Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm 3 of oleic acid per cm 3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm 2 by considering 100 spherical drops of radius ((3/40 π))(1/3) × 10-3 cm . Then the thickness of oleic acid layer will be x × 10-14 m. Where x is .

Question

Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm 3 of oleic acid per cm 3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm 2 by considering 100 spherical drops of radius ((3/40 π))(1/3) × 10-3 cm . Then the thickness of oleic acid layer will be x × 10-14 m. Where x is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

4 t T =100 × (4/3) π r 3 =100 × (4 π/3) × (3/40 π) × 10-9=10-8 cm 3 t T =25 × 10-10 cm =25 × 10-12 m t 0=0.01 t T =25 × 10-14 m =25

4tT​=100×34​πr3 =100×34π​×40π3​×10−9=10−8cm3 tT​=25×10−10cm =25×10−12m t0​=0.01tT​=25×10−14m =25

$4 t_{T} = 100 \times \frac{4}{3} π r^{3}$
$= 100 \times \frac{4 π}{3} \times \frac{3}{40 π} \times 1 0^{- 9} = 1 0^{- 8} c m^{3}$
$t_{T} = 25 \times 1 0^{- 10} c m$
$= 25 \times 1 0^{- 12} m$
$t_{0} = 0.01 t_{T} = 25 \times 1 0^{- 14} m$
$= 25$