0.027 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)`

30 g of acetic acid is dissolved in 1 dm^3 of a solvent. The molality of the solution will be (Given, density of solvent = 1.25 g cm^-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 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]

2.52 g of oxalic acid dehydrate was dissolved in 100 ml of water, 10 mL of this solution was diluted to 500 mL. The normality of the final solution and the amount of oxalic acid (mg/mL) in the solution are respectively-

0.63 g of dibasic acid was dissolved in water. The volume of the solution was made 100 mL. 20 mL of this acid solution required 10 mL N/5 NaOH solution. What is the normality of acid.

6.3 g of oxalic acid dihydrate have been dissolved in water to obtain a 250 ml solution. How much volume of 0.1 N NaOH would be required to neutralize 10 mL of this solution ?

0.63 g of diabasic acid was dissolved in water. The volume of the solution was made 100 mL . 20 mL of this acid solution required 10 mL of N//5 NaOH solution. The molecular mass of acid is: