Question Video: Calculating the Solubility of Glucose from Experimental Data Chemistry

A student prepares a saturated solution of glucose in water at 20°C. 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°C? Give your answer as a whole number.

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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.

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