Question:

The figures (I, II, III) given below schematically represent variation of surface tension of three different aqueous solutions with increasing concentration of each of the solutes (surfactant, sodium chloride, and n-propanol). Match the figures with appropriate solutes and choose the correct option. 

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Surfactants reduce surface tension, sodium chloride increases it, and alcohols like n-propanol decrease surface tension as concentration increases.
Updated On: Sep 8, 2025
  • I – surfactant, II – sodium chloride, III – n-propanol
  • I – sodium chloride, II – n-propanol, III – surfactant
  • I – surfactant, II – n-propanol, III – sodium chloride
  • I – n-propanol, II – sodium chloride, III – surfactant
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the behavior of solutes.
- Surfactants: Increase the surface area by reducing surface tension initially as concentration increases and then level off after a critical concentration (Critical Micelle Concentration).
- Sodium chloride: As concentration increases, surface tension typically increases due to the ionic nature and interaction with water molecules.
- n-Propanol: Surface tension decreases with increasing concentration as it reduces hydrogen bonding between water molecules.
Step 2: Matching the figures.
- Figure I: Corresponds to n-propanol because the surface tension decreases continuously with concentration.
- Figure II: Corresponds to sodium chloride because surface tension increases with increasing concentration.
- Figure III: Corresponds to surfactant because the surface tension decreases initially and then levels off.
Step 3: Conclusion.
The correct match is I – n-propanol, II – sodium chloride, III – surfactant.
Final Answer: \[ \boxed{\text{I – n-propanol, II – sodium chloride, III – surfactant}} \]
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