Hello! I\'m in a chemistry course and I have some learning difficulties. I am ha

Question

Hello!

I'm in a chemistry course and I have some learning difficulties. I am having trouble with this problem. I am eager to learn and any help is greatly greatly greatly appreciated!!

No need to solve all parts of this problem (no complaints if you decide too lol). Showing me how to solve part A and/or B is more than enough! Thank you very much for your time!!

5. You have been given 1,000 grams of an aqueous solution that is 10 percent glucose by weight. Answer the following questions about this solution. A. How many grams of glucose are in this solution? B. C. D. E. How many moles of glucose are in this solution? How many grams of water are in this solution? How many liters of water are in this solution? What is the molality of this solution?

Explanation / Answer

Ans. Given, [Glucose] = 10% (w/w). A 10% (w/w) concentration means that there is 10.0g of solute per 100.0 g of solution.

#A. Mass of glucose = [Glucose] x Mass of solution in kg

                                    = 10 % (w/w) x 1000.0 g solution

                                    = (10 g glucose / 100.0 g solution) x 1000.0 g solution

                                    = 100.0 g glucose

#B. Moles of glucose = Mass of glucose / Molar mass

                                    = 100.0 g / (180.15768 g/mol)

                                    = 0.5551 mol

#C. Mass of water = Mass of solution – Mass of glucose

                                    = 1000.0 g – 100.0 g

                                    = 900.0 g

#D. Assuming density of water to be 1.000 g/ mL –

Volume of water = Mass / Density

                                    = 900.0 g / (1.000 g/ mL)

                                    = 900.0 mL

                                    = 0.900 L

#E. Assumptions- It’s assumed that the addition of 100.0 g glucose does not affect the volume of water because glucose is completely soluble in water. So, volume of final solution remains to be 900.0 mL but mass becomes 1000.0 g

# Molarity of glucose = Moles of glucose / Volume of solution in liters

                                    = 0.5551 mol / 0.900 L

                                    = 0.6168 mol/ L

                                    = 0.6168 M

Note: The values may vary depending upon assumptions taken. The best result could be obtained if density of glucose solution were provided.

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