Question

A food
scientist is using a standard solution of fructose to analyze the monosaccharide
composition of an unknown extract. He used 17 g of fructose and 126 g of water to
make the solution. The density of the solution is 1.0476 g/cc and the molecular weight
of fructose is 180.16 g/mol. Using a graph express how the density of the solution
would change as function of amount of fructose in solution (i.e. if the amount of fructose
is increased in solution in increments of 10 g, up to 57 g). Hint: Assume that the
volume of the solution would not change with the increase in density. cc is the
volume unit, which is equal to cm3
.
the graph into the solution sheet.

Answer 1

Calculate the volume of the solution. It is mentioned that the volume of the solution stays constant throughout the experiment; hence, we will use the same volume for all our calculations.

The solution contains 17 g fructose and 126 g water; therefore, the mass of the solution is (17 + 126) g = 143 g.

The density of the solution is 1.0476 g/cm³; consequently, the volume of the solution is (mass of solution)/(density of solution) = (143 g)/(1.0476 g/cm³) = 136.5025 cm³.

Next, we increase the amount of fructose in the solution in increments of 10 g upto 57 g. We can calculate the density of the solution as below.

Trial	Mass of fructose (g)	Mass of solution (g)	Volume of solution (cm3)	Density of solution (g/cm3)
1	17	143	136.5025 (volume stays the same throughout the experiment)	143/136.5025 = 1.0476
2	27	(126 + 27) = 153		153/136.5025 = 1.1208
3	37	(126 + 37) = 163		163/136.5025 = 1.1941
4	47	(126 + 47) = 173		173/136.5025 = 1.2674
5	57	(126 + 57) = 183		183/136.5025 = 1.3406

Plot a graph between the densities of the solution vs the amounts of fructose in the solution as below. The plot is shown below.

Plot of density vs grams of fructose added. The plot is linear with a slope of 0.0073 g/cm³