0
I'll straightly go to the equation. KCL molar mass is 74.5513 g/mol.
To prepare 0.02 M KCL solution,
*0.02 M is also known as 0.02 mol/litre.
74.5513 g/mol x 0.02 mol/litre x 1 litre = 1.4910 g
*note that in the equation we can cancel the mol and litre leave out the gram alone. So we need only 1.4910 g of KCL in 1 litre. You can change the volume, depend on your convenience but must in litre unit.
What else can I help you with?
How you prepare 1 M KCl solution?
To prepare a 1 M (molar) KCl solution, you would dissolve 74.55 grams of KCl (potassium chloride) in water and then dilute to a final volume of 1 liter. This solution will have a concentration of 1 mole of KCl per liter of water.
How many grams are KCl are required to prepare 250 ml of a 4M solution?
By definition, 1 liter of a 4 M solution must contain 4 moles of the solute. Because solutions are homogeneous mixtures and 250 ml is one quarter of a liter, the amount of KCl required is 1 mole. The gram molar mass* of KCl is 74.55 grams; this is therefore the amount of KCl required.** __________________________________ *Because KCl is an ionic rather than a molecular compound, its "gram molar mass" should preferably be called its "gram formula unit mass". **Strictly according to the rules of significant digits, this answer should be written as "7 X 10" grams, because the datum "4M" needed to calculate the answer contains only one significant digit.
What is specific conductance of 0.02 molar KCl solution?
The specific conductance of a 0.02 M KCl solution would depend on the concentration of KCl and the temperature. Specific conductance is measured in Siemens per meter (S/m) and is directly proportional to the concentration of the electrolyte solution.
How many grams of KCl are required to prepare 2.25L of a 0.75m KCl solution?
To calculate the grams of KCl needed, first determine the molar mass of KCl (74.55 g/mol). Then use the formula: grams = molarity x volume (L) x molar mass. Plugging in the values, you get: grams = 0.75 mol/L x 2.25 L x 74.55 g/mol = 126.60 grams of KCl.
How many grams of KCl would you need to make 250 ml of a 0.5-M Solution w work?
Molar mass of KCl = 39 g/mol (K) + 35.5 g/mol (Cl) = 74.5 g/mol. A 0.5 M solution is required (0.5 mol/L or 0.5 moles per litre). 0.5 moles of KCl is 0.5 mol x 74.5 g/mol = 37.25 g. Dissolving this 37.25 g of KCl in a litre of water would give a 0.5 M solution. If 1 L or 1000 mL of 0.5 M solution contains 0.5 moles then 1 mL of the same concentration solution would contain 0.5/1000 moles and 250 mL would contain 250 x 0.5/1000 moles = 0.125 moles. 0.125 moles of KCl is 0.125 mol x 74.5 g/mol = 9.31 g.
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 is needed to make 2 molar solution in 1 liter of water?
To prepare a 2 M solution of KCl in 1 liter of water, you would need to dissolve 149.5 grams of KCl. This is because the molar mass of KCl is approximately 74.5 g/mol, and 2 moles of KCl are needed to prepare a 2 M solution in 1 liter of water.
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 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 grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution?
To calculate the amount of KCl needed, we first need to find the number of moles of KCl required using the formula: moles = Molarity x Volume (in L). Then, we convert moles to grams using the molar mass of KCl, which is 74.55 g/mol. Finally, we use the formula: grams = moles x molar mass to find that approximately 6.33 grams of KCl are needed to prepare 125 mL of a 0.720 M solution.
How will you prepare 3 liters of 2.5 M solution of KCl solution in water?
To prepare 3 liters of a 2.5 M solution of KCl, you would need to dissolve 187.5 grams of KCl (3 liters * 2.5 moles/liter * 74.55 g/mole) in enough water to reach a final volume of 3 liters. First, measure out 187.5 grams of KCl, add it to a suitable container, and then add water gradually while stirring until the final volume reaches 3 liters. Finally, mix the solution thoroughly to ensure the KCl is completely dissolved.
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
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
Which mass of kcl would you add to water to make a 2.00 l of a 1.20 m solution?
To prepare a 1.20 M potassium chloride (KCl) solution in 2.00 L of water, you would need to add 144 grams of KCl. This calculation can be determined using the formula: moles = Molarity x volume (in liters). Then, convert moles to grams using the molar mass of KCl.
How you prepare 1 M KCl solution?
To prepare a 1 M (molar) KCl solution, you would dissolve 74.55 grams of KCl (potassium chloride) in water and then dilute to a final volume of 1 liter. This solution will have a concentration of 1 mole of KCl per liter of water.
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 ------------------
How many milliliter a of 0.500M KCl solution would you need to dilute to make 100.0 mL of 0.100M KCl?
To find the volume of the 0.500 M KCl solution needed to prepare 100.0 mL of 0.100 M KCl, you can use the dilution equation (C_1V_1 = C_2V_2). Here, (C_1 = 0.500 , \text{M}), (C_2 = 0.100 , \text{M}), and (V_2 = 100.0 , \text{mL}). Rearranging the equation to solve for (V_1) gives: [ V_1 = \frac{C_2V_2}{C_1} = \frac{0.100 , \text{M} \times 100.0 , \text{mL}}{0.500 , \text{M}} = 20.0 , \text{mL}. ] Thus, you would need 20.0 mL of the 0.500 M KCl solution.
