Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm³ of oleic acid per cm³ of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm² by considering 100 spherical drops of radius ( 3/40π )^(1/3) × 10⁻³ cm. Then the thickness of oleic acid layer will be x × 10⁻¹⁴ m. Where x is ____________.

Solution

Physics sir · 4 steps

Find the volume of pure oleic acid in the drops, then spread it over 4 cm² and read off the thickness.

1. Volume of one drop. With r³ = 3/40π × 10⁻⁹ cm³, the drop volume is 4/3π r³ = 4/3π × 3/40π × 10⁻⁹ = 10⁻¹⁰ cm³.
2. Oleic acid only. 100 drops hold 100 × 10⁻¹⁰ = 10⁻⁸ cm³ of solution. Only 0.01 of that is oleic acid, so the acid volume is V = 10⁻¹⁰ cm³.

   t = V/A = (10⁻¹⁰ cm³)/(4 cm²) = 2.5 × 10⁻¹¹ cm

3. Convert to metres. 1 cm = 10⁻² m, so t = 2.5 × 10⁻¹³ m = 25 × 10⁻¹⁴ m.
4. So. The thickness of the oleic acid layer is 25 × 10⁻¹⁴ m, so x = 25.

The common slip: 2500. Using the full solution volume and forgetting the 0.01 concentration gives a thickness 100 times too large, so x comes out as 2500, not 25.

The idea to keep. A film's thickness is just volume divided by area. Use only the volume of the substance that forms the film, not the whole solution, and convert units at the end.

Answer from the NTA final answer key. Solution by PrepWiser.