0
Unfortunately sugar is an imprecise term. It is better to specify glucose (usually means dextrose) or sucrose or fructose etc.
A 5 percent solution of one of these sugars would contain 5 grams weight dissolved in 100mL of water (or could be another solvent).
Read more: What_does_a_5_percent_sugar_solution_mean
Still curious? Ask our experts.
Chat with our AI personalities
Jordan
Fran
Professor
Add your answer:
Earn +20 pts
How you can convert 10 percent sugar solution in 5 percent sugar solution?
To convert a 10% sugar solution into a 5% sugar solution, you would need to dilute the original solution by adding a calculated amount of water. For example, to create 1 liter of 5% sugar solution from a 10% solution, you would mix 500ml of the 10% solution with 500ml of water.
How much solvent must you add to 1 ml of a 50 percent solution to make a 10 percent solution?
To make a 10 percent solution, you would need to dilute the 50 percent solution by adding 4 ml of solvent to 1 ml of the 50 percent solution. This will result in a total volume of 5 ml with a 10 percent concentration.
If 5 liters of a 4 percent silver iodide solution must be mixed with a 10 percent solution to get a 6 percent solution How many liters of the 10 percent solution are needed?
Let x be the number of liters of the 10% solution needed. The amount of silver iodide in the 5-liter solution is 0.04 * 5 = 0.2 liters. The final amount of silver iodide in the mixture would be 5 * 0.06 = 0.3 liters. Set up the equation: 0.1x + 0.2 = 0.3, solve for x, x = 1 liter. Hence, 1 liter of the 10% solution is needed.
How would you prepare 100 ml of a 5 percent dextrose solution from a 50 percent dextrose solution?
To prepare 100 ml of a 5% dextrose solution from a 50% dextrose solution, you would use the formula: C1V1 = C2V2. You will need 10 ml of the 50% solution (C1) and dilute it with 90 ml of water (V1) to obtain the desired 100 ml of 5% dextrose solution.
The molarity of a 5 percent solution of NH3 in water?
The molarity of a 5% solution of NH3 in water depends on the density and molecular weight of NH3. Without this information, it is not possible to calculate the molarity.
