At 50 degrees, 100 grams of water is saturated with cerium (III) sulfate. How many grams of cerium (III) sulfate must be added to saturate the solution at zero degrees celsius?
To determine how many grams of cerium (III) sulfate must be added to saturate the solution at zero degrees Celsius, we need to use the solubility data for cerium (III) sulfate.
At 50 degrees Celsius, 100 grams of water is saturated with cerium (III) sulfate. We can assume that all of the 100 grams of water is the solvent, as the cerium (III) sulfate is the solute in this case.
To obtain the amount of cerium (