1.25 moles of Lithium Chloride
This is from the website linked to the left of this answer under Web Links:Solubility in water, g/100 ml at 20°C: 74.5Therefore, in 1 liter, 745.0 grams of CaCl2 will dissolve to make a saturated solution.
To determine the molarity of a potassium chloride solution, you need to know the moles of potassium chloride dissolved in a liter of solution (mol/L). It can be calculated by dividing the number of moles of potassium chloride by the volume of the solution in liters.
Molarity is calculated as moles of solute divided by volume of solution in liters. In this case, you have 2 moles of sodium chloride in a 0.5 liter solution. So the molarity would be 2 moles / 0.5 L = 4 M.
To calculate the total amount of sodium chloride needed for a 13 L solution at 4 grams per liter, multiply the concentration by the volume of the solution: 4 grams/L x 13 L = 52 grams of sodium chloride. Therefore, you will need 52 grams of sodium chloride to make the 13 L solution.
To prepare 0.1N mercuric chloride solution, you would dissolve 2.72 grams of mercuric chloride (HgCl2) in 1 liter of water. 0.1N means the solution contains 0.1 moles of mercuric chloride in 1 liter of solution. Be cautious when working with mercuric chloride as it is toxic and should be handled with proper safety precautions.
This is from the website linked to the left of this answer under Web Links:Solubility in water, g/100 ml at 20°C: 74.5Therefore, in 1 liter, 745.0 grams of CaCl2 will dissolve to make a saturated solution.
it would be the solute
There would be 0.1 moles of NaCl present in 1 liter of a 0.1M solution of sodium chloride. This is based on the definition of molarity which is moles of solute per liter of solution.
To determine the molarity of a potassium chloride solution, you need to know the moles of potassium chloride dissolved in a liter of solution (mol/L). It can be calculated by dividing the number of moles of potassium chloride by the volume of the solution in liters.
Molarity is calculated as moles of solute divided by volume of solution in liters. In this case, you have 2 moles of sodium chloride in a 0.5 liter solution. So the molarity would be 2 moles / 0.5 L = 4 M.
To make normal saline, you would need to add 9 grams of sodium chloride to one liter of water. This is equivalent to approximately 0.9% saline solution.
In chemistry, the concentration of a substance in solution is determined by molarity, which is symbolized by "M". This indicates the number of moles of a substance dissolved in one liter of a solvent (usually water). For example: - 1 mole of sodium chloride = 58 grams - If 116 grams of sodium chloride are dissolved in 1 liter of water, then that solution is a 2-molar (2 M) solution of sodium chloride. - If 232 grams of sodium chloride are dissolved in 1 liter of water, then that solution is a 4-molar (4 M) solution of sodium chloride.
In a 1M solution of sodium chloride, there would be 1 mole of sodium ions and 1 mole of chloride ions in 1 liter of the solution. This is because each formula unit of sodium chloride dissociates into one sodium ion and one chloride ion in solution.
To calculate the total amount of sodium chloride needed for a 13 L solution at 4 grams per liter, multiply the concentration by the volume of the solution: 4 grams/L x 13 L = 52 grams of sodium chloride. Therefore, you will need 52 grams of sodium chloride to make the 13 L solution.
To prepare 0.1N mercuric chloride solution, you would dissolve 2.72 grams of mercuric chloride (HgCl2) in 1 liter of water. 0.1N means the solution contains 0.1 moles of mercuric chloride in 1 liter of solution. Be cautious when working with mercuric chloride as it is toxic and should be handled with proper safety precautions.
It depends on the volume, if we consider 1 liter of the solution 500 mg of sodium chloride is needed.
1 L of water weights 1000 grams: Suppose you need X grams of calcium chloride.X grams CaCl2 / [X + 1000] grams solution = 0.35 = (35%/100%) and than solve the XX = 0.35 * (X + 1000)= 0.35X + 350X - 0.35X = 3500.65X = 350X = 350 / 0.65 = 538.46 = 538 grams of calcium chlorideAdd 538 grams of calcium chloride to 1 Litre waterand you'll getabout 1.54 kg of the 35% CaCl2 solution (this is less than 1.54 Liter!!)