Q.24 The molarity of water in a water : ethanol mixture (15 : 85, v/v) is approximately
(A) 0.85 (B) 5.55
(C) 8.5 (D) 55.5
Step-by-Step Calculation
- Assume 100 mL of mixture for simplicity
- Water volume: 15 mL × density 1 g/mL = 15 g
- Moles of water: 15 / 18 = 0.833 mol (molar mass H₂O = 18 g/mol)
- Ethanol volume: 85 mL × density ≈0.79 g/mL = 67.15 g
- Moles of ethanol: 67.15 / 46 ≈ 1.46 mol (molar mass C₂H₅OH = 46 g/mol)
- Total volume: 100 mL = 0.1 L
- Water molarity: 0.833 / 0.1 = 8.33 M ≈ 8.5 M
Why Option (A) 0.85 is Incorrect
0.85 M implies treating the mixture volume as 1 L with only 15 g water, ignoring volume fractions. This underestimates moles per liter by a factor of 10, confusing volume percent with mass.
Why Option (B) 5.55 is Incorrect
5.55 M reflects pure water molarity scaled incorrectly (55.5 M for 1000 g/L water halved somehow). It ignores ethanol’s volume dominance, treating water as if in a lower-density or partial solvent.
Why Option (D) 55.5 is Incorrect
55.5 M is pure water’s molarity (1000/18). In this ethanol-rich (85%) mixture, water occupies only 15% volume, diluting it to about 1/6.5 of pure value.
Exam Tips for Similar Problems
- Always use 100 mL basis for v/v mixtures to simplify ratios
- Verify densities: water 1 g/mL, ethanol 0.789 g/mL at 25°C
- For precise work, note actual solution density <1 g/mL due to contraction, but approximation suffices here
