Lesson video

This lesson is the effect of sugar concentration on the mass of plant tissue, data analysis and is from the unit coordination and control, maintaining a constant internal environment.

Hi there, my name's Mrs. McCready and I'm here to guide you through today's lesson.

So thank you very much for joining me today.

In our lesson today, we're going to interpret a graph showing the effect of sugar concentration on the mass of potato tissue and we're going to explain our results.

So in our lesson today, we're gonna come across a number of keywords and they're listed up here on the screen for you now.

You may wish to pause the video and make a note of them, but I will introduce them to you as we come across them.

So in our lesson today, we're going to analyse our data before we can explain the results.

So are you ready to go? I certainly am.

Let's get started.

So water is an essential component of life and is the major component in cell cytoplasm.

Now, water moves in and out of cells by the process of osmosis.

And osmosis is the net movement of water molecules through a selectively-permeable membrane from a more dilute solution to a more concentrated solution.

So if we look at the diagram there, we can see that the more concentrated solution has more of the big purple hexagons and the more dilute solution has fewer of those big purple hexagons.

That means there is more water where there are fewer purple hexagons and therefore that solution is more dilute.

Conversely, where there are more hexagons, there are fewer water molecules and that makes that solution more concentrated.

Therefore, water will move from the dilute side on the left where there are fewer purple hexagons to the more concentrated side on the right where there are more purple hexagons.

Now, in an experiment investigating the effects of different concentrations of sugar solution on the mass and length of potato chips, we can see that the length and the mass of potato chips changes according to the concentration of the sugar solution in which the potato chip was placed.

And the sample data here shows that both mass and length have changed for each of those potato chips.

Now we can calculate the change of mass or length by subtracting the initial length from the final length.

So, in the data table, we can see that the initial mass is 2.

59 grammes.

The final mass is 3.

07 grammes.

If we take the initial mass away from the final mass, it results in a change in mass of plus 0.

48 grammes.

Similarly, we can do that with the length.

So its initial length was 4 centimetres and its final length was 4.

3 centimetres.

So if we subtract the initial length from the final length, we can see that the change in length is plus 0.

3 centimetres.

So that's calculating the change in the mass or the length, but we also need to calculate the percentage change in mass.

So to calculate the percentage change, we subtract the initial value from the final value, then we divide that by the initial value.

Then because we're calculating it as a percentage, we times it by 100.

So, in this case, we would take 3.

07 grammes as the final mass, subtract 2.

59 grammes as the initial mass, divide that value by the initial mass, 2.

59 grammes and then times it by 100 to convert it into a percentage, meaning that there is a percentage change in mass of plus 18.

5%.

Similarly, we can do this with the length measurements.

So we can take 4.

3 centimetres of the final length, subtract the initial length of 4 centimetres, divide that value by the initial length, and then times it by 100 to calculate the percentage, which gives us a plus 7.

5% increase in length.

So let's quickly check our understanding.

If the initial mass is 5 grammes and the end mass is 4.

5 grammes, what is the change in mass? Is it minus 10%, plus.

5 grammes, or minus.

5 grammes? I'll give you five seconds to think about it.

So the change in mass going from 5 to 4.

5 grammes is minus 0.

5 grammes.

Well done if you got that correct.

So let's just work this through for the percentage change in mass.

So percentage change in mass is final minus initial divided by the initial measurement, all times by 100.

So that's 4.

5 grammes minus 5 grammes divided by 5 grammes times by 100.

So 4.

5 minus 5 is minus 0.

5, divide that by 5 and then times it by 100 gives us minus 0.

1 times by 100, which gives us a percentage change in mass of minus 10%.

So that's my example.

What I would like you to do is to calculate the percentage change in mass if the initial mass is 6.

1 grammes and the end mass is 4.

0 grammes.

So I'll give you five seconds to calculate it.

Okay, let's work our way through this calculation then.

So firstly, you should have written the equation.

Always worth writing that at each occasion because that helps to create some muscle memory for you, ready for your exams as well.

So percentage change is final minus initial divided by initial times by 100.

So final minus initial is 4.

