The freezing point (Tf) for t-butanol is 25.508C and Kf is 9.18C/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 10.0-g sample t-butanol is measured as 24.598C, how many grams of water are present in the sample?
[A] 10 g
[B] 1.8 g
[C] 0.10 g
[D] 0.018 g
[E] 18. g
The Correct Answer and Explanation is:
To solve this problem, we use freezing point depression, a colligative property described by the equation: ΔTf=Kf⋅m\Delta T_f = K_f \cdot m
Where:
- ΔTf\Delta T_f = freezing point depression (°C)
- KfK_f = freezing point depression constant (°C/m)
- mm = molality of solute (mol/kg of solvent)
Step 1: Determine ΔTf\Delta T_f
ΔTf=Tf,pure−Tf,solution=25.508∘C−24.598∘C=0.910∘C\Delta T_f = T_{f,\text{pure}} – T_{f,\text{solution}} = 25.508^\circ C – 24.598^\circ C = 0.910^\circ C
Step 2: Solve for molality (m)
m=ΔTfKf=0.9109.18=0.09913 mol/kgm = \frac{\Delta T_f}{K_f} = \frac{0.910}{9.18} = 0.09913\ \text{mol/kg}
Step 3: Calculate moles of water in the sample
We are told that the total mass of t-butanol is 10.0 g. Water is the impurity (solute), and t-butanol is the solvent.
Molality mm is defined as: m=moles of solute (water)kg of solvent (t-butanol)m = \frac{\text{moles of solute (water)}}{\text{kg of solvent (t-butanol)}}
So: moles of water=m×kg of t-butanol=0.09913×10.01000=9.913×10−4 mol\text{moles of water} = m \times \text{kg of t-butanol} = 0.09913 \times \frac{10.0}{1000} = 9.913 \times 10^{-4}\ \text{mol}
Step 4: Convert moles of water to grams
Molar mass of water = 18.015 g/mol mass of water=9.913×10−4×18.015=0.01784 g\text{mass of water} = 9.913 \times 10^{-4} \times 18.015 = 0.01784\ \text{g}
Final Answer:
0.018 g⇒Choice [D]\boxed{0.018\ \text{g}} \Rightarrow \text{Choice [D]}
Explanation (Summary)
This problem deals with freezing point depression, where the presence of solute particles (in this case, water) lowers the freezing point of a solvent (here, t-butanol). When exposed to air, t-butanol absorbs water, acting as a solute that disrupts the freezing process. The key to solving the problem is calculating how much water is needed to lower the freezing point by 0.910°C using the known constant Kf=9.18°C/mK_f = 9.18°C/m. After finding the molality, it’s used with the mass of the t-butanol to determine how many moles (and then grams) of water were present. The precise calculation yields approximately 0.018 g, confirming that Choice [D] is correct.