Questions: What is the experimental MW of isopropanol as derived from the freezing point depression, in g/mol? Report in significant figures with appropriate rounding Room Temperature: 20.7°C Volume of distilled water used: 15.1 mL Freezing point temperature, pure water (average across two runs): -0.1°C Volume of isopropanol used: 2.9 mL Isopropanol density: 0.785 g / mL Freezing point temperature, water/isopropanol solution (average across two runs): -4.9°C Molal freezing point depression constant: 1.86 kg · °C / mol Experimental MW of isopropanol: 60.095 g / mol

What is the experimental MW of isopropanol as derived from the freezing point depression, in g/mol? Report in significant figures with appropriate rounding

Room Temperature: 20.7°C
Volume of distilled water used: 15.1 mL
Freezing point temperature, pure water (average across two runs): -0.1°C
Volume of isopropanol used: 2.9 mL
Isopropanol density: 0.785 g / mL
Freezing point temperature, water/isopropanol solution (average across two runs): -4.9°C
Molal freezing point depression constant: 1.86 kg · °C / mol
Experimental MW of isopropanol: 60.095 g / mol

Transcript text: What is the experimental MW of isopropanol as derived from the freezing point depression, in g/mol? Report in significant figures with appropriate rounding \begin{tabular}{ll} \hline Room Temperature: & $20.7^{\circ} \mathrm{C}$ \\ Volume of distilled water used: & 15.1 mL \\ \begin{tabular}{ll} Freezing point temperature, pure water \\ (average across two runs): \end{tabular} & $-0.1^{\circ} \mathrm{C}$ \\ Volume of isopropanol used: & 2.9 mL \\ Isopropanol density: & $0.785 \mathrm{~g} / \mathrm{mL}$ \\ \hline Freezing point temperature, water/isopropanol & $-4.9^{\circ} \mathrm{C}$ \\ solution (average across two runs): & $1.86 \mathrm{~kg} \cdot{ }^{\circ} \mathrm{C} / \mathrm{mol}$ \\ Molal freezing point depression constant: & $60.095 \mathrm{~g} / \mathrm{mol}$ \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Calculate the Mass of Isopropanol

First, we need to calculate the mass of isopropanol used. Given the volume of isopropanol is 2.9 mL and its density is 0.785 g/mL, the mass $ m $ can be calculated as:

\[ m = \text{volume} \times \text{density} = 2.9 \, \text{mL} \times 0.785 \, \text{g/mL} = 2.2765 \, \text{g} \]

Step 2: Calculate the Freezing Point Depression

The freezing point depression $ \Delta T_f $ is the difference between the freezing point of pure water and the freezing point of the solution:

\[ \Delta T_f = -0.1^{\circ} \mathrm{C} - (-4.9^{\circ} \mathrm{C}) = 4.8^{\circ} \mathrm{C} \]

Step 3: Calculate the Molality of the Solution

Using the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m \]

where $ K_f = 1.86 \, \text{kg} \cdot {}^{\circ} \mathrm{C} / \mathrm{mol} $ is the molal freezing point depression constant, and $ m $ is the molality. Solving for $ m $:

\[ m = \frac{\Delta T_f}{K_f} = \frac{4.8^{\circ} \mathrm{C}}{1.86 \, \text{kg} \cdot {}^{\circ} \mathrm{C} / \mathrm{mol}} = 2.5806 \, \text{mol/kg} \]

Step 4: Calculate the Moles of Isopropanol

The molality $ m $ is defined as moles of solute per kilogram of solvent. The mass of the solvent (water) is 15.1 mL, which is approximately 15.1 g or 0.0151 kg. Therefore, the moles of isopropanol $ n $ is:

\[ n = m \times \text{mass of solvent} = 2.5806 \, \text{mol/kg} \times 0.0151 \, \text{kg} = 0.03896 \, \text{mol} \]

Step 5: Calculate the Experimental Molar Mass of Isopropanol

Finally, the molar mass $ M $ of isopropanol can be calculated using the mass and moles:

\[ M = \frac{\text{mass of isopropanol}}{\text{moles of isopropanol}} = \frac{2.2765 \, \text{g}}{0.03896 \, \text{mol}} = 58.43 \, \text{g/mol} \]