0 minus 6.

1 divided by the initial, which is divided by 6.

1 times by 100.

So 4 minus 6.

1 makes minus 2.

1 divided by 6.

1 times by 100, that's minus 0.

344 times by 100, which gives us a percentage change in mass of minus 34.

4%.

Did you get that correct? Just check your workings through if you didn't and see if you did get it wrong, whereabouts in the methodology did you get it wrong? If you've laid all of your workings out, that should be very, very easy to do.

If all you've done is write the answer down, then it will be impossible to work out where you went wrong and it will also be impossible for an examiner to do that too.

So always show your workings, please.

So we've just been calculating percentage change in mass, but why? Now, when we're handling data, it's really important that we process it properly and thoroughly before we go on and draw conclusions using it.

Otherwise, our conclusions may well be incorrect.

Now, you will have seen in the data that the potato chips had different masses at the start of the experiment, and this is due to factors beyond our control.

We cannot change the mass of each of those potato chips without affecting the length and therefore causing a different change, a different difference between the potato chips.

And therefore, what this does is introduce what's called a random error into our results because the potato chips are all starting off at slightly different masses from each other because of factors that we cannot change and therefore the changes that we're seeing in mass may be due to other factors that are different other than the one that we are choosing to change.

So we need to account for and handle this random error and we can do that by calculating the percentage change in mass instead because what we are then doing by using the percentage change in mass is comparing each potato directly with each other.

So that means we can then draw valid conclusions using our data based on direct comparisons where only the sugar solution has changed because we have calculated the percentage change in mass and the percentage change in length rather than starting with the slightly varied starting masses at the beginning of the experiment instead.

Now, we can plot the percentage change in mass as a line graph.

So we'll plot the concentration of sugar solution in moles per decimeter cubed along the x-axis, 0, 0.

25, 0.

5, 0.

75, and 1 moles per decimeter cubed.

And we'll plot percentage change in mass along the y-axis, going to just above the greatest change in mass positively to just below the greatest change in mass negatively.

And that's where I've got my values plus 20 to minus 40 from by looking at my data.

Your data may differ, so your maximum on the y-axis and your minimum on the y-axis may well differ.

Once we've got our axes sorted and labelled including the units, we can then plot the data.

So we can plot those data points, the percentage change in masses on the graph as appropriate, and then draw a line of best fit through those data points.

Remember, it's a line of best fit, so you don't have to wiggle between each of those data points.

You need to draw a smooth line which best fits the data rather than being a dot to dot.

So, where on the graph will the concentration of sugar solution be plotted? Is it A, the x-axis, B, the y-axis, or C, as data points? I'll give you five seconds to think about it.

Okay, so you should have recalled that the concentration of sugar solution, this is our independent variable and this gets plotted on the x-axis.

All independent variables get plotted on the x-axis.

Okay, what I'd like you to do now is to go ahead with your data and calculate the change in the percentage change in mass and length for the data that you have got.

And then I would like you to plot a graph using the concentration of sugar solution against the percentage change in mass.

Now take your time, make sure that both your calculations are correct and that your graph is plotted accurately.

So pause the video and come back to me when you are ready.

Okay, so you should have analysed your data and plotted it as a graph.

So your changes and percentage changes in mass and length should be recorded within your table and you should have drawn a graph with the concentration of sugar solution on the x-axis, the percentage change in mass along the y-axis, and your data points plotted accurately.

Well done.

That was a lot of data analysis to complete.

Okay, let's see if we can explain our results.

So when we're talking about water moving by osmosis, we're talking about water moving down a water concentration gradient from where there is a high water concentration to where there is a low water concentration.

But we can also talk about this in terms of a dilute or a concentrated solution.

So a dilute solution is where there is a lot of water and much less dissolved substance, in this case, sugar and a concentrated solution is one where there is a lot of dissolved substance and much less water.

So you can think about this a bit like squash drink for instance, like blackcurrant squash or orange squash.

So when it's in the bottle before you pour it out into your glass, it's concentrated.

