Solution :
Assume that the diameter of the watch glass is 9.5 cm.
There are 10 drops of hexagon in 1 mL.
V = volume of hexagon in per drop = [tex]$\frac{1}{10}$[/tex] mL per drop
V = 0.1 mL per drop
N = number of drops to form phospholipid layer
N = 39 (assumed)
[tex]$\bar V$[/tex] = volume of hexagon required to form phospholipid layer
[tex]$\bar V$[/tex] = V x N
[tex]$\bar V$[/tex] = 0.1 x 39
= 3.9 mL
Since 1 mL = 1 [tex]$cm^3$[/tex]
∴ [tex]$\bar V$[/tex] = 3.9 [tex]$cm^3$[/tex]
d = diameter of the watch glass = 9.5 cm
A = area of one molecule
[tex]$A=\frac{\bar V }{\pi d}$[/tex]
[tex]$A=\frac{3.9 }{3.14 \times 9.5}$[/tex]
[tex]$A=0.1307 \ cm^2$[/tex]
