Soft drinks contain a variety of sugars - Scottish Highers Chemistry - Question 11 - 2016

The student could use a calibrated pipette to accurately dispense 10.0 cm³ of the sucrose solution into a clean, dry weighing boat or container. The mass of the empty container should be measured on a digital balance first. After adding the solution, the combined mass should be measured again. The mass of the solution can be found by subtracting the mass of the empty container from the mass of the full container.

The soft drink needed to be flat to ensure that any gas bubbles present would not affect the measurement of density. Bubbles could cause inaccuracies in the volume measurement since they would give a false reading of the liquid volume.

Using the equation provided, we can substitute the density of the soft drink into:

ext{density of sugar in g cm}^{-3} = (0.0204 imes ext{% concentration of sugars}) + 1.00

Thus, substituting 1.07 for density:

$1.07 = (0.0204 imes ext{concentration}) + 1.00$

Rearranging gives: $0.07 = 0.0204 imes ext{concentration}$

So, ext{concentration} = rac{0.07}{0.0204} = 3.43 ext{% concentration of sugars.}

The total mass of sugar in the drink can be calculated using the concentration: Given that there are 10⁶ grams of sugar in 100 cm³, we can set up a proportion: $ext{Total sugar} = ext{Concentration} imes ext{Volume}$ Now substituting values: ext{Total sugar} = rac{10^6}{100} imes 330 = 3.3 imes 10^6 ext{g}

The functional group for an aldehyde is represented as:

  O
  ||
H-C

Where the carbon atom is double bonded to an oxygen atom and single bonded to a hydrogen atom.

The ion-electron equation can be represented as follows:

The colour change observed would be from blue (due to Cu²⁺ ions) to brick red or brown, indicating the reduction of copper ions.

To find the concentration of reducing sugars in the diluted sample: Cup reaction requires using the titration results: Using the formula: $ext{C}_1 ext{V}_1 = ext{C}_2 ext{V}_2$ Where:

• C₁ is the concentration of the Cu²⁺ solution.
• V₁ is the volume of Cu²⁺ used.

We know:

• $C_1 = 0.0250 ext{ mol dm}^{-3}$
• $V_1 = 19.8 ext{ cm}^3 = 0.0198 ext{ dm}^3$
• $V_2 = 0.025 ext{ dm}^3$

Thus, $0.0250 imes 0.0198 = C_2 imes 0.025$

Solving gives: $C_2 = 0.0198 ext{ mol dm}^{-3}$