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.


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Answer 1

Area of the film, A=πr2cm2

=25πcm2 [∵ Radius of the film, r=5cm]

Volume of the drop, V=43π(Radius of oil drop)3

=43×π×(0.25mm)3

=43×π×(2.5×10−1)3mm3

=43×π×15.6×10−3mm3

=43×π×15.6×10−6cm3

Molecular size = Thickness of the film

=VA

=4π×15.6×10−63×25π

=0.83×10−6 cm

=8.3×10−9 m

Answer 2

Answer:

Here,r=0.25mm=0.25×10−3m,

R=10cm=10−1m

Thickness of oile film =volume of oil droparea of the film

43πr3πR2=43r3R2=43(0.25×10−3)3(10−1)2

=2.08×10−9m

If we assume the oil film to be one molecule

thick, then size of molecule of oleic acid

=2.08×10−9m