Question

0.500 grams of an unknown alcohol is added to 4.71 grams of tert - butyl alcohol. the freezing point of the solution is measured to be 6.52. use 18.016 for the molar mass of di water, 25.5 °c for the freezing point of pure tert - butyl alcohol, and 8.37 °c kg mol⁻¹ for the freezing point constant (kf) for tert - butyl alcohol. what is the molar mass of the unknown alcohol? type answer:

Explanation:

Step1: Calculate freezing point depression

The freezing point depression $\Delta T_f$ is the difference between the freezing point of pure solvent and the solution. So $\Delta T_f = T_f^{\circ} - T_f = 25.5^{\circ}\text{C} - 6.52^{\circ}\text{C} = 18.98^{\circ}\text{C}$

Step2: Use freezing point depression formula to find molality

The formula for freezing point depression is $\Delta T_f = K_f \times m$, where $m$ is molality. Rearranging for $m$: $m = \frac{\Delta T_f}{K_f}$. Substituting values: $m = \frac{18.98^{\circ}\text{C}}{8.37^{\circ}\text{C kg mol}^{-1}} \approx 2.2676 \text{ mol kg}^{-1}$

Step3: Find moles of unknown alcohol

Molality $m$ is moles of solute per kg of solvent. Mass of solvent (tert - butyl alcohol) is $4.71 \text{ g} = 0.00471 \text{ kg}$. Let moles of unknown alcohol be $n$. Then $m=\frac{n}{\text{mass of solvent (kg)}}$, so $n = m\times\text{mass of solvent (kg)} = 2.2676 \text{ mol kg}^{-1} \times 0.00471 \text{ kg} \approx 0.01068 \text{ mol}$

Step4: Calculate molar mass of unknown alcohol

Molar mass $M = \frac{\text{mass of solute}}{\text{moles of solute}}$. Mass of unknown alcohol is $0.500 \text{ g}$, moles is $0.01068 \text{ mol}$. So $M=\frac{0.500 \text{ g}}{0.01068 \text{ mol}} \approx 46.8 \text{ g mol}^{-1}$ (rounded to appropriate significant figures)

Answer:

$\approx 46.8 \text{ g mol}^{-1}$ (or more precise value depending on calculation steps, e.g., if we use more precise values during calculation: $m=\frac{18.98}{8.37}\approx2.2676$, $n = 2.2676\times0.00471\approx0.01068$, $M=\frac{0.5}{0.01068}\approx46.8$)