Dissolve 100 g glucose in 1 L distilled water (or 10 g glucose in 100 mL disitlled water).
Yes, during process of osmoses the solvent from higher concentration to lower concentration moves through semipermeable membrane, the 2% solution has lower concentration of solute therefore higher concentration of solvent.
Dissolve 10 g pure glucose in 100 mL distilled water.
To find the molality of a solution, you need the mass of the solvent (usually water) in kilograms and the number of moles of solute (glucose). Given that the solution is 7.80% glucose by weight, you can calculate the mass of glucose in the solution and then convert it to moles using the molar mass of glucose. From there, you can find the molality by dividing the moles of glucose by the mass of the solvent in kilograms.
I don't think you can. The maximum solubility of glucose in water is 91% w/v. That would mean dissolving 91g of glucose in 100ml of water. for a 100% solution you would need to dissolve 100g in 100ml, and you cant do it under normal conditions.
No. Everything below 0.9% of NaCl is hypotonic and every solution with concentration over 0.9% is hypertonic solution. Isotonic solution (to blood) is the one that has 0.9% of NaCl, or some other concentration of another compound.
A 30% glucose solution is purely glucose and water, though it is actually impossible to keep other contaminants out of it. To create a 30% solution of glucose, you take a fixed volume of water and add 30% of that value of glucose to the water. The amount of glucose is in grammes. For example, 3g of glucose would be added to 10ml of water.
Yes, during process of osmoses the solvent from higher concentration to lower concentration moves through semipermeable membrane, the 2% solution has lower concentration of solute therefore higher concentration of solvent.
The balloon will contain a mixture of the 10% and 5% glucose solutions. Since water can pass through but not glucose, the glucose concentration inside the balloon will decrease over time as water moves from the lower concentration in the beaker to the higher concentration in the balloon through osmosis.
A 5 percent dextrose solution contains 5 grams of dextrose (glucose) per 100 milliliters of solution. Therefore, in one liter (1000 milliliters) of a 5 percent dextrose solution, there would be 50 grams of glucose.
Dissolve 10 g pure glucose in 100 mL distilled water.
No,5percent glucose is an isotonic solution. 0.9 percent is for NaCl.
To find the molality of a solution, you need the mass of the solvent (usually water) in kilograms and the number of moles of solute (glucose). Given that the solution is 7.80% glucose by weight, you can calculate the mass of glucose in the solution and then convert it to moles using the molar mass of glucose. From there, you can find the molality by dividing the moles of glucose by the mass of the solvent in kilograms.
The administration route for norcuron is intravenously only. It can be mixed with lactated ringers, 0.9 percent NaCl solution, 5 percent glucose in saline, 5 percent glucose in water, and sterile water for injection.
To determine the mass of the solution, we need to calculate the total mass of the solution when 81g of glucose is added. Since the solution is 15.0% glucose by mass, the remaining 85.0% is water. Therefore, the total mass of the solution can be calculated using the mass of glucose added and the percentage of water. This would result in a total mass of solution greater than 81g due to the addition of water to dissolve the glucose.
Let's say the total solution is 100 liters. 50 of the liters is glucose and 50 is water. We want to make the 50 glucose equal to 10% of the total solution. For that to happen, we need to make the total solution 500 liters (50 of the 500 would be a 10% solution). So we add 400 liters of water to the original 100 liter (50/50) solution. Take the total number of units and multiply by 4. Add that much in water.
200 grams/1,000 mL x 100= 20%
Glucose is the solute; water is the solvent.