Problem 118 - Chemistry Matter and Change

Question

Compare Which of the following solutions has the highest concentration? Rank the solutions from the greatest to the smallest boiling point depression. Explain your answer. $$\begin{array}{l}{\text { a. } 0.10 \mathrm{mol} \mathrm{NaBr} \text { in } 100.0 \mathrm{mL} \text { solution }} \\ {\text { b. } 2.1 \mathrm{mol} \mathrm{KOH} \text { in } 1.00 \mathrm{L} \text { solution }} \\ {\text { c. } 1.2 \mathrm{mol} \mathrm{KMnO}_{4} \text { in } 3.00 \mathrm{L} \text { solution }}\end{array}$$

Step-by-Step Solution

Answer

KOH has the highest concentration and boiling point depression, followed by NaBr, and then KMnO4.

Step 1: Determine Molarity of Each Solution

Molarity is defined as the number of moles of solute per liter of solution. To find the molarity \( M \), use the formula \( M = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume of the solution in liters.

Step 2: Calculate Molarity for Solution (a)

For NaBr: \( n = 0.10 \, \text{mol} \) and \( V = 0.100 \, \text{L} \).Molarity \( M = \frac{0.10}{0.100} = 1.0 \, \text{M} \).

Step 3: Calculate Molarity for Solution (b)

For KOH: \( n = 2.1 \, \text{mol} \) and \( V = 1.00 \, \text{L} \).Molarity \( M = \frac{2.1}{1.00} = 2.1 \, \text{M} \).

Step 4: Calculate Molarity for Solution (c)

For KMnO extsubscript{4}: \( n = 1.2 \, \text{mol} \) and \( V = 3.00 \, \text{L} \).Molarity \( M = \frac{1.2}{3.00} = 0.4 \, \text{M} \).

Step 5: Determine Boiling Point Depression with Van 't Hoff Factor

Boiling point depression \( \Delta T_b \) depends on molality and the Van 't Hoff factor \( i \), given by: \( \Delta T_b = i \cdot K_b \cdot \, m \), where \( m \) is the molality. Since all solutions are assumed to have the same solvent, compare primarily on molarity and \( i \).

Step 6: Compare and Rank Each Solution

NaBr \( i = 2 \), KOH \( i = 2 \), KMnO extsubscript{4} \( i = 1 \). Use the product of molarity and \( i \) to find effective concentration: - NaBr: \( 1.0 \times 2 = 2.0 \)- KOH: \( 2.1 \times 2 = 4.2 \)- KMnO extsubscript{4}: \( 0.4 \times 1 = 0.4 \)Hence, rank from greatest to smallest: KOH, NaBr, KMnO extsubscript{4}.

Key Concepts

Molarity

Molarity is a measure of the concentration of a solution. It's defined as the number of moles of solute present per liter of solution. When determining molarity (M), you use the formula: M = \frac{n}{V}, where n is moles of solute, and V is the volume in liters.
To get the molarity of a solution:

• First, calculate the moles of solute, often provided in the problem as in our exercise.
• Then, determine the volume of the solution in liters; it's crucial to convert milliliters to liters if needed.
• Finally, divide the moles by volume to get molarity.

In the provided examples: NaBr has a molarity of \(1.0 \, \text{M}\), KOH has a molarity of \(2.1 \, \text{M}\), and KMnO\(_4\) has a molarity of \(0.4 \, \text{M}\). These values help in determining the concentration and are critical when applying boiling point depression.

Van 't Hoff Factor

The Van 't Hoff factor, i, represents the number of particles a compound dissociates into in solution. It plays a pivotal role in colligative properties, such as boiling point depression.
When a compound dissolves:

• Electrolytes (like NaBr and KOH) dissociate into ions, increasing the number of effective particles in the solution. For instance, NaBr dissociates into Na\(^+\) and Br\(^-\), so i = 2\>.
• Nonelectrolytes don’t dissociate, with i = 1\>.

In our comparison, NaBr has i = 2\>, KOH also has i = 2\>, and KMnO\(_4\), being a relatively stable compound in water, has i = 1\>. The Van 't Hoff factor is crucial when calculating changes in colligative properties because it affects equations like \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the boiling point elevation.

Molality

Molality, often confused with molarity, measures solution concentration but in a distinct way. It's defined as the number of moles of solute per kilogram of solvent. This measurement is important specifically in colligative properties, like boiling point depression, because it remains constant with changes in temperature and pressure.
Here's why it's different:

• Molality (m) uses mass (kg) for the solvent, not volume like molarity.
• This makes molality particularly useful in boiling point depression calculations, as volume can change with temperature, but mass stays constant.
• Molality is used in conjunction with the Van 't Hoff factor and the ebullioscopic constant (K_b) in the formula \( \Delta T_b = i \cdot K_b \cdot m \).

Although the original problem didn’t require direct calculation of molality, understanding it helps explain how boiling point depression is mathematically modeled and why concentration methods might affect boiling point changes.