Problem 8 - Chemistry and Chemical Reactivity

Question

You want to prepare an aqueous solution of ethylene glycol, 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 g of water? What is the molality of the solution?

Step-by-Step Solution

Answer

469.95 g of ethylene glycol, molality is 7.93 mol/kg.

Step 1: Understand Mole Fraction

The mole fraction \( X_i \) of a component in a solution is defined as the ratio of the moles of that component to the total moles of all components in the solution. In this case, the mole fraction of ethylene glycol is \( 0.125 \). This implies that for every mole of ethylene glycol, there are \( 7 \times 0.125 = 0.875 \) moles of water.

Step 2: Set Up Mole Ratio Equation

We know that the mole fraction of ethylene glycol \( X_{E} = 0.125 \), and \( X_{W} \) for water is \( 0.875 \). The total moles \( n_T \) can be represented by the equation: \[ X_{E} = \frac{n_{E}}{n_{E} + n_{W}} = 0.125 \] where \( n_{E} \) is the moles of ethylene glycol and \( n_{W} \) is the moles of water.

Step 3: Calculate Moles of Water

The molar mass of water is \( 18.015 \text{ g/mol} \). With 955 g of water, you calculate the moles of water:\[ n_{W} = \frac{955 \text{ g}}{18.015 \text{ g/mol}} \approx 53.01 \text{ moles} \].

Step 4: Solve for Moles of Ethylene Glycol

Using the equation from Step 2 and the calculated moles of water, solve for \( n_{E} \):\[ 0.125 = \frac{n_{E}}{n_{E} + 53.01} \] Rearrange to find \( n_{E} \): \[ n_{E} = 0.125(n_{E} + 53.01) \] Solving, \[ 0.875n_{E} = 6.62625 \] \[ n_{E} \approx 7.57 \text{ moles} \].

Step 5: Calculate Mass of Ethylene Glycol

The molar mass of ethylene glycol is \( 62.068 \text{ g/mol} \). Multiply the moles of ethylene glycol by its molar mass to find the mass:\[ m = 7.57 \text{ moles} \times 62.068 \text{ g/mol} \approx 469.95 \text{ g} \].

Step 6: Determine Molality of the Solution

Molality (\( m \)) is defined as the moles of solute per kilogram of solvent. Using the moles of ethylene glycol calculated in Step 4 and the mass of water (converted to kilograms):\[ m = \frac{7.57 \text{ moles}}{0.955 \text{ kg}} \approx 7.93 \text{ mol/kg} \].

Key Concepts

Molality

Molality is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality does not change with temperature because it is based on mass, not volume. This makes it particularly useful in situations involving temperature fluctuations.

To calculate molality, you use the formula:

• \( m = \frac{n_{solute}}{m_{solvent}} \)
• where \( n_{solute} \) is the moles of the solute, and \( m_{solvent} \) is the mass of the solvent in kilograms.

For instance, if you have calculated that there are 7.57 moles of ethylene glycol dissolved in 955 grams (or 0.955 kilograms) of water, the molality of the solution would be approximately 7.93 mol/kg. This tells you how concentrated the ethylene glycol is in the water.

Molar Mass

Molar mass is a fundamental concept that allows us to translate between the mass of a substance and the number of moles. It is defined as the mass of one mole of a substance and is expressed in grams per mole (g/mol).

Every element on the periodic table has a specific molar mass, which helps us calculate the molar mass of compounds. For example:

• To find the molar mass of water (H2O), you sum the molar masses of two hydrogen atoms (2 x 1.008 g/mol) and one oxygen atom (15.999 g/mol) to get approximately 18.015 g/mol.
• Similarly, the molar mass of ethylene glycol (C2H6O2) is the sum of the molar masses of its constituent atoms: carbon (2 x 12.011 g/mol), hydrogen (6 x 1.008 g/mol), and oxygen (2 x 15.999 g/mol), totaling 62.068 g/mol.

Understanding molar mass is crucial for converting between grams and moles in chemistry, which is necessary for computations involving mole ratios and concentrations.

Mole Ratio

A mole ratio is a proportion used in chemistry that relates the amounts of substances involved in a chemical reaction or present in a mixture. It is derived from the coefficients of a balanced chemical equation or based on given concentrations within a solution.

In the context of our problem, we are working with the mole ratio given by the mole fraction of a solution. For the ethylene glycol solution:

• The mole fraction of ethylene glycol tells us that out of the total number of moles, 12.5% is ethylene glycol, and the rest (87.5%) is water.
• From this information, you can set up equations to represent the mole ratio, such as \( X_{E} = \frac{n_{E}}{n_{E} + n_{W}} \).

These equations help calculate the exact amount of ethylene glycol needed for specific mole ratios, which are essential for creating solutions with specific properties.

Solute

In a solution, the solute is the substance that is dissolved in the solvent, forming a homogeneous mixture. The solute is usually present in a smaller amount compared to the solvent. In our problem, the solute is ethylene glycol, and it is dissolved in water, which is the solvent.

Understanding the role of the solute is key to determining solution properties like molality and mole fraction. The quantity of solute, whether measured in mass or moles, influences how concentrated the solution is:

• The greater the moles of solute (like ethylene glycol), the higher the molality of the solution.
• The mole fraction also helps understand the proportion of solute relative to the entire solution.

By preparing a solution with a known mass of solute, like the ethylene glycol in this exercise, we can precisely control the concentration and achieve desired solution characteristics.