30 grams
40 grams, this is the 1M NaOH standard laboratory solution.
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
In this instance, 50 mol of sodium chloride is needed and molar mass of NaCl is 58.5 g/mol. Hence the mass we need is 29250 g. But this amount of salt could not be dissolved in 500 ml of water, so we cannot prepare this solution practically.
3.8 grams of EDTA salt in 1 liter of DI water made up using a volumetric flask will give you 0.02n or 0.01m of EDTA solution. normality*eq.wt*volume rqrd weight= 1000 then will get weight of the compound required for that normality
The answer is 6,71 g dried KCl.
40 grams, this is the 1M NaOH standard laboratory solution.
See the two Related Questions to the left for the answer.The first is how to prepare a solution starting with a solid substance (and dissolving it). The second question is how to prepare a solution by diluting another solution.
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
4314.9 grams
8.9g
It depends on the final solution Volume you want to prepare. For 100ml of a 6M NaCL solution, you add 35.1g of NaCl to water until you reach 100ml. Dissolve and autoclave for 15 mins.
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
14.575 round is 14.6 g
1% solution = 1 gram per 100 mL, 10 grams per liter 20 grams
1.17 grams :)
1 percent = 10 grams 2 % = 20 grams x 3 liters = 60 grams
3.33