Freezing Point Depression
0.100 kg water contains 0.010 mol glucose (i=1). Kf(water)=1.86 K kg mol−¹. ΔTf is:
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About This Question
- Subject
- chemistry
- Chapter
- solutions
- Topic
- freezing point depression
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
0.186 K
- Step 1: Calculate Molality (m):
- m=mass of solvent in kgmoles of solute
- m=0.100 kg0.010 mol=0.10 mol/kg.
- Step 2: Use Colligative Equation:
- ΔTf=i×Kf×m
- For Glucose (non−electrolyte), i=1.
- Calculation:
- ΔTf=1×1.86 K kg mol−1×0.10 mol kg−1
- ΔTf=0.186 K.
- Note: The freezing point would be 0∘C−0.186K=−0.186∘C.
This medium difficulty chemistry question is from the chapter solutions, covering the topic of freezing point depression. It appeared in the 2025 exam.
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