JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Units & Measurement

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

(\frac{3}{40 π})^{\frac{1}{3}} \times 1 0^{- 3} cm

. Then the thickness of oleic acid will be

x \times 1 0^{- 14} m

. Where

x

is ______

Answer: 25

Solution

Volume of one drop:

V_{d r o p} = \frac{4}{3} π r^{3} = \frac{4}{3} π [(\frac{3}{40 π})^{1/3} \times 1 0^{- 3}]^{3} = \frac{4}{3} π \times \frac{3}{40 π} \times 1 0^{- 9} = \frac{4}{40} \times 1 0^{- 9} = 1 0^{- 10}

cm³. Total volume of 100 drops:

V_{t o t a l} = 100 \times 1 0^{- 10} = 1 0^{- 8}

cm³. Volume of oleic acid in solution:

V_{o l e i c} = 0.01 \times V_{t o t a l} = 0.01 \times 1 0^{- 8} = 1 0^{- 10}

cm³. This oleic acid spreads over area

A = 4

cm². Thickness:

t = \frac{V _{o l e i c}}{A} = \frac{1 0 ^{- 10}}{4} = 2.5 \times 1 0^{- 11}

= 2.5 \times 1 0^{- 13}

= 25 \times 1 0^{- 14}

m. So

x = 25

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer