The freezing point for t-butanol is 25.50 degrees C and kf is 9.1 degrees C/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 10.0 g sample of t-butanol is measured as 24.59 degrees C, how many grams of water are present in the sample?
The Correct Answer and Explanation is:
To determine how many grams of water are present in the t-butanol sample, we need to apply the freezing point depression formula. Freezing point depression occurs when a solute (in this case, water) is dissolved in a solvent (t-butanol). The formula for freezing point depression is: ΔTf=Kf×m\Delta T_f = K_f \times m
Where:
- ΔTf\Delta T_f is the freezing point depression,
- KfK_f is the cryoscopic constant of the solvent (in this case, t-butanol),
- mm is the molality of the solution.
Step 1: Calculate the freezing point depression (ΔTf\Delta T_f)
The freezing point of pure t-butanol is 25.50°C, and the measured freezing point of the sample with water is 24.59°C. Therefore, the freezing point depression is: ΔTf=25.50∘C−24.59∘C=0.91∘C\Delta T_f = 25.50^\circ C – 24.59^\circ C = 0.91^\circ C
Step 2: Use the freezing point depression formula
Now that we know the freezing point depression (ΔTf=0.91∘C\Delta T_f = 0.91^\circ C) and the cryoscopic constant (Kf=9.1∘C/mK_f = 9.1^\circ C/m), we can use the formula to solve for the molality (mm) of the solution: m=ΔTfKf=0.91∘C9.1∘C/m=0.1 mol/kgm = \frac{\Delta T_f}{K_f} = \frac{0.91^\circ C}{9.1^\circ C/m} = 0.1 \, \text{mol/kg}
Step 3: Determine the amount of water in the sample
The molality (mm) is defined as the number of moles of solute (water) per kilogram of solvent (t-butanol). Since we know the sample of t-butanol weighs 10.0 g, or 0.0100 kg, we can calculate the moles of water: m=moles of watermass of t-butanol in kgm = \frac{\text{moles of water}}{\text{mass of t-butanol in kg}}
Rearranging for moles of water: moles of water=m×mass of t-butanol in kg=0.1 mol/kg×0.0100 kg=0.001 mol\text{moles of water} = m \times \text{mass of t-butanol in kg} = 0.1 \, \text{mol/kg} \times 0.0100 \, \text{kg} = 0.001 \, \text{mol}
Now, we can convert moles of water to grams. The molar mass of water is 18.015 g/mol: grams of water=moles of water×molar mass of water=0.001 mol×18.015 g/mol=0.018 g\text{grams of water} = \text{moles of water} \times \text{molar mass of water} = 0.001 \, \text{mol} \times 18.015 \, \text{g/mol} = 0.018 \, \text{g}
Final Answer:
The sample of t-butanol contains 0.018 grams of water.
Explanation:
Freezing point depression is a colligative property, meaning it depends on the number of solute particles, not their identity. When water dissolves in t-butanol, it lowers the freezing point, which is the principle behind this calculation. By using the freezing point depression formula and knowing the mass of t-butanol and the cryoscopic constant, we can calculate the amount of water present based on the observed change in freezing point.