0.012mol KCl x (1L/0.25mol KCl) x (1000ml/1L) = 48 mL KCl
I'm guessing you meant KCl or potassium chloride.
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
No. Potassium chloride (KCl) is soluble in water.
KCl and CCl4 do they form solution
moles KCl = ( M solution ) ( V solution in L )moles KCl = ( 2.2 mol KCl / L solution ) ( 0.635 L of solution )moles KCl = 1.397 moles KCl
0.012mol KCl x (1L/0.25mol KCl) x (1000ml/1L) = 48 mL KCl
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
I'm guessing you meant KCl or potassium chloride.
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
KCl is soluble in DMF
KCl is a compound not an element.
I did not know that you could get a concentration of 75.66 M KCl, but; Molarity = moles of solute/Liters of solution 75.66 M KCl = moles KCl/1 liter = 75.66 moles of KCl 75.66 moles KCl (74.55 grams/1 mole KCl) = 5640 grams KCl that is about 13 pounds of KCl in 1 liter of solution. This is why I think there is something really wrong with this problem!
No. Potassium chloride (KCl) is soluble in water.
KCl is highly soluble in water and some other solvents.
KCl and CCl4 do they form solution
The answer is of course 0,9 M.