Question

As a food sweetener, often sugar is replaced with the low calorie replacement aspartame. The trade names for aspartame are numerous and include Equal, NutraSweet and others Aspartame is also used as a sweetener for diet sodas. Determine the solubility of aspartame in water as grams of aspartame per 100 g of water at 5 °C, 25 C, and 60 °C using the data in the table below. 1. T,°C | Solubility, Mole Fraction 3.38 x104 7.01 x 104 2.74 x 10-3 25 60 2. Determine the solubility of sugar 5 °C, 25 C and oo Ho does the solublity Solubility of Salt and Sugar aspartame at 5 °C, 25 C, and 60 °C compare to the solubility of sugar (sucrose) at those temperatures? 500 E 450 400 350 300 250 Sugar 3. One packet of Equal is 1 g, and only 3.87 mass %aspartame. The rest of the mass in a packet of Equal is dextrose and maltodextrin, used as "bulkening agents" to dilute very potent sweeteners. What is the mass of aspartame in 1 packet of Equal? 100 50 Salt 0 20 40 60 80 100 Temperature (°C) 4. Is the low solubility of aspartame a problem if you are using Equal instead of sugar (i.e. 3, and explain your will it all dissolve in your drink)? Use your answers from Q's 1 and 5. The sugar substitute Equal daims that "1 packet [of Equal] sweetens like 2 teaspoons of sugar". (Two teaspoons of sugar is approximately 9 g of granulated sugar (sucrose).) Approximately how much sweeter is aspartame than sucrose? 6. If you added 2 teaspoons of pure sugar (sucrose) to a 12 oz drink: a. What would be the boiling point of the drink? b. What would be the freezing point of the drink?

Answer 1

"Given solubility of as ""mole factor"" at 5, 25460 5 degree c = 3.38 times 10^-4 25 degree c = 7.01 times 10^-4 60 degree c - 2.74 times 10^-3 mol net of H_2O = 18 g Mol net of = 294 g To find solubility in term germ of Aspartame in 100 g water at 5 degree rightarrow mole factor as aspartame = mole of aspartame/ mole of water + moles of aspartame 3.38 times 10^-4 = x/100/18 + x x - (3.78 times 10^-4 x = 100/18 times 3.38 times 10^-4 x(1 - 3.382 times 10^-4) = 100 squareroot 3.38 times 10^-4/18 therefore x =\ 100 times 3.38 times 10^-4/18 = 18.77 times 10^-4 moles net of aspartame = mol net time moles = 294 times 18.77 times 10^-4 = 0.552 g similarly at 25 degree c x = 100 times 7.01 times 10^-4/18 = 38.94 times 10^-4 therefore net of aspartame in 100 g water "