Question

7a The density of sugar solutions at various concentrations is tabulated below. Using the chan construct the graph density y-axis) versus % weight (x-axis). Note: Must use Excel to graph the below data % by weight (w/v) Density (g/mL) 0.998 1.018 1.038 1.059 1.081 1.104 1.127 1.151 1.176 7. A student determines the density of degassed (flat soda). The density is found to be 1.040 g/ mL. Using the graph you constructed in 7a, determine the sugar content in a 12-oz can of Sprite®. (12 fl oz can equals 335 mL). (Hint: % weight by volume is the mass of sugar in 100 mL of drink, e.g. if the density is 1.018 g/mL there are 5 grams of sugar in 100 mL of soda). Y = |x10x + 0.0039x +0:9981

Answer 1

7a.

Plotting the given data by taking density in y axis and % weight in x axis, we get the following best fit plot

  

Density vs % weight 1.2 y = 1E-05x2 + 0.0039x+0.9981 1.18 R2-1 1.16 1.14 1.12 1.1 1.08 1.06 1.04 1.02 1 0.98 0 10 15 20 25 30 35 40 45 % weight Density (g/mL) LO

7b.

The best fit equation we get from the plot is

  $y = 1\times 10^{-5} x^2 + 0.0039x + 0.9981$

where y denotes the density and x is the % weight of sugar (w/v)

Now, the density of the drink is found to be 1.040 g/mL

Hence, if we set y = 1.040, we can calculate the weight percent of sugar in the drink by calculating the value of x using the best fit equation

Hence,

$y = 1\times 10^{-5} x^2 + 0.0039x + 0.9981 \\ \Rightarrow 1.040 = 1 \times 10^{-5} x^2 + 0.0039 x + 0.9981 \\ \Rightarrow 10^{-5} x^2 + 0.0039x + 0.9981-1.040 = 0 \\ \Rightarrow 10^{-5} x^2 + 0.0039x - 0.0419 = 0$

Now, to solve for the value of x, we have a quadratic equation.

Hence, we can use the quadratic formula to solve for x

For a generic quadratic equation  $ax^2 + bx +c = 0$

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Our equation is

$10^{-5} x^2 + 0.0039 x - 0.0419 = 0$

Hence,

$a = 10^{-5}$

$b=0.0039$

$c=-0.0419$

Hence, the value of x can be calculated as

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \Rightarrow x = \frac{-0.0039 + \sqrt{0.0039^2 - 4 \times 10^{-5} \times (-0.0419)} }{2\times 10^{-5}} \ or \ \frac{-0.0039 - \sqrt{0.0039^2 - 4 \times 10^{-5} \times (-0.0419)} }{2\times 10^{-5}}\\ \\ \Rightarrow x = \frac{-0.0039 + 0.00411}{2 \times 10^{-5}} \ or \ \frac{-0.0039 - 0.00411}{2 \times 10^{-5}} \\ \\ \Rightarrow x \approx 10.5\ or -400.5$

Note that % weight can never be negative  Hence, correct value of x is 10.5.

Hence, the weight percent of sugar in the drink is 10.5 % approximately.

Now, it is given that the volume of the drink is 335 mL.

Hence, the amount of sugar in the drink can be calculated as

$\% \ weight = \frac{weight \ of \ sugar \ in \ g}{Volume \ of \ sugar \ drink \ in \ mL} \times 100 \\ \Rightarrow 10.5 = \frac{weight \ of \ sugar}{335 } \times 100 \\ \Rightarrow weight \ of \ sugar = \frac{10.5 \times 335}{100}\ g \approx 35.2 \ g$

Hence, the amount of sugar in the drink is about 35.2 g (rounded to three significant figures).