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Dissolve 0.01 mole of KCl in upto 1 Liter.
0.01 mole = 0.01 (mol) * MKCl (g/mol KCl) = (0.01*MKCl) gram KCl per Liter
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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
Can KCl and CCl4 form a solution?
KCl and CCl4 do they form solution
How much kcl require for prepaire 3M kcl?
Molar mass of KCl = 74.55g/mol.ie, if you dissolve 74.55g KCl in 1litre (1000 ml) of water, it will be 1M KCl solution.If you want to make 3M KCl solution,Dissolve 3 ×74.55 = 223.65g KCl in 1litre (1000 ml) of water.If you want to make different molar solutions of KCl, just calculate as per below given equation.Weight of KCl to be weighed =Molarity of the solution needed × Molecular weight of KCl (ie, 74.55) × Volume of solution needed in ml / 1000.To prepare 3M KCl in 1 litre, it can be calculated as follows,3 mol × 74.55 g/mol × 1000 ml / 1000 ml = 223.65gByPraveen P Thalichalam, Kasaragod (Dist), Kerala.
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%
How do you prepare 3.5M KCl solution?
3.5M means 3.5 moles of KCl. 1 mole is the combined molecular weight of the compound per litre. Molecular weight of K (potassium) = 39.10g Molecular weight of Cl (chlorine) = 35.45g So molecular weight of KCl = (39.10 + 35.45) = 74.55g That means that 1 mole of KCL = 74.55 grams per litre If 1 mole of KCL contains 74.55g then 3.5M of KCL will contain 74.55g x 3.5 and so 3.5M of KCL = 260.925g/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 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 you prepare 3 M KCl solution?
Weigh 22.35 grams of KCl and Dissolve in 100 mL of Distilled Water
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 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 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
What mass of KCl is required to prepare 383.00 mL of 0.500 M solution?
I'll go straight to the equation, to prepare 0.5 M of KCl in 383 ml, simply calculate it this way, *Molar mass of KCl is 74.5513 g/mol *0.5 M is also known as 0.5 mol/Litre 74.5513 g/mol x 0.5 mol/litre x 0.383 Litre = 14.2766 g So, you need 14.2766 g of KCl to prepare 0.5 M in 383 ml solution.
How will prepare the 0.1N of KCL solution?
0.1 N KCl is the same as 0.1 M KCl. This requires one to dissolve 0.1 moles per each liter of solution. The molar mass of KCl is 74.6 g/mol. So 0.1 moles = 7.46 gDissolve 7.46 g KCl in enough water to make 1 liter (1000 ml)Dissolve 3.73 g KCl in enough water to make 0.5 liter (500 ml)Dissolve 0.746 g KCl in enough water to make 0.1 liter (100 ml)etc., etc.
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 do you prepare 0.1n kcl from 0.3n kcl?
use this V1*M1=V2*M2
