JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Surface Chemistry

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0.27 g

of a long chain fatty acid was dissolved in

100 cm^{3}

of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer?
[Density of fatty acid

= 0.9 g cm^{- 3}, π = 3

]

(A) $1 0^{- 4} m$
(B) $1 0^{- 8} m$
(C) $1 0^{- 2} m$
(D) $1 0^{- 6} m$ ✓ Correct

Answer: (D)

Solution

First find how much fatty acid actually reaches the water. The

10 mL

taken from the

100 cm^{3}

solution carries

\frac{0.27 \times 10}{100} = 0.027 g

, whose volume is

\frac{0.027}{0.9} = 0.03 cm^{3}

. This spreads into a monolayer over the watch-glass area

π r^{2} = 3 \times (10)^{2} = 300 cm^{2}

. The film thickness is its volume divided by its area,

\frac{0.03}{300} = 1 0^{- 4} cm

, which is

1 0^{- 6} m

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Surface Chemistry

Solution

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