Question

drop of oil olive oil of radius 0.30 mm spread into rectangular of 13cm × 15 CM on the surface of water .estimate the molecular size of olive oil

Answer 1

here drop is a form of sphere so volume of a drop is
0.30mm = 0.03 cm
v = 4πr³/3
v = 4×22/7×(0.03)³/3
v = 88×27×10^-6/3×7
v = 88×9×10^-6/7
v = 113.14×10^-6
v = 11314×10^-8cm³

here volume of spread area is equal to the volume of drop
volume of drop = volume of spread area of rectangular shape
1134×10^-8 = l×b×h
1134×10^-8cm³ = 13cm×15cm×h
1134×10^-8cm³ = 195cm²×h
1134×10^-8 cm³/195cm² = h
5.83 × 10^-8cm = h

here molecular size is equal to height of one molecul

so molecular size of olive oil is 5.83×10^-8 cm

here value is approximate