There is more of the squash and the sugars and the colours and flavourings and so on in the bottle than there is water when you pour tiny bit of it into your glass and then top it up with water and therefore make it more dilute because you're adding water into the solution.

So it goes from being concentrated when it's inside the bottle to being dilute in your glass before you drink it.

And essentially, we're talking about the same sort of thing here.

So the dilute solution is outside the cell.

There is a little bit of sugar and a lot of water.

Whereas inside the cell, it is much more concentrated.

There is a lot of sugar and a lot less water.

So inside the cell in this example, it is like the squash inside the bottle before it's poured, whereas outside the cell, it's like it's in the glass and has been topped up with water ready for you to drink.

Now, water moves down the concentration gradient of water molecules and therefore will move from a dilute solution where there's lots of water and not very much of the dissolved substance to a position where there is not very much water and a lot of the dissolved substances.

And in this scenario where there is a dilute solution outside of the cell and a more concentrated solution inside the cell, water will move down the concentration gradient and therefore into the cell.

This will increase the mass of the cell because it's taking on water and because it's taking on water, it's swelling up, the cells are swelling up and therefore the potato chip overall will increase in length.

So, have a look at your data, where did this occur? At which concentrations did the potato chips increase in length? So you should be able to see that the increase in length occurred when there was lots of water and not very much sugar.

In other words, in the dilute sugar solutions or in the distilled water solution.

This is where there is low sugar concentration, it is more dilute.

So water has moved into the cell by osmosis from a dilute solution outside the cell where there is more water molecules to inside the cell where it is more concentrated and there is less water.

This means that there was more water outside the cell in the dilute solution than inside the cell in the more concentrated solution.

And because water is moving down the concentration gradient of water molecules, it means that the mass and the length of the potato chip has increased because the cells have taken on water, swollen up, and therefore gained mass and length.

So, that's when there was low sugar solution outside the potato chips.

What about the opposite scenario? So what happens when there is a concentrated solution outside the cell compared to inside the cell? So in scenarios like this, water will move out of the cell.

Again, water is moving down its concentration gradient from where there is a lot of water to where there is much less water.

Now, it's dilute comparatively inside the cell from outside where it is much more concentrated.

So water will move down the concentration gradient from inside the cell to outside the cell.

So water will leave the cells, this will decrease the mass, make the cell shrivel up a bit, and that will reduce the length of the potato chip as well.

So just check your results, at which concentrations did this scenario occur? So you should have seen that this occurred at solutions where there was a high concentration of sugar, so perhaps at 0.

75 and 1 molar sugar solution where water left the cells.

So let's just review this in a bit more detail.

Water is leaving the cell, moving out of the cell by osmosis because there is more water inside the cell than outside and therefore water is moving down its concentration gradient.

Therefore, there was less water outside the cell than inside.

And so water has moved down the concentration gradient of water molecules from a dilute solution inside to a concentrated solution outside, and this has decreased the mass and the length of the potato chip as the potato chips lost water.

Okay, so let's check our understanding.

Which diagram shows a more dilute solution inside the cell? I'll give you five seconds to think about it.

So dilute means more water and less dissolved substance inside the cell.

That's diagram B.

Did you get that right? Well done if you did.

Now, what do you think of this? In which diagram would the net movement of water be into the cell? I'll give you five seconds to think about it.

So the net movement of water into the cell will happen where there is less water inside the cell than outside, and that is scenario A.

Lots of water outside the cell, much less inside the cell.

In other words, the cell is quite concentrated.

Now, the concentration of sugar inside the potatoes can be determined from the graph.

So the point at which the graph line of best fit crosses the x-axis is where there was no change in mass, neither up nor down.

The overall change in mass was 0%.

So, in other words, as much water left the cell as entered the cell.

That's what net movement is about, as much water left the cell as entered and so there was no gain in mass and length and there was no loss in mass and length.

In other words, there was as much sugar outside the cell as inside the cell.

So the concentrations of sugar on both sides of the cell membrane were the same and therefore there was no net movement of water.

We can see that in the diagram, there is as much sugar outside the cell as inside the cell.

There is as much water outside the cell as inside the cell.

