Problem 8 - Chemistry and Chemical Reactivity

Question

You want to prepare an aqueous solution of ethylene glycol, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH},\) in which the mole fraction of solute is \(0.125 .\) What mass of ethylene =glycol, in grams, should you combine with \(955 \mathrm{g}\) -of water? What is the molality of the solution?

Step-by-Step Solution

Answer

Mass of ethylene glycol: ~470 g. Molality: ~7.93 mol/kg.

Step 1: Understand the Problem

We are tasked with preparing a solution where ethylene glycol is the solute and water is the solvent. Given the mole fraction of ethylene glycol is 0.125, we need to find the mass of ethylene glycol required for this solution and its molality.

Step 2: Write the Relations for Mole Fraction

The mole fraction of a solute is defined as the ratio of the moles of the solute to the total moles in the solution. Mathematically, it is represented as:\[x_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = 0.125\]where \(x_{solute}\) is the mole fraction of ethylene glycol, \(n_{solute}\) is the moles of ethylene glycol, and \(n_{solvent}\) is the moles of water.

Step 3: Calculate Moles of Water

Given mass of water is 955 g, we convert this to moles using the molar mass of water (18.015 g/mol):\[n_{solvent} = \frac{955 \, \text{g}}{18.015 \, \text{g/mol}} \approx 53.02 \, \text{mol}\]Thus, there are approximately 53.02 moles of water.

Step 4: Set up Equation for Moles of Ethylene Glycol

Using the mole fraction relation, solve for \(n_{solute}\):\[0.125 = \frac{n_{solute}}{n_{solute} + 53.02}\]Rearrange to find \(n_{solute}\):\[n_{solute} = 0.125(n_{solute} + 53.02)\]\(n_{solute} = 0.125n_{solute} + 6.6275\) \(0.875n_{solute} = 6.6275\) \[n_{solute} \approx 7.574\, \text{mol}\]Thus, approximately 7.574 moles of ethylene glycol are needed.

Step 5: Calculate Mass of Ethylene Glycol

Convert moles of ethylene glycol to mass using its molar mass (62.07 g/mol):\[m_{solute} = n_{solute} \times \text{molar mass of ethylene glycol} = 7.574 \, \text{mol} \times 62.07 \, \text{g/mol} \approx 470 \, \text{g}\]Thus, approximately 470 grams of ethylene glycol are needed.

Step 6: Calculate Molality of the Solution

Molality is defined as moles of solute per kilogram of solvent:\[m = \frac{n_{solute}}{\text{mass of solvent in kg}} = \frac{7.574 \, \text{mol}}{0.955 \, \text{kg}}\approx 7.93\, \text{mol/kg}\]Thus, the molality of the solution is approximately 7.93 mol/kg.

Key Concepts

Molality

Molality is a measure of the concentration of a solution in terms of moles of solute per kilogram of solvent. It provides a way to quantify how concentrated a solution is.
To calculate molality, the formula used is:

• \[ m = \frac{n_{solute}}{\text{mass of solvent in kg}} \]

Here, \( n_{solute} \) represents the moles of the solute, and the mass of solvent is given in kilograms. This measure is used instead of molarity when temperature changes are involved, because molality is independent of temperature as it does not involve volume, which can expand or contract with temperature fluctuations.
In the exercise, it's calculated as \( \frac{7.574 \text{ mol}}{0.955 \text{ kg}} \approx 7.93 \text{ mol/kg} \), which tells us how many moles of ethylene glycol are present per kilogram of water.

Aqueous Solution

An aqueous solution is simply a solution in which water is the solvent. This type of solution is very common in chemistry due to water's polar nature, which helps dissolve many substances.
Water's ability to dissolve so many substances, including salts and organic compounds like ethylene glycol (as in our problem), makes it known as the "universal solvent."
In the given exercise, since ethylene glycol is dissolved in water, we call it an aqueous solution. The goal is to determine how much of the glycol solute is needed to result in a certain mole fraction and corresponding molality.

Molar Mass

Molar mass, often expressed in grams per mole, is the mass of one mole of a given substance. It tells us how much one mole of a chemical species weighs and is crucial for converting between moles and grams in chemical calculations.
For instance, in the problem, the molar mass of ethylene glycol is 62.07 g/mol. This information is essential for calculating the exact mass needed once the number of moles is determined. After calculating the moles of ethylene glycol needed as approximately 7.574 mol, you use its molar mass to find the weight:

• \[ 7.574 \text{ mol} \times 62.07 \text{ g/mol} \approx 470 \text{ g} \]

Thus, understanding molar mass is key to moving from theoretical calculations to practical use.

Chemical Calculation

Chemical calculations involve a series of steps to convert between different units related to chemical substances. Visualizing relationships through equations is crucial for accurate results.
In solving the exercise, it's demonstrated through calculating mole fraction and molality:

• Understanding mole fraction \[ x_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = 0.125 \]
• Calculating moles: converting mass to moles as done with water: \[ n_{solvent} = \frac{955 \text{ g}}{18.015 \text{ g/mol}} = 53.02 \text{ mol} \]

These steps help to precisely determine the mass required and the solution's concentration. Thus, understanding both the mathematical and chemical fundamentals solidifies one's ability to tackle real-world chemical problems effectively.