### Video Transcript

A student prepares a saturated
solution of glucose in water at 20 degrees Celsius. The student then dries the solution
to remove the solvent and obtain the solid. By using the experimental data in
the table provided, what is the solubility of glucose at 20 degrees Celsius? Give your answer as a whole
number. Mass of evaporating dish is 43.56
grams. Mass of evaporating dish plus
saturated solution is 56.33 grams. And mass of evaporating dish plus
solid or glucose is 49.61 grams.

We are using the provided data to
determine the solubility of glucose. Solubility can be defined as the
maximum amount of solute that can dissolve in a given amount of solvent at a given
temperature. Solubility is often expressed as an
amount of solute in grams that can dissolve per 100 grams of water at a given
temperature. To determine the solubility, the
student has prepared a saturated solution, which is a solution that contains the
maximum amount of solute dissolved in a solvent at a given temperature. In other words, when a solution is
saturated, the limit of solubility has been reached. And at that temperature no more
additional solute can be dissolved in the solvent. We can determine the solubility of
our compound glucose using the data collected from the saturated solution
prepared.

Our solute in this question is
glucose. Our solvent in which the glucose is
dissolved is water. And the given temperature is 20
degrees Celsius. Let’s have a look at the data. We are given the mass of an
evaporating dish only. The student prepared the saturated
solution in the evaporating dish, the combined mass of which we are also given, the
solution was then dried to remove the solvent likely by heating, leaving the
remaining solid which was glucose in the evaporating dish. We are also given this combined
mass of the evaporating dish and the glucose. We can use these three masses to
find the mass of just the solute and to find the mass of only the solvent, which
will allow us to calculate solubility.

Let’s first find the mass of our
solute. To find the mass of just our solute
glucose, we can take the mass of the evaporating dish and the solid glucose and
subtract the mass of the evaporating dish only, which will give us the mass of our
solid glucose alone. So the mass of our glucose will be
49.61 grams minus 43.56 grams, which equals 6.05 grams. Now that we have found the mass of
our solute alone, let’s find the mass of our solvent. We can find the mass of our solvent
water by taking the mass of the evaporating dish plus the saturated solution, which
contains both the dissolved glucose and water and subtracting the mass of the
evaporating dish plus the solid glucose. This will give us the mass of our
solvent, so the mass of our solvent will be 56.33 grams minus 49.61 grams, which
equals 6.72 grams.

Let’s now take the mass of our
solute and mass of our solvent to solve for the solubility. The experimental data shows us that
6.05 grams of glucose can be dissolved in 6.72 grams of water at 20 degrees
Celsius. However, this is not our final
answer, since solubility is generally expressed in grams of solute per 100 grams of
water. By setting up a proportion using
our experimental data of 6.05 grams of glucose dissolved per 6.72 grams of water, we
can solve for the amount in grams of glucose dissolved per 100 grams of water. Using our proportion, we can cross
multiply to help us solve for 𝑥.

Let’s begin by multiplying 6.05 by
100. For now, we will ignore the units
of these measurements. The result is 605. Next, let’s multiply 𝑥 by 6.72,
which gives us 6.72𝑥. The resulting equation is 605
equals 6.72𝑥. To isolate 𝑥, we can divide both
sides of the equation by 6.72. When we divide both sides by 6.72,
we get that 𝑥 equals 90.0298. Now we need to round our answer to
the nearest whole number. This gives us 𝑥 equals 90. Since 𝑥 was the mass of our
solute, glucose, we have found from the experimental data that the solubility of
glucose at 20 degrees Celsius given to a whole number is 90 grams per 100 grams of
water.