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 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]

Question

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 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) 10-8 m (B) 10-6 m (C) 10-4 m (D) 10-2 m

Tardigrade Admin · Accepted Answer

Mass of fatty acid = 0.027 g Radius of plate = 10 cm Density of fatty acid = 0.9 g/cm3 Volume of fatty acid = (0.027/0.9) = 0.003 cm3 Area of plate = π r2 = 3× 102 = 300 cm2 Height of fatty acid = (Volume/Area) = (0.03/300) = 10-4cm = 10-6 m

$1 0^{- 8} m$

$1 0^{- 6} m$

$1 0^{- 4} m$

$1 0^{- 2} m$

10−8m

10−6m

10−4m

10−2m

Mass of fatty acid =0.027g Radius of plate =10cm Density of fatty acid =0.9g/cm3 Volume of fatty acid = 0.90.027​=0.003cm3 Area of plate = πr2=3×102=300cm2 Height of fatty acid = AreaVolume​=3000.03​=10−4cm=10−6m

$1 0^{- 8} m$

$1 0^{- 6} m$

$1 0^{- 4} m$

$1 0^{- 2} m$