Problem 74 - Chemistry The Central Science

Question

Using data from Table \(13.3,\) calculate the freezing and boiling points of each of the following solutions: (a) \(0.25 \mathrm{~m}\) glucose in ethanol; (b) \(20.0 \mathrm{~g}\) of decane, \(\mathrm{C}_{10} \mathrm{H}_{22}\), in \(50.0 \mathrm{~g} \mathrm{CHCl}_{3} ;\) (c) \(3.50 \mathrm{~g}\) \(\mathrm{NaOH}\) in \(175 \mathrm{~g}\) of water, (d) 0.45 mol ethylene glycol and \(0.15 \mathrm{~mol} \mathrm{KBr}\) in \(150 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

3 for ethanol. \(ΔT_f = K_f * m = 1.99 * 0.25 = 0.498 K\) \(ΔT_b = K_b * m = 1.22 * 0.25 = 0.305 K\) New freezing point = Original freezing point - ΔT_f = -114.14 - 0.498 = -114.638°C New boiling point = Original boiling point + ΔT_b = 78.37 + 0.305 = 78.675°C The freezing and boiling points of the 0.25 m glucose in ethanol solution are -114.638°C and 78.675°C, respectively.

Step 1: (a) Solution Properties

First, we need to find the molality of the glucose solution (m): Given mass of glucose = 0.25 mol Mass of solvent (ethanol) = 100 g (we assume 1 molal solution) molality (m) = moles of solute/mass of solvent (in kg) = 0.25 / (100/1000) = 0.25 mol/kg

Step 2: (a) Calculate the Freezing and Boiling Points

We need to find the change in the freezing and boiling points using the respective depression and elevation constants provided in Table 13._makeConstraints

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating colligative property that happens when a non-volatile solute is added to a solvent. This effect occurs because the solute particles disrupt the orderly crystal formation of the solvent, requiring a lower temperature to reach the freezing point. The formula used to calculate the change in freezing point is:\[\Delta T_f = i \cdot K_f \cdot m\]Where:

• \(\Delta T_f\) is the change in freezing point
• \(i\) is the van 't Hoff factor (the number of particles the solute splits into)
• \(K_f\) is the freezing point depression constant
• \(m\) is the molality of the solution

For example, in the calculation for 0.25 m glucose in ethanol, assuming glucose doesn’t ionize, the van 't Hoff factor \(i\) is 1. With the necessary constants from the table, you can compute the depression effect by plugging in these values into the formula.

Boiling Point Elevation

Boiling point elevation is another key colligative property. It occurs when a solute is added to a solvent, which raises the temperature necessary for the solvent to boil. Essentially, the presence of solute particles lowers the vapor pressure of the solvent, and more energy is required to raise the vapor pressure to match the atmospheric pressure. Hence, the boiling point increases. The formula used for calculating boiling point elevation is:\[\Delta T_b = i \cdot K_b \cdot m\]Where:

• \(\Delta T_b\) is the change in boiling point
• \(i\) is the van 't Hoff factor
• \(K_b\) is the boiling point elevation constant
• \(m\) is the molality of the solution

For instance, if you are analyzing a solution of glucose in ethanol, and glucose remains intact in solution (producing an \(i\) of 1), the elevation constant \(K_b\) and molality \(m\) guide the calculation, showing how much higher the boiling point will become.

Molality Calculations

Molality is a concentration measurement often used in colligative property calculations because it remains unaffected by temperature changes. It is defined as the moles of solute per kilogram of solvent. To find molality, the following formula is used:\[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\]When calculating molality, follow these steps:

• Determine the number of moles of the solute involved.
• Measure the mass of the solvent used, making sure it’s in kilograms for consistency.
• Apply the formula to find your molality result.

In the given problems, such as finding the molality of glucose, decane or NaOH in their respective solvents, you adjust for the mass of the solvent and naturally any dissociation of compounds (like NaOH splitting into Na⁺ and OH^-) to get an accurate molality calculation. This value is crucial when further determining changes in freezing or boiling points.