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Barium chloride is added to potassium chloride What will be colour of the final solution?
BaCl2- and KCl-solutions have no color, and also none when mixed.
What volume of 0.060 M KCl solution contains 2.39g of KCl?
MW KCl = 74.6 g/mol2.39 gKCl * (1 mol KCL/74.6 g KCl)*(1 L solution/0.06 mol KCL) = 0.534 L
How will you prepare 3 liters of 2.5 M solution of KCl solution in water?
Dissolve 566.25g of KCl in 3L.
How much KCL is required for 500mM KCL solution?
0.745*0.5 g
How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution?
The answer is 6,71 g dried KCl.
How many moles of KCl are contained in 0.635L of 2.2M KCl?
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
Can KCl and CCl4 form a solution?
KCl and CCl4 do they form solution
How many moles of KCl are contained in 1.7L of a 0.83M KCl solution?
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
Barium chloride is added to potassium chloride What will be colour of the final solution?
BaCl2- and KCl-solutions have no color, and also none when mixed.
What volume of 0.060 M KCl solution contains 2.39g of KCl?
MW KCl = 74.6 g/mol2.39 gKCl * (1 mol KCL/74.6 g KCl)*(1 L solution/0.06 mol KCL) = 0.534 L
How many moles of KCl are needed to prepare 1.00 L of a 1.0M KCl solution?
M= moles in solution/liters so plug in what you know 3.0M of KCl solution = moles in solution/ 2.0L multiply both sides by 2.0L moles solute = 1.5 moles KCl so you need 1.5 moles KCl to prepare the solution
How will you prepare 3 liters of 2.5 M solution of KCl solution in water?
Dissolve 566.25g of KCl in 3L.
How much 75.66MKCL is required toprepare 1L of solution?
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!
How many grams of KCl are required to prepare 2.25L of a 0.75m KCl 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
How much KCL is required for 500mM KCL solution?
0.745*0.5 g
What is the molarity of a solution made by dissolving 4.88 g of KCl in 423 mL of solution?
Need mole KCl first. 4.88 grams KCl (1 mole KCl/74.55 grams) = 0.06546 moles KCl =======================now, Molarity = moles of solute/Liters of solution ( 423 ml = 0.423 Liters ) Molarity = 0.06546 moles KCl/0.423 Liters = 0.155 M KCl ------------------
A solution was prepared by dissolving 40.0 g of KCl in 225 g of water. Calculate the mole fraction of KCl in the solution?
18%
