5. Percent Solutions a) Calculate the weight of powdered glucose required to pre

Question

5. Percent Solutions a) Calculate the weight of powdered glucose required to prepare of a ten percent (10%) solution of glucose. b) Sucrose (table sugar) is a disaccharide composed of a molecule of glucose and a molecule of fructose and is exactly twice the mass of a single glucose molecule. If you were asked to prepare 1 L of a 10% solution of sucrose, how much sucrose would you use? c) A nurse prepared an I.V. saline solution by diluting 1.7g NaCl into a final volume of 2 liters of water. Determine the percentage of the solution he prepared.

Explanation / Answer

5a. In order to prepare 1% solution of glucose, 1 gm of glucose is dissolved in final volume of 100 ml of solution. Hence, 10 g is dissolved in final volume of 1L of solution.

For a 10% solution, 10 g are dissolved in final volume of 100 ml of solution. Therefore, amount of glucose required to prepare 1L of 10% solution is 10 g X 10= 100 g. Weigh 100 g of glucose, add small amount of water to dissolve it and then make up volume to 1000 ml or 1L.

5b. Sucrose is twice the mass of a single glucose molecule. However, the amount of sucrose required to prepare 1L of 10% solution is still 100g. When preparing a percent solution, amount of compound does not depend on the molecular mass. Molecular mass/weight is only considered for molar solution. Only amount of water added will change, as less water is required to make 1L of solution due to the heavy mass.

For a 10% solution, 10 g are dissolved in final volume of 100 ml of solution. Therefore, amount of sucrosee required to prepare 1L of 10% solution is 10 g X 10= 100 g. Weigh 100 g of sucrose, add small amount of water to dissolve it and then make up volume to 1000 ml or 1L.

5c. 1.7 g of Nacl is dissolved in final volume of 2 L of water.

1 g of NaCl in 100 ml is 1% solution.

1L of solution has 1.7/2=0.85 g of NaCl

100 ml of solution will have 0.85g/10= 0.085 g of NaCl.

Hence, the percentage solution prepared is 0.085% solution.

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