Question

Calculate the molality of each of the following solutions: (a) \(14.3 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(685 \mathrm{~g}\) of water, (b) 7.15 moles of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) in \(3505 \mathrm{~g}\) of water.

Short answer

(a) 0.0610 mol/kg; (b) 2.040 mol/kg.

Step-by-step solution

Understand the Concept of Molality

Molality ( \(m\)) 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. The formula for molality is \(m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\).

Calculate the Moles of Sucrose

First, we need to calculate the moles of sucrose. The molar mass of sucrose \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) is calculated as follows: \( (12 \times 12.01) + (22 \times 1.01) + (11 \times 16.00) = 342.30 \text{ g/mol}\).Thus, moles of sucrose = \( \frac{14.3 \text{ g}}{342.30 \text{ g/mol}} \approx 0.0418 \text{ mol}\).

Convert Water Mass to Kilograms

For part (a), convert the mass of water from grams to kilograms: \( 685 \text{ g} = 0.685 \text{ kg}\).

Calculate Molality of Sucrose Solution

Using the moles of sucrose and the mass of water in kilograms, calculate the molality: \( m = \frac{0.0418 \text{ mol}}{0.685 \text{ kg}} \approx 0.0610 \text{ mol/kg}\).

Moles of Ethylene Glycol

For part (b), the moles of ethylene glycol are already given as 7.15 mol.

Convert Water Mass to Kilograms

For part (b), convert the mass of water from grams to kilograms: \( 3505 \text{ g} = 3.505 \text{ kg}\).

Calculate Molality of Ethylene Glycol Solution

Using the moles of ethylene glycol and the mass of water in kilograms, calculate the molality: \( m = \frac{7.15 \text{ mol}}{3.505 \text{ kg}} \approx 2.040 \text{ mol/kg}\).

Key concepts

Concentration of Solutions

When talking about solutions, knowing how concentrated they are is key to understanding their properties and potential uses. Concentration tells us how much solute is present in a given quantity of solvent or solution. One common way to express concentration is molality.
A solute is the substance that is dissolved in a solvent to make a solution. The solvent is usually present in a larger amount and is the medium in which the solute is dissolved.
Molality is particularly useful because it doesn't change with temperature. Unlike molarity, which is concentration measured in moles per liter of solution, molality is measured in moles per kilogram of solvent. This makes it ideal for situations where temperature changes, such as in experimental chemistry.

• Molality is denoted by lowercase 'm'.
• The formula for molality is: \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \).
• This measure is useful when assessing physical properties like freezing point depression or boiling point elevation.

Understanding the concentration of solutions is essential for predicting and modeling chemical behaviors in both theoretical and practical applications.

Moles of Solute

The concept of "moles of solute" is central to chemistry. It expresses the amount of a substance. A mole is a number unit used to count very small things like atoms or molecules. Avogadro's number, \(6.022 \times 10^{23}\), tells us how many particles are in a mole.
To find the moles of a given solute, you divide the mass of the solute by its molar mass.

• The molar mass of a compound is the sum of the masses of all the atoms in its formula. For instance, the molar mass of sucrose \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\) is 342.30 g/mol.
• Example: If we have 14.3 g of sucrose, the moles of sucrose is calculated by dividing the mass by its molar mass: \( \frac{14.3 \text{ g}}{342.30 \text{ g/mol}} \approx 0.0418 \text{ mol} \).

Understanding how many moles of a solute are present helps in figuring out reactions and solution concentrations in chemical processes.

Kilograms of Solvent

In the context of molality, kilograms of solvent is a crucial component. It refers to the weight of the liquid in which the solute is dissolved, measured in kilograms. Converting grams to kilograms is simple: divide the mass in grams by 1000.
For example, when determining the kilograms of water in a solution, if you have 685 grams of water, the equivalent in kilograms is 0.685 kg.

• This step is important because molality uses the mass of the solvent (not the solution) as its denominator. This consistency makes molality unaffected by temperature, differing from molarity which uses volume.
• Example: In a scenario where you have 3505 grams of water, convert to kilograms: \( 3505 \text{ g} = 3.505 \text{ kg} \).

Converting to kilograms ensures accurate calculations for the molality and determines how the solution will behave under varying conditions.