The question is a little ambiguous so I'll answer it the best way I can. Concentration (M) is the number of moles of a substance divided by the volume that it is dissolved in. This leads to the expression c = n/V Where n is the number of moles, V is the volume in litres and c is the concentration. In answer to the question, 500 ml of 0.25M solution can be prepared as follows: c = n/V Therefor: 0.25 = n/0.5L n = 0.125 mol 500ml of an 0.25M solution can be prepared by dissolving 0.125 mol of the substance in 500ml (providing it dissolves).
pipette out 8.5 ml perchloric acid in to 500 ml acetic acid and add 21 ml of acetic anhydride make up to volume to 1000 ml with acetic acid*. * Let this solution stand for one day and check the water content
The normality of 98 g of sulfuric acid in 500 mL of solution is 4 N
0.0125
combine 100 mL 6 M HCl with 500 mL H2O
It's 1.5 M
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
pipette out 8.5 ml perchloric acid in to 500 ml acetic acid and add 21 ml of acetic anhydride make up to volume to 1000 ml with acetic acid.Stand iday this solution and and check the water content not exceedsto 0.025 to 0.5% then standardize the solution by PHP
pipette out 8.5 ml perchloric acid in to 500 ml acetic acid and add 21 ml of acetic anhydride make up to volume to 1000 ml with acetic acid*. * Let this solution stand for one day and check the water content
222.223 ml @ 20% solution = 44.444 ml 277.777 ml @ 65% solution = 180.555 ml total = 225 ml out of 500 ml = 45%
The normality of 98 g of sulfuric acid in 500 mL of solution is 4 N
6 molar
sdfa
0.0125
combine 100 mL 6 M HCl with 500 mL H2O
The answer is 1,5 moles.
0.6m
It's 1.5 M