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Current time:0:00Total duration:6:45

AP Bio: ENE‑2 (EU), ENE‑2.H (LO), ENE‑2.H.1 (EK), ENE‑2.I (LO), ENE‑2.I.1 (EK), ENE‑2.I.2 (EK), ENE‑2.J (LO), ENE‑2.J.1 (EK)

- [Instructor] We're
told that six identical potato core cubes were
isolated from a potato. The initial weight of
each cube was recorded. Each cube was then placed
in one of six open beakers, each containing a
different sucrose solution. The cubes remained in
the beakers for 24 hours at a constant temperature
of 23 degrees Celsius. After 24 hours, the cubes were removed from the beakers, blotted, and reweighed. The percent change in mass, due to a net gain or loss of water, was calculated for each cube, and the results are shown
in the graph to the right, so, this graph right over here. A straight line is drawn on the graph to help estimate results from other sucrose
concentrations not tested. Using the straight line on the graph, calculate the water potential, in bars, of the potato core cubes
at 23 degrees Celsius. Give your answer to one decimal place. So, pause this video and see
if you can work that out. All right, so, first, let's just make sure we're understanding what's going on here. So, there was a potato. We took six cubes from that potato and we stuck those six cubes into six different sucrose
molarity solutions, and so, this data point right over here, this was the situation where
we took one of the cubes, so, this was a sucrose solution. This was a solution, actually,
that contained no sucrose, and so, when we put the
cube in that solution, we saw a net gain of mass. It looks like it's about 22% gain in mass, and so, that would have happened because water would have
flowed into the cube. Now, at the other
extreme, right over here, this is a solution that
has a lot of sucrose. It had a very high sucrose concentration, and when we put a cube in there, we see that the mass of
that cube went down by 25%, and that would have been because of the net outflow
of water from that cube. So, how do we figure
out the water potential of the core cubes at 23 degrees Celsius? Well, we could think about a situation where there's some sucrose
concentration where, if the cube and the sucrose solution have the same water potential, then you're not going to have
any net inflow or outflow, and so, where do we see that on the graph? Well, what we'd wanna
do, we have that line where they're trying
to fit the data points, and so, where would we expect
to see 0% change in mass? So, we would go right over
here to 0% change in mass. We would go to the line right over there, and then, we see that this line would say that there's a 0% change in mass. See, if this is 0.4 right over here, this is 0.5 right over here, so, this is about a 0.44
molar sucrose solution, 0.44 molar solution. So, if we can figure
out the water potential of this 0.44 molar sucrose solution, well, that's also going to be the water potential of the potato cubes. Well, how do we do that? Well, we've seen the equation before where we introduced ourselves to the idea of water potential, that water potential,
using the Greek letter psi, is going to be equal
to the solute potential plus the pressure potential. Now, we're dealing with
all open containers. We don't have anything that's some piston or something that's pressing
down on these containers, and so, because of that, the pressure potential is
going to be equal to zero, and so, we just have to figure
out the solute potential. So, the solute potential,
we have introduced ourselves to this formula in previous videos. It's negative i times C times R times T. This i right over here, this
is our ionization constant. This is, since we're dealing
with sucrose solutions, it says okay, if I took
sucrose and put it into water, every one of those sucrose molecules, does it stay one molecule,
or does it disassociate? Well, sucrose doesn't disassociate at all. It just stays one molecule,
so, this would be one. If we were dealing with,
say, sodium chloride, each sodium chloride
molecule would disassociate into a sodium ion and a chloride ion, and so, then, this would be two, but this was one for sucrose. C is the molarity of our solution, and so, we estimated that to be 0.44, so, let me write this down. Our solute water potential is going to be equal to negative one times 0.4, 0.44, I should say, and
that's going to be moles. I'll write out all the units. Moles per liter times,
it's sometimes called the pressure constant in this context, but this is also the
universal gas constant, and if you were doing
something like the AP exam, they would give you what this is. So, this is 0.0831 liters times bars, all of that over mole Kelvin. If you're used to seeing
other values of this, it's probably because they're dealing with other units right over here, but this is the universal gas constant. And then, we have to multiply that times the temperature that we're
dealing with in Kelvin. Now, it's 23 degrees Celsius. To convert to Kelvin, we just add to 273, so, 273 plus 23 is going
to be 296, 296 Kelvin, and so, this is going to be equal to, we have a negative here, and
we could look at the units. We have the liter canceling out liters, moles canceling out with moles, Kelvin canceling out with Kelvin, so, we're going to get something
in bars, which makes sense. That is the unit for our water potential. And then, we get the calculator out. So, we have 0.44 times
0.831 times 296, 296, is equal to, and they want us to round our answer to one decimal place, so, approximately 10.8, and we already had that
negative out front, so, negative 10.8 bars, and we're done.

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