4 gram molecular weights (moles): However many grams four moles* of the solute is.
* Hint: four moles of sodium chloride weighs less than four moles of sucrose.
You need 50 g of this drug.
To prepare a 2% solution in 3 liters, you would need 60 grams of the drug. This is calculated by multiplying the volume (3 liters) by the percentage (2%) and converting the result to grams. 3 liters x 2% = 60 grams.
To prepare 0.5L of D5W (5% dextrose in water), you need 25 grams of dextrose. This is because 5% of 0.5L is 25 grams.
To prepare a 0.25 M solution of ammonium sulfate (NH₄)₂SO₄, you first need to determine the molar mass of ammonium sulfate, which is approximately 132.14 g/mol. For a 0.25 M solution, you need 0.25 moles per liter. Therefore, to make 1 liter of a 0.25 M solution, you would need 0.25 moles × 132.14 g/mol = 33.04 grams of ammonium sulfate. If you want to prepare a solution at a concentration of 6 M instead, you would need a significantly higher amount, specifically 6 moles × 132.14 g/mol = 792.84 grams for 1 liter.
To prepare a 0.05 M disodium EDTA solution, you would need to dissolve 3.72 grams of disodium EDTA dihydrate (Na2C10H14N2Na2·2H2O) in enough water to make 1 liter of solution.
To prepare a 3% solution of sulfosalicylic acid, you would need 30 grams of sulfosalicylic acid for every 1 liter of solution.
4314.9 grams
To prepare a 0.01N solution of sodium metabisulfite, you would need 2.31 grams of sodium metabisulfite per liter of solution.
To prepare a 5% NaCl solution, you will need 200 grams of NaCl for 4000 mL (4 L) of solution. This is calculated as 5% of 4000 mL, which equals 200 grams.
You need 50 g of this drug.
To prepare a 2% solution in 3 liters, you would need 60 grams of the drug. This is calculated by multiplying the volume (3 liters) by the percentage (2%) and converting the result to grams. 3 liters x 2% = 60 grams.
To prepare a 0.20 M solution of I2 in CCl4, you would need to calculate the moles of I2 required first. Molarity (M) = moles of solute / liters of solution. Since you know the molarity and volume of the solution, you can calculate the moles of I2 required and then convert that to grams using the molar mass of I2.
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
To prepare 160 grams of potassium acetate with 5% w/w concentration, you will need 8 grams of potassium acetate (160 grams x 5% = 8 grams). Since potassium acetate is the solute in this case, you only need to consider the weight of the solute. The remaining weight will come from the water.
You need 116,88 g dried and pure sodium chloride.
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.
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.