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Q.

0.54 g of a long chain fatty acid was dissolved in 200 $c m^{3}$ of hexane. 10ml. 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 to centre of the watch glass is 10cm. What is the height of the monolayer? (Density of fatty acid = 0.9 $c m^{- 3}$ ; $π$ = 3)

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a

$10^{- 8} c m$

b

$10^{- 4} c m$

c

$10^{- 4} m$

d

$10^{- 6}$ m

answer is A, C.

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Detailed Solution

In 200 ml gm of fatty acid = 0.54 gm
1ml gm ……. = $\frac{0.54}{200}$
10 ml gm …… = $\frac{0.54}{200}$ x 10 = 0.027
$d = \frac{m}{v}$
d x v = m
$0.9 (\frac{g m}{c m^{3}})$ x area x height = 0.027 gm
0.9 x (3) x (10)2 x h = 0.054
h = 10-4 cm
h = 10-6 m

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