Question

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

Answer 1

Here is ur answer:

Area of film,

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

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

[∴Radius of the film, r=5]

Volume of the drop,

$v = \frac{4}{3} \pi {(radius \: of \: oil \: drop)}^{3}$

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

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

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

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

Molecular size= Thickness of the film

$= \frac{v}{a}$

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

=0.83× 10^-6 cm

=8.3 × 10^-9m ✔

Hope this will help you ❤

Answer 2

Explanation:

Given area of film = 0.25mm

A = πr²cm²

A = 25πcm²

radius of film (r) = 5cm

Volume = 4/3π(radius)³

V = 4/3 × π × (0.25mm)³

V = 4/3 × π×(15.6 × 10^-1)

V = 4/3 × π × 15.6 × 10^-3

V = 4/3 × π × 15.6 × 10^-6 cm³

Molecular size = 0.83 × 10^-6 cm

= 8.3 × 10^-9 cm

= 83 × 10^-10m