The answer is 20,15 mL.
This depends on your experiment.
Use V1*c1 = V2*c2 with similar units on each side.V1* 2.5(mol/L) = 100.0(mL) * 0.50(mol/L)So V1 = 100*0.5/2.5 = 20 mL
Take 333 milliliters of your stock solution and dilute it to 1L with water.
0.26
0.01 molar
This depends on your experiment.
Use V1*c1 = V2*c2 with similar units on each side.V1* 2.5(mol/L) = 100.0(mL) * 0.50(mol/L)So V1 = 100*0.5/2.5 = 20 mL
Take 333 milliliters of your stock solution and dilute it to 1L with water.
To prepare a 0.1 N glacial acetic acid solution, calculate the required mass by multiplying 0.1 moles by the molar mass of glacial acetic acid (60.05 g/mol). Weigh out the calculated mass and add it to a clean container. Dissolve the glacial acetic acid completely by stirring it with distilled water. Transfer the solution to a 1-liter volumetric flask and dilute it to the 1-liter mark with distilled water. Mix thoroughly, label, and store the solution properly, taking necessary safety precautions when handling glacial acetic acid.
0.26
0.01 molar
Iodine monochloride is soluble in alcohol, ether, acetic acid. ICl is prepared from iodine and chlorine.
Dissolve 16.235 g of ICl into 1000 ml solution with glacial acetic acid. The solution will be 0.1N Wij's Solution. However weighing is a very difficult task you need to have the required skill and sound analytical knowledge and detailed MSDS of ICl. Man Katuwal
0.13 is the concentration of the acetic acid solution.
add 2.4ml glacial aetic acid in one liter of d.m water
.26
Dehydration of acetic acid to prepare acetic anhydride occur at 8oo oC.