Fig. 9.1 shows the apparatus used to find the vitamin C content of a blackcurrant drink. The drink was added from a syringe, drop by drop, into 1.0 cm3 of D...

Assessment: Biology 0610 | Paper 5 Mock 01 | Practical Test Subject: Biology - 0610

Question 1 Report

0610-p5-molecules-dcpip-titration-9

Fig. 9.1 shows the apparatus used to find the vitamin C content of a blackcurrant drink. The drink was added from a syringe, drop by drop, into 1.0 cm3 of DCPIP until the blue colour was just lost. To find the best way to store the drink, samples were kept at three temperatures for 5 days before testing. A standard vitamin C solution of 1.0 mg per cm3 needed 0.50 cm3 to decolourise the DCPIP. Vitamin C concentration = (0.50 / volume of drink needed) x 1.0. The results are shown in Table 9.1.

Storage temperature / degrees CVolume of drink needed / cm3Vitamin C concentration / mg per cm3
40.80?
201.20?
302.40?

(a) Name the apparatus in Fig. 9.1 used to add the drink drop by drop, and name a piece of apparatus that would measure the 1.0 cm3 of DCPIP. [2]
(b) State why the volume of drink added is read from the syringe. [1]
(c) Measure, using Table 9.1, the extra volume of drink needed for the 30 degrees C sample compared with the 4 degrees C sample. [1]
(d) Measure, using Table 9.1, how many times more drink is needed at 30 degrees C than at 4 degrees C. [1]
(e) Calculate the vitamin C concentration of the drink stored at 4 degrees C. Show your working. [2]
(f) Complete Table 9.1 by calculating the vitamin C concentration for the 20 degrees C and 30 degrees C samples. [2]
(g) Describe the effect of storage temperature on the vitamin C content of the drink. [2]
(h) Explain why the drink stored at 30 degrees C needed the most drink to decolourise the DCPIP. [3]
(i) A student says a fridge is the best place to store the drink. State whether the results support this and give a reason. [2]
(j) Suggest two variables that must be kept constant. [2]
(k) Describe two ways to make judging the end point more reliable. [2]

Answer Details

Labelled answer diagram:

0610-p5-molecules-dcpip-titration-9 labelled answer

This is a titration-style vitamin C investigation with DCPIP. Fewer molecules of vitamin C mean more drink is needed to decolourise a fixed amount of DCPIP, so a larger volume needed means less vitamin C.

(a) The drink is added drop by drop from a syringe (a burette is also acceptable) [1]; the 1.0 cm3 of DCPIP would be measured with a measuring cylinder or a graduated pipette [1].

(b) The volume is read from the syringe so that the exact amount of drink needed to decolourise the DCPIP can be measured, which quantifies how much vitamin C is present [1].

(c) Extra volume needed at 30 °C compared with 4 °C \( = 2.40 - 0.80 = 1.60 \text{ cm}^3 \) [1].

(d) Number of times more \( = \dfrac{2.40}{0.80} = 3 \) times [1].

(e) Using concentration \( = \dfrac{0.50}{\text{volume needed}}\times1.0 \): for 4 °C \( \dfrac{0.50}{0.80}\times1.0 = 0.625 = 0.63 \text{ mg cm}^{-3} \). Working [1], answer [1].

(f) Same method: 20 °C \( \dfrac{0.50}{1.20}\times1.0 = 0.42 \text{ mg cm}^{-3} \) [1]; 30 °C \( \dfrac{0.50}{2.40}\times1.0 = 0.21 \text{ mg cm}^{-3} \) [1].

(g) As storage temperature increases, the vitamin C content decreases [1]; more drink is needed to decolourise the DCPIP at higher storage temperatures [1].

(h) The 30 °C sample needed the most drink because warmth speeds up the oxidation / breakdown of vitamin C [1], so fewer vitamin C molecules remain in the drink [1]; a larger volume of the weaker drink is then required to decolourise the same DCPIP [1].

(i) Yes, the results support storing it in a fridge [1]: the sample kept at 4 °C (fridge temperature) had the most vitamin C and needed the least drink, showing the lowest temperature preserves vitamin C best [1].

(j) Any two controlled variables: storage time (5 days), volume / concentration of DCPIP, the size of each drop, or the same drink [2].

(k) Any two ways to judge the end point more reliably: view the tube against a white background; use the same person to judge the colour; add drop by drop with swirling near the end point; repeat and take a mean [2].

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