0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and 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; 
= 3]
(a)
10−8
(b)
10−4
(c)
10−2
(d)
10−6
0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and 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; = 3]
(a)
10−8
(b)
10−4
(c)
10−2
(d)
10−6
Question ID - 52489 | SaraNextGen Top Answer
0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and 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; = 3]
(a)
10−8
(b)
10−4
(c)
10−2
(d)
10−6
|
0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and 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; |
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|
(a) |
10−8 |
(b) |
10−4 |
|
(c) |
10−2 |
(d) |
10−6 |
0.27 gm in 100 ml of hexane
∴ in 10 ml of aqueous solution only 0.027 gm acid is present
Volume of 0.027g acid =
ml
∴ 
r2h = 
(given r = 10 cm, 
= 3)
∴ h = 10−4cm
= 10−6m