Dissolve 566.25g of KCl in 3L.
Take 37.25g of KCl exactly and dissolve in 500ml of water
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
Volumetric Flask
0.745*0.5 g
Dissolve 566.25g of KCl in 3L.
In a 3.4 M solution, there are 3.4 moles per liter. If you want to make 3 liters of solution, you'll need 3 liter * 3.4 moles/liter = 10.2 moles The molar mass of KCl is 39.098 g/mole K + 35.453 g/mole Cl = 74.551 g/mole KCl To get the number of grams, multiply the number of moles by the molar mass: 10.2 moles * 74.551 g/mole KCl = 760.4202 g = 0.760 kg
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
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.
For a 10 % solution you need 250 g KCl.
KCl and CCl4 do they form solution
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
Weigh 22.35 grams of KCl and Dissolve in 100 mL of Distilled Water
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.
Take 37.25g of KCl exactly and dissolve in 500ml of water
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
MW 74.5, 1M solution requires 74.5g in 1L, 5M solution requires 74.5x5 = 372.5g in 1 L. You would need 93.1g KCL to make 250mL 5M solution