AQA A Level Biology Paper 1 2018

(2.2) The loss in mass shown in Figure 3 is due to osmosis. The rate of osmosis between 0 and 40 minutes is faster in B (the eight small cubes) than in A (single large cube)

Is the rate of osmosis per mm2 per minute different between A and B during this time? Use appropriate calculations to support your answer. (3)

How would you work out the calculations for this question please?
https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FBiology%2FA-level%2FPast-Papers%2FAQA%2FPaper-1%2FQP%2FJune%25202018%2520QP.pdf
Question 2.2

(edited 11 months ago)

Reply 1

pirateguy27

5

Personally, I would calculate the rate of osmosis by doing rate = change in mass/time for both A and B. So it would be 62-57/40 = 0.125gmin^-1for the large cube and for the small cubes it would be 62-53/40 = 0.225gmin^-1. Then I would say as the rate of osmosis is different, for B it is 0.225gmin^-1and for A it is 0.125gmin^-1this means aning that the rate of osmosis is quicker in B than in A.

Reply 2

wwwwwwwkobo

1

Original post by pirateguy27

Personally, I would calculate the rate of osmosis by doing rate = change in mass/time for both A and B. So it would be 62-57/40 = 0.125gmin^-1for the large cube and for the small cubes it would be 62-53/40 = 0.225gmin^-1. Then I would say as the rate of osmosis is different, for B it is 0.225gmin^-1and for A it is 0.125gmin^-1this means aning that the rate of osmosis is quicker in B than in A.

ah I c thxx

(edited 3 months ago)

Reply 3

sisterr

5

Original post by pirateguy27

Personally, I would calculate the rate of osmosis by doing rate = change in mass/time for both A and B. So it would be 62-57/40 = 0.125gmin^-1for the large cube and for the small cubes it would be 62-53/40 = 0.225gmin^-1. Then I would say as the rate of osmosis is different, for B it is 0.225gmin^-1and for A it is 0.125gmin^-1this means aning that the rate of osmosis is quicker in B than in A.

is this enough to get all the marks

just asking as mark scheme talks about the surface areas

Reply 4

KBIOCHEMENG

1

Original post by pirateguy27

Personally, I would calculate the rate of osmosis by doing rate = change in mass/time for both A and B. So it would be 62-57/40 = 0.125gmin^-1for the large cube and for the small cubes it would be 62-53/40 = 0.225gmin^-1. Then I would say as the rate of osmosis is different, for B it is 0.225gmin^-1and for A it is 0.125gmin^-1this means aning that the rate of osmosis is quicker in B than in A.

You can use that rate alongside the surface areas for the answer. For cube A, each side of the cube is 35mm so 35 x 35= 1225 which is the area of 1 side of the cube, so x by 6 = 7350 which is the total surface area( which is what they want since they asked for mm2).For B, the also used the same lengths cube so the total surface area is the same, but because they cut it again you x 7350 by 2 , which gives you 14700mm2 for B. They said that the RATE is PER mm2 min-1, so you do the Rate which was found divided by the area. So for A its 0.125gmin-1 divided by 7350 = 1.7 x 10-5 and for B its 0.225gmin-1 divided by 14700= 1.5 x 10-5

AQA A Level Biology Paper 1 2018

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