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
Put 100 grams in a beaker and and around 500 mls of water until it dissolves, then top up the beaker to a liter. That is your 10% solution. The percentage solution is a ratio of the weight of the compound to the weight of the final solution.
Take 5 grams of calcium chloride and dissolve it in 100ml of solution to get a 5% solution of calcium chloride. The standard way to make a weight-volume solution is to take grams of the dry substance in 100ml of volume.
To make a 5% (w/v) solution, you would dissolve 25 grams of sodium chloride (5% of 500 grams) in water to make a final volume of 500 mL. Sodium chloride has a molecular weight of 58.5 g/mol.
It depends on the final solution Volume you want to prepare. For 100ml of a 6M NaCL solution, you add 35.1g of NaCl to water until you reach 100ml. Dissolve and autoclave for 15 mins.
To make a 0.25 M solution of ammonium sulfate from a stock solution of 6 M, you would need to dilute the stock solution. The dilution equation is C1V1 = C2V2 where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the concentration and volume of the final solution. You would need to set up this equation to calculate the volume of the stock solution needed and then convert that volume to grams using the molar mass of ammonium sulfate.
To make a 3.7% EDTA solution, you would add 3.7 grams of EDTA to 100 mL of solution.
To prepare a 0.01N solution of sodium metabisulfite, you would need 2.31 grams of sodium metabisulfite per liter of solution.
To make a 10% NaOH solution, you would need 100 grams of NaOH per liter of water. So to make 1 liter, you would need 100 grams of NaOH.
20% salt solution is the equivalent of adding 200gr salt in a 800 ml (1000ml -200ml) of water. you now have one liter of a 20% solution.
You need 841,536 g NaCl.
The Molecular Weight of NaCl = 58.5 So to make 1L of 4M NaCl solution you need 4*58.5=234g of NaCl So to make 100mL of the above solution you need 23.4 grams of NaCl
290 grams
26.8125 g
To make a 4M solution in 20 ml, you would need 0.32 grams of LiCl. This can be calculated using the formula: moles = molarity x volume (in L), then converting moles to grams using the molar mass of LiCl.
Put 100 grams in a beaker and and around 500 mls of water until it dissolves, then top up the beaker to a liter. That is your 10% solution. The percentage solution is a ratio of the weight of the compound to the weight of the final solution.
To make a 2 molar solution of hydrochloric acid, you would need to know the volume of the solution you want to make. Once you have the volume, you can use the molarity formula (M = moles of solute / liters of solution) to calculate the grams of hydrochloric acid needed.
Take 5 grams of calcium chloride and dissolve it in 100ml of solution to get a 5% solution of calcium chloride. The standard way to make a weight-volume solution is to take grams of the dry substance in 100ml of volume.