So whilst water will move back and forth across the cell membrane, the same number of molecules will leave as will enter and therefore there will be no net movement and therefore no mass and length change.

Now we can identify this on the graph by looking at where the line of best fit crosses the x-axis because this is where there was no mass change.

So, on the x-axis, you can see that I've highlighted it on the screen there with a big pink cross and an arrow pointing to it.

There is no mass change up or down and therefore, that is the concentration of sugar inside our potato chip.

However, we have to exercise a little bit of caution because this isn't a concentration of sugar solution that we used within our experiment.

It is an interpolation on the line of best fit and therefore we are relying upon us having drawn our line of best fit accurately and our data points following precisely on that line as well.

But assuming all is equal, then we can read that value off the graph and see that the concentration of sugar solution inside the potato is approximately 0.

275 moles per decimeter cubed.

So what does that mean then? There was about 0.

3 moles per decimeter cubed of sugar within the cytoplasm of the cells within the potato chip, 0.

3 moles per decimeter cubed.

Now, a fizzy drink contains approximately 39 grammes of sugar, which is equivalent to 0.

321 moles per decimeter cubed.

So compare that to the potato chip and you can see that potatoes are almost as sugary as a fizzy drink.

However, potatoes have actually got a lot more sugar stored within them because most of the sugar within a potato is stored as starch.

And we turn starch, digest starch into sugar in our digestive system before we absorb it into our blood when we eat.

So, potatoes are actually really sugary.

Okay, let's quickly check our understanding.

So, the sugar concentration inside the potato can be found on the graph where A, the line on the graph is steepest, B, the line on the graph crosses the y-axis, or C, the line on the graph crosses the x-axis.

I'll give you five seconds to think about it.

So you should have said that we can find the concentration of the potato by looking at where the line crosses the x-axis.

Well done if you got that correct.

So, what I'd like you to do now is to use your graph to identify the approximate sugar concentration inside the potato chips.

And then, I would like you to use the graph data and your understanding of osmosis to explain the change in mass at the following concentrations, at naught moles per decimeter cubed, at the concentration where there was no mass change, whatever that is for your potato, and at 1 moles per decimeter cubed as well.

So pause the video and come back to me when you are ready.

Okay, so firstly, you should have used your graph to identify the approximate sugar concentration inside the potato chip by identifying where the line of best fit crosses the x-axis.

And it's at this point where there was no net movement of water inside and outside of the cell, and therefore no change in mass.

Then I asked you to explain these changes in mass.

So at naught moles per decimeter cubed, you should have said that the potato chip has gained mass.

This means that water has moved into the potato and this is because the solution outside the potato was more dilute than the solution inside the potato.

At the concentration at which there was no mass change, you should have said that there was no mass change and this means that there was no net movement of water into or out of the potato.

And this is because the concentration was equal inside and outside the potato.

Then, for the 1 mole per decimeter cubed concentration, you should have said that the potato chip lost mass, which means water moved out of the potato.

And this is because the solution outside the potato was more concentrated than the solution inside the potato.

So just review your work.

Did you get your explanations correct talking about dilute and concentrated solutions and their movement into or out of the potato? I hope that's clear.

And well done, that was quite hard to do.

So, we've come to the end of our lesson today.

And today, we have seen that water moves by osmosis into and out of cells through the selectively-permeable cell membrane, down a concentration gradient of water molecules.

Now, potato chips, which are immersed in a concentrated sugar solution lose mass because water moves by osmosis out of the cells into the external solution, whereas potato chips immersed in distilled water gain mass because water moves by osmosis into the cells from the external solution.

Now, we can calculate the percentage change in mass and by doing so, this removes the effects of random error caused by things that are beyond our control.

And we've also seen that the point at which there is no change in mass is the point at which there was equal concentration of sugar solution inside the potato and outside the potato.

Therefore, no net movement of water occurs at this concentration, and we can see this from the graph where the line of best fit crosses the x-axis.

So I hope you've enjoyed our lesson today.

Thank you very much for joining me, and I hope to see you again soon.

Bye.