Investigating Cells Revision

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1.In an investigation, three 25 g samples of sultanas were put into separate beakers of distilled water, as shown below:

After 24 hours, the sultanas were removed from the water, blotted on filter paper and reweighed. The results are shown in the table:

|Sample |Mass after 24 hrs (g) |Percentage change in mass | |1 |32.5 |30.0 | |2 |32.2 |28.8 | |3 |32.4 |29.6 |

a)Complete the table with the percentage change in mass of the sultanas in sample 3.

32.4 g – 25 g = 7.4g

(7.4 / 25) = 0.296

0.296 x 100 = 29.6 %

b)The change in mass of the sultanas was caused by the movement of water.

i)Name this process.


ii)Explain the results in terms of water concentrations.

The water moved from an area of high water concentration outside the cell to an area of low water concentration inside the cell. 1.Continued

c)Which of the following is the best reason for blotting the sultanas before reweighing? Tick the correct box.

| |To stop them sticking together | |To remove external sugar solution | | | | | | |X |To remove external water | |To make sure the sultanas were dried |

2.Plant cells and animal cells were left in water or 10% sucrose solution for 10 minutes. The cells were then examined under the microscope: The appearance of three individual cells is shown below.

Cell R Cell S Cell T

a)Which two of the cells had been placed in 10% sucrose solution?

Cell S and Cell T.

b)The change in the cells was caused by the movement of water into or out of the cells.

What is the name of this process?


c)With reference to the cells placed in water, what is meant by the term “concentration gradient”?

The concentration gradient is the difference between the high water concentration outside the cells and the lower water concentration inside the cells.

3.Potato cylinders of equal mass were placed in separate test tubes, as shown in the diagram.


The tubes contained salt solutions of 0%, 0.5%, 1.5%, 10%, 15% concentrations.

After two hours the change in mass of each cylinder was measured. Results are shown in the table.

|Tube |Change in mass (g) |Salt solution (%) | |A |-0.6 |15 | |B |-0.5 |10 | |C |-0.2 |1.5 | |D |+0.1 |0.5 | |E |+0.2 |0 |

a)Complete the table by adding the correct concentration of the salt solution in each tube.

b)Which tube contained a solution with a water concentration closest to that of the potato cell sap?

Tube D

c)The original mass of each potato cylinder was 5g. Calculate the percentage change in mass for the cylinder in tube D. Space for calculation

(0.1 / 5) x 100 = 2

2 %
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