One mole of NaOH is 40g. So 40g dissolved in 1 litre will give a 1M solution.
Molarity:1.0 mol (NaOH) /L(solution) = 1.0 M NaOHThe conversions of molality, b, to and from the molarity , c,for one-solute solutions are:c = ρ.b / [1 + b.M]andb = c / [ρ -c.M]where ρ is the mass density of the solution, b is the molality, and M is the molar mass of the solute.
You prepare a solution by dissolving a known mass of solute into a specific amount of solvent. In solutions, M is the molarity, or moles of solute per liter of solution. For 300 ml of a 0.1 M Na CL solution from a solid Na CL solution and water you need water and sodium chloride.
If you concentrate a solution, the molarity (moles/liter) will increase.
The molarity is 0,025.
The molarity is 0,041
Molarity = moles solute/Liters solution get moles NaOH 0.240 grams NaOH (1 mole NaOH/39.998 grams) = 0.0060 moles NaOH ----------------------------------as one to one OH- has this many moles also Molarity = 0.0060 moles OH-/0.225 Liters = 0.0267 M OH- ----------------------- -log(0.0267 M OH-) = 14 - 1.573 = 12.4 pH -------------
Molarity:1.0 mol (NaOH) /L(solution) = 1.0 M NaOHThe conversions of molality, b, to and from the molarity , c,for one-solute solutions are:c = ρ.b / [1 + b.M]andb = c / [ρ -c.M]where ρ is the mass density of the solution, b is the molality, and M is the molar mass of the solute.
Molarity = moles of solute / liters of solution. First, calculate the moles of NaOH using its molar mass. Then, divide the moles by the final volume in liters (3.00 L) to find the molarity.
The hydroxide ion concentration in a 4.0 x 10^-4 M solution of Ca(OH)2 can be calculated by first finding the molarity of OH- ions from Ca(OH)2, which is twice the molarity of the overall solution. Therefore, the [OH-] is 8.0 x 10^-4 M.
The molarity of Na+ ions in a 0.25 M solution of Na3PO4 is 0.75 M. This is because each formula unit of Na3PO4 dissociates into 3 Na+ ions in solution. Therefore, the concentration of Na+ ions is three times the molarity of the Na3PO4 solution.
6 kg = 6000 grams and density of water = 1.00 grams/milliliters. 1.00 g/ml = 6000 grams/X ml = 6000 ml which = 6 liters ======================== Molarity = moles of solute/Liters of solution Molarity = 2 moles NaOH/6 Liters = 0.3 M NaOH solution -----------------------------
To calculate the molarity, first convert the mass of sodium nitrate to moles using its molar mass. Then, divide the number of moles by the volume of solution in liters (265 mL = 0.265 L) to get the molarity. The molarity of the solution is about 0.68 M.
The relationship between weight and molarity in a solution is that weight is directly proportional to molarity. This means that as the molarity of a solution increases, the weight of the solute in the solution also increases. Conversely, as the molarity decreases, the weight of the solute in the solution decreases.
If you raise a solution temperature the molarity will decrease.
To find the molarity of a solution, divide the number of moles of solute by the volume of the solution in liters. Molarity (M) moles of solute / liters of solution.
To calculate the molarity of a solution, you divide the number of moles of solute by the volume of the solution in liters. The formula is: Molarity (M) moles of solute / liters of solution.
To calculate the molarity of a solution, you divide the number of moles of solute by the volume of the solution in liters. The formula is: Molarity (M) moles of solute / liters of solution.