 \text{ g/mol} \). \[ 1.271 \times 105.99 = 134.68 \text{ grams} \].

Key Concepts

Sodium carbonate

Sodium carbonate, commonly known as soda ash or washing soda, is a white, powdery salt with the chemical formula \( \mathrm{Na}_2 \mathrm{CO}_3 \). It is an important substance in chemistry with various industrial applications, such as in glass manufacturing, soapmaking, and water softening. The structure of sodium carbonate includes 2 sodium atoms, one carbon atom, and 3 oxygen atoms. Each of these atoms contributes to its molar mass, which is essential when performing calculations related to the quantity needed for solutions, such as the problem at hand involving molality.Understanding the molar mass of compounds like sodium carbonate is crucial for accurately measuring the number of grams needed to achieve desired chemical properties in solutions. When working with sodium carbonate in a lab setting or real-world application, it's important to handle it with care, as it can be slightly irritating to the skin and eyes.

Solution preparation

Solution preparation is a fundamental task in chemistry, involving the accurate mixing of substances to achieve a specific concentration. In this context, we are preparing a solution of sodium carbonate with a desired molality. Molality is a measure of the number of moles of solute (in this case, sodium carbonate) per kilogram of solvent.To prepare this solution:

• First, ensure you have an accurate scale to measure the mass of the sodium carbonate and water.
• Convert the mass of your solvent from grams to kilograms as precision is key in molality calculations. For example, 155 grams becomes 0.155 kg.
• Use the molality formula \( m = \frac{n}{\text{kg of solvent}} \) where \( n \) is the moles of solute, in this case, the sodium carbonate.

This step-by-step approach ensures the solution created has the precise concentration needed for experiments or industrial processes. Proper preparation prevents errors that could affect the outcome of a chemical reaction or application.

Moles calculation

Moles calculation is pivotal in creating solutions with specific concentrations. To begin, understand that moles express quantity in chemistry, referring to the number of atoms, molecules, or ions in a given substance. Here, moles of sodium carbonate are calculated to determine how much is needed for a particular solution.To calculate moles, use the formula as derived from molality:

• Given the desired molality (8.20 mol/kg), and the mass of water (0.155 kg), employ the equation: \( n = m \times \text{kg of solvent} \).
• Substitute the values: \( n = 8.20 \times 0.155 = 1.271 \) moles of \( \mathrm{Na}_2 \mathrm{CO}_3 \).

Once you have the moles, multiply this by the molar mass of the compound to find the grams needed for the solution. Moles calculations are critical as they lay the foundation for determining how much substance interacts in a reaction and ensuring that you achieve the intended solution concentration.