Problem 52 - Chemistry The Molecular Science

Question

You want to prepare a \(1.0 \mathrm{~mol} / \mathrm{kg}\) solution of ethylene glycol, \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{OH})_{2},\) in water. Calculate the mass of ethylene glycol you would need to mix with \(950 . \mathrm{g}\) water.

Step-by-Step Solution

Answer

Approximately 59.0 g of ethylene glycol is required.

Step 1: Understanding Molality

We need to prepare a solution with a molality of \(1.0 \text{ mol/kg}\). Molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solvent is water and we want the molality to be \(1.0\).

Step 2: Determine Moles of Ethylene Glycol Required

Since the desired molality is \(1.0\) mol/kg and the mass of the water (solvent) is \(950\) grams or \(0.950\) kg, we calculate the moles of ethylene glycol needed using the equation: \[ \text{moles of ethylene glycol} = \text{molality} \times \text{mass of solvent in kg} = 1.0 \times 0.950 = 0.950 \text{ mol}\]

Step 3: Calculating Molar Mass of Ethylene Glycol

The molar mass of ethylene glycol \(\text{C}_{2}\text{H}_{4}(\text{OH})_{2}\) can be calculated by adding the atomic masses: \(2\times12.01\) (Carbon) + \(4\times1.01\) (Hydrogen) + \(2\times16.00\) (Oxygen) = \(62.08\) \text{ g/mol}.

Step 4: Mass of Ethylene Glycol Needed

Using the molar mass of ethylene glycol, calculate the mass needed: \[ \text{mass} = \text{moles} \times \text{molar mass} = 0.950 \times 62.08 = 58.976 \text{ g}\]

Step 5: Conclusion

You need approximately \(59.0\) g of ethylene glycol to make a \(1.0 \text{ mol/kg}\) solution with \(950\,\text{g}\) of water.

Key Concepts

Ethylene Glycol

Ethylene glycol is a chemical compound commonly used as an antifreeze and in the production of polyester fibers. It is a colorless, odorless liquid with the chemical formula \( \text{C}_2\text{H}_4(\text{OH})_2 \). Ethylene glycol is known for its ability to lower the freezing point of water, which is why it is so effective as an antifreeze. In chemistry, it often appears in problems related to solutions and molality.

When using ethylene glycol in laboratory experiments or for solution preparation, understanding its properties is crucial. These include its molar mass, boiling point, and toxicity. It is important to handle it with care and follow safety guidelines.

In educational exercises like the one at hand, ethylene glycol serves as the solute for which students learn how to calculate molality and prepare specific concentrations in solutions.

Molar Mass Calculation

Calculating the molar mass is essential when dealing with chemical compounds. It helps in determining the mass of a substance required for any given chemical reaction or solution preparation. Molar mass is calculated by summing up the atomic masses of all the atoms present in a molecule.

For ethylene glycol \( \text{C}_2\text{H}_4(\text{OH})_2 \), the calculation proceeds as follows:

• Carbon (C): 2 atoms, each with an atomic mass of 12.01, gives \( 2 \times 12.01 = 24.02 \) g/mol.
• Hydrogen (H): 6 atoms (4 from \( \text{H}_4\) and 2 from \((\text{OH})_2\)), with an atomic mass of 1.01, gives \( 6 \times 1.01 = 6.06 \) g/mol.
• Oxygen (O): 2 atoms, each with an atomic mass of 16.00, gives \( 2 \times 16.00 = 32.00 \) g/mol.

Adding these together, the total molar mass is \( 62.08 \) g/mol. Knowing this value is crucial for converting moles to grams in solution preparation.

Solution Preparation

Solution preparation involves creating a mixture with a specific concentration of solute in a solvent. In this exercise, ethylene glycol is the solute, and water is the solvent.

To achieve the desired molality of \( 1.0 \text{ mol/kg} \), we need to use the relationship:

• Molality (\(m\)) = moles of solute / kg of solvent.

Given \(950\) g of water, which is equivalent to \(0.950\) kg, we need \(0.950\) moles of ethylene glycol. Using the calculated molar mass of \(62.08\) g/mol, the necessary mass of ethylene glycol can be found as:

\[ \text{mass\ of ethylene glycol} = \text{moles} \times \text{molar mass} = 0.950 \times 62.08 = 58.976 \text{ g} \]

Thus, approximately \(59.0 \text{ g} \) of ethylene glycol is needed to prepare the solution. Always ensure measurements are precise, and note that temperature can affect solution preparation in practical settings.