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 0.05 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 1000 ppm (parts per million) solution of KMnO4 (potassium permanganate), you need 1000 mg of KMnO4 per liter of solution. Since 1 gram equals 1000 mg, you would need 1 gram of KMnO4 dissolved in enough water to make a final volume of 1 liter. Therefore, to prepare a 1000 ppm solution, dissolve 1 gram of KMnO4 in 1 liter of water.
4314.9 grams
To prepare a 3% solution of sulfosalicylic acid, you would need 30 grams of sulfosalicylic acid for every 1 liter 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 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 a 5% w/w concentration, you would need to calculate the mass of the potassium acetate in the solution. Since the concentration is given as a percentage by weight, 5% of 160 grams is 8 grams of potassium acetate. The remaining mass in the solution would be water. Therefore, you would need 152 grams of water to prepare 160 grams of potassium acetate with a 5% w/w concentration.
You need 116,88 g dried and pure sodium chloride.
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 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.