Question

$0.27\, g$ of a long chain fatty acid was dissolved in $100 \,cm^3$ of hexane. $10\, mL$ 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 $10\, cm$. What is the height of the monolayer?
[Density of fatty acid = $0.9\, g$ cm$^{-3}$, $\pi$ = 3]

• A. $10^{-8} \,m$
• B. $10^{-6}\,m$
• C. $10^{-4}\,m$
• D. $10^{-2}\,m$

Answer 1

Answer: (B)

Mass of fatty acid $= 0.027\, g$
Radius of plate $= 10 \,cm$
Density of fatty acid $= 0.9 g/cm^3$
Volume of fatty acid = $\frac{0.027}{0.9} = 0.003 cm^{3}$
Area of plate = $\pi r^{2} = 3\times 10^{2} = 300\,cm^{2}$
Height of fatty acid = $\frac{Volume}{Area} = \frac{0.03}{300} = 10^{-4}cm = 10^{-6} m$