KCl has a molar mass of 39+35.5=74.5g/mol, and 48/74.5=0.64mol.
The answer is 0,644 moles.
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
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
The answer is of course 0,9 M.
12.57 grams KCl (1mol KCl/74.55g ) = 0.1686 moles
Molarity = moles of solute/Liters of solution ( 50.0 ml = 0.05 Liters ) 0.552 M KCl = moles/.0.05 liters = 0.0276 moles of 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
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
Molarity is moles of solute per L of solution.moles KCl = ( 1.68 M ) ( 0.121 L ) = ( 1.68 mol/L ) ( 0.121 L )moles KCl = 0.203 moles KCl
The answer is of course 0,9 M.
12.57 grams KCl (1mol KCl/74.55g ) = 0.1686 moles
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
Molarity = moles of solute/Liters of solution ( 50.0 ml = 0.05 Liters ) 0.552 M KCl = moles/.0.05 liters = 0.0276 moles of KCl
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
2 KClO3 ------ 2KCl + 3O2 so 2 moles of KClO3 produces two mole of KCl. Therefore 0.440 moles of potassium chlorate will produce 0.44 moles of KCl - potassium chloride.
279.56
2.3M means 2.3 moles per litre. Number of moles = 2.3mol/litre x 0.630litre = 1.449mols
0.012mol KCl x (1L/0.25mol KCl) x (1000ml/1L) = 48 mL KCl