JEE Main20198 Apr 2019Evening ShiftChemistrySome Basic Concepts of ChemistryActual
0.27 g of a long chain fatty acid was dissolved in 100 c m 3 of hexane. 10 m L of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 c m . What is the height of the monolayer? [Density of fatty acid = 0.9 g c m - 3 ; π = 3 ]
Options
- A10 - 4 m
- B10 - 6 m
- C10 - 8 m
- D10 - 2 m
Correct answer
B. 10 - 6 m
Step-by-step solution
Mass of fatty acid = 0.027   g in 10   m l solution Density of fatty acid 0.9   g/cc Volume of fatty acid = 0.027 0.9 = 0.03   c c Area of plate = π r 2 = 3 × 10 2 = 300    c m 2 Height of fatty acid layer = v o l u m e a r e a = 0.03 300 = 10 - 4   c m = 10 - 6 m