Question

Answer this: A drop of olive oil of radius 0.25mm spreads into a circular film of radius 5cm on the water surface. Estimate the molecular size of he olive oil.

Answer 1

$\huge \mathfrak {Answer:-}$

Area of the film,

$A = \pi \: r ^{2} cm ^{2}$

$A = 25\pi \: cm^{2}$

$\bold{Since,}$

Radius of the film,

$r = 5 \: cm$

$\bold{Now,}$

Volume of the drop, V

$= \frac{4}{3} \pi$ (Radius of the oil drop)$^{3}$

$= \frac{4}{3} \times \pi \times (0.25mm)^{3}$

$= \frac{4}{3} \times \pi \times (2.5 \times 10 ^{ - 1} ) ^{3} mm ^{3}$

$= \frac{4}{3} \times \pi \times 15.6 \times 10^{ - 3} mm ^{3}$

$= \frac{4}{3} \times \pi \times 15.6 \times 10 ^{ - 6} cm^{3}$

Molecular size = Thickness of the film

$= \frac{V}{A}$

$= \frac{4\pi \times 15.6 \times 10 ^{ - 6} }{3 \times 25\pi}$

$= 0.83 \times 10 ^{ - 6} cm$

$= 8.3 \times 10 ^{ - 9} m$

#Be Brainly❤️

Answer 2

Hello user

The question is quite interesting but if you will focus in the major theme of it then it will came to your daily life which you experience.

Well the molecular size of the olive oil is equal the thickness which it do form on the film.

Let us suppose the diameter to be D

So , ATQ,

D= volume of the oil drop/ area of the film

= 4/3 *pi* (0.025cm)^3/(pi* (5cm)^2

= 2.08X10^-7 /4 cm

= 5.2 × 10^-8 cm
Hope it works