Question

Peeled potatoes were cut into two sizes of cubes $–1{\text{cm}}^3(1\text{cm}\times 1\text{cm}\times 1\text{cm})$ and $27{\text{cm}}^3(3\text{cm}\times 3\text{cm}\times 3\text{cm})$ and weighed. Then, 27 of the $1{\text{cm}}^3$ cubes were placed in one beaker of distilled water, and one $27{\text{cm}}^3$ cube was placed in another beaker. One hour later the cubes were removed and their surfaces carefully dried. They were then reweighed. The conclusions drawn from the experiment are as follows:
(i) The 27 small cubes were used as they have the same total volume and surface area as the large cube.
(ii) The lower water potential of the potato compared to the surrounding solution results in the mass increase of the potato.
(iii) The difference in percentage increase in mass of the large and small cubes is due to the larger volume of the large cube as compared with one of the individual small cubes.
(iv) If the potatoes were left in the water long enough for the system to reach equilibrium the large cube would display percentage increase in mass equal to that of the small cubes.
Which of the following statement (s) is (are) correct?

• A. i, ii and iii
• B. ii only
• C. iii only
• D. ii and iv

Answer 1

The correct option is B ii only

Peeled potatoes were cut into two sizes of cubes – $1{\text{cm}}^3(1\text{cm}\times 1\text{cm}\times 1\text{cm})$ and $27{\text{cm}}^3(3\text{cm}\times 3\text{cm}\times 3\text{cm})$ and weighed. Then, 27 of the $1{\text{cm}}^3$ cubes were placed in one beaker of distilled water, and one $27{\text{cm}}^3$ cube was placed in another beaker. One hour later the cubes were removed and their surfaces carefully dried. They were then reweighed. The conclusions drawn from the experiment were:

i. The 27 small cubes were used as they have the same total volume and surface area as the large cube.
Total volume of 27 small cubes $=27{\text{cm}}^3$
Surface area of the 27 small cubes $=162{\text{cm}}^2$
Total volume of 1 large cube $=27{\text{cm}}^3$
Surface area of the 1 large cube $=54{\text{cm}}^2$
So, they do not have same surface area and total volume. That makes statement i wrong.

ii. The lower water potential of the potato compared to the surrounding solution results in the mass increase of the potato.
The potato cell has concentrated cell sap than the surrounding distilled water. So due to this, the endosmosis occurs which results in the transfer of the solvent from higher water potential area (outside) to the lower water potential area (here potato), so the cubes become firm and increase in size. Hence, statement ii is correct.

iii. The difference in percentage increase in mass of the large and small cubes is due to the larger volume of the large cube as compared with one of the individual small cubes.
Total volume of 1 small cube $=1{\text{cm}}^3$
Surface area of the 1 small cube $=6{\text{cm}}^2$
Surface area: volume ratio of the 1 small cube $=6$:1
Total volume of 1 large cube $=27{\text{cm}}^3$
Surface area of the 1 large cube $=54{\text{cm}}^2$
Surface area: volume ratio of the 1 large cube $=2$:1
So, there would be a difference in percentage increase in mass of the large and small cube due to large volume of the large cube in comparison to one of the small cubes. But in comparison to 27 small cubes volume of 27 cubes will be larger. That makes statement iii incorrect.

iv. If the potatoes were left in the water long enough for the system to reach equilibrium the large cube would display percentage increase in mass equal to that of the small cubes.
Even if the potatoes were left in water for a long time for the system to reach equilibrium large cube’s percentage increase in mass would not be equal to that of the small cubes, as there is a difference between their surface area to volume ratios. That makes statement iv incorrect.