dissolve KOH powder in ethanol
ethanolic KOH can precipitate the impurities in solution
By addition of 10% ethanolic KOH, water and HCl
I assume you mean ethanol + KOH? if so, in order to prevent co-precipitation of piperine and the resin acids, dilute ethanolic KOH is added to the concentrated extract to keep acidic materials in solution as their potassium salts (acid/base rxn occurs). basically it keeps impurities in solution, while piperine precipitates out.
I assume you mean 7.25 M KOH, but the process is the same anyway.Molarity = moles of solute/Liters of solution ( 400 ml = 0.4 Liters )7.25 M KOH = X moles KOH/0.4 Liters= 2.9 moles KOH (56.108 grams/1 mole KOH)= 163 grams potassium hydroxide needed===============================
K+ and OH- is KOH -----------
ethanolic KOH can precipitate the impurities in solution
By addition of 10% ethanolic KOH, water and HCl
I assume you mean ethanol + KOH? if so, in order to prevent co-precipitation of piperine and the resin acids, dilute ethanolic KOH is added to the concentrated extract to keep acidic materials in solution as their potassium salts (acid/base rxn occurs). basically it keeps impurities in solution, while piperine precipitates out.
Ethanolic extraction will give 4hydroxyisoleucine.......
A perfume is a homogeneous mixture, generally an ethanolic solution.
I assume you mean 7.25 M KOH, but the process is the same anyway.Molarity = moles of solute/Liters of solution ( 400 ml = 0.4 Liters )7.25 M KOH = X moles KOH/0.4 Liters= 2.9 moles KOH (56.108 grams/1 mole KOH)= 163 grams potassium hydroxide needed===============================
1% solution of KOH contains 1g of KOH in 100g of solution. This means that you need to mix 1g of KOH and 99g of water.
K+ and OH- is KOH -----------
When KOH reacts with HCl, these products are formed. This is a neutralization reaction. KOH is a base while HCl is an acid.
0.800 M KOH (1mol/2.5L)(56.1g KOH/1mol)It's set up stoichiometrically, but hard to show that here... 0.800 M KOH / 2.5 L x 56.1gYour answer should come out 17.952 g KOH. If you're following sig figs, then your answer should come out 18. g KOH
Molar mass of KOH = 39.1+16.0+1.0 = 56.1Amount of KOH = mass of sample / molar mass = 32/56.1 = 0.570mol
An 11 M KOH solution indicates there are 11 moles of KOH per liter of water. 1 mole of KOH has a volume of 27.4 mL, so to account for the added volume it is necessary to add 15.75 moles of KOH per liter of water, or 884 grams per liter.