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A student investigated the effect of different concentrations of sugar solution on pieces of potato - AQA - GCSE Biology Combined Science - Question 1 - 2020 - Paper 1

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A student investigated the effect of different concentrations of sugar solution on pieces of potato. This is the method used. 1. Cut five pieces of potato. 2. Reco... show full transcript

Worked Solution & Example Answer:A student investigated the effect of different concentrations of sugar solution on pieces of potato - AQA - GCSE Biology Combined Science - Question 1 - 2020 - Paper 1

Step 1

What is the independent variable?

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Answer

The independent variable in this experiment is the concentration of the sugar solution. This is the factor that is intentionally changed to observe its effect on the mass of the potato pieces.

Step 2

Explain why the potato in 0.0 mol/dm² sugar solution increased in mass.

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Answer

The potato in the 0.0 mol/dm² sugar solution increased in mass because the surrounding solution was pure water. Water entered the potato cells through osmosis, moving from an area of higher water concentration (the solution) to an area of lower water concentration (inside the potato). As a result, the potato gained water, which resulted in an increase in mass.

Step 3

Complete Figure 1.

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Answer

To complete Figure 1, plot all data points from Table 1 on the graph with 'Concentration of sugar solution in mol/dm²' on the x-axis and 'Change in mass in grams' on the y-axis. Then, draw a line of best fit through these points, ensuring it reflects the trend of the data.

Step 4

Determine the concentration of sugar solution inside the potato cells.

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Answer

From Figure 1, read the concentration at which the potato mass remains unchanged. This value is around 0.23 to 0.24 mol/dm², indicating the concentration of sugar solution inside the potato cells.

Step 5

Calculate the percentage change in mass for the potato in 0.2 mol/dm² sugar solution.

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Answer

Using Table 1, the change in mass for the potato in 0.2 mol/dm² is 0.25 grams and the starting mass is 7.96 grams.

Using the formula:

extpercentagechangeinmass=(0.257.96)×100 ext{percentage change in mass} = \left( \frac{0.25}{7.96} \right) \times 100

Calculating this gives:

(0.257.96)×1003.14%\left( \frac{0.25}{7.96} \right) \times 100 \approx 3.14\%

Thus, the percentage change in mass to three significant figures is 3.14%.

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