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[0.27g" of a long chain fatty acid was d...

[0.27g" of a long chain fatty acid was dissolved in "],[100cm^(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 edge "],[" to centre of the watch glass is "10cm." What is "],[" the height of the monolayer? "],[" [Density of fatty acid "=0.9gcm^(-3);pi=3]],[" [JEE Main "2019,8" April Shift-II] "],[[" (a) "10^(-6)m," (b) "10^(-4)m," (c) "10^(-8)m," (d) "10^(-2)m]]

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0.27 g of a long chain fatty acid was dissolved in 100cm^(3) of hexane. 10ml of this solution was added dropwise to the surface of water in a round watch glass . Hexane evaporates and a monolayeer is formed. The distance from edge to center of the watch glass is 10 cm . What is the height of the monolayers? ["Density of fatty acid "=0.9 g cm^(-3),pi=3]

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. The height of the monolayer is 10 ^( -x ) m. what is numerical value of x ? [Density of fatty acid = 0.9 g cm ^ ( -3) , pi = 3 ]

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. The height of the monolayer is 10 ^( -x ) m. what is numerical value of x ? [Density of fatty acid = 0.9 g cm ^ ( -3) , pi = 3 ]

0.27 g of fatty acid is dissolved in 100 ml of solvent, 10 ml such solution is taken & placed over round plate. Distance from the centre of edge of round plate is 10 cm. Now solvent is evaporated & only fatty acid is remained. Density of fatty acid is 0.9g//c . Determine height of fatty acid layer. (pi=3)

0.27 g of fatty acid is dissolved in 100 ml of solvent, 10 ml such solution is taken & placed over round plate. Distance from the centre of edge of round plate is 10 cm. Now solvent is evaporated & only fatty acid is remained. Density of fatty acid is 0.9g//c . Determine height of fatty acid layer. (pi=3)

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0.71 g of a sample of bleaching powder (CaOCl_2) is dissolved in 100 " mL of " water. 50 " mL of " this solution is titrated with KI solution. The I_2 so liberated required 10 mL 0.1 M Na_2S_2O_3 (hypo) solution in acidic medium for complete neutralisation. Calculate the percentage of available Cl_2 from the sample of bleaching power.

One gram of serum albumin dissolved in 1000cm^(3) of water gives a solution height of 3.9 mm of Ht of 25^(@)C . What is the molar mass of serum albumin? Density of the solution can be taken as 1.0 g cm^(-3) .