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 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 ]
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 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:
10.78 g of H_3 PO_4 in 550 ml solution is 0.40 N. Thus this acid:
If 1.34g Na_(2)C_(2)O_(4) is dissolved in 500 mL of water and this solution is titrated with M/10 KMnO_(4) solution in acidic medium, the volume of KMnO_(4) used is :
3.0 g of oxalic acid [(CO_2H)_2. 2H_2O] is dissolved in a solvent to prepare a 250 mL solution. The density of the solution is 1.9g/mL. The molality and normality of the solution, respectively, are cloest to
1.17 g an impure sample of oxalic acid dihydrate was dissolved and make up of 200 ml with water. 10 ml of this solution in acidic medium required 8.5 ml of a solution of potassium permanganate containing 3.16 g per litre of solution. The percentage purity of oxalic acid will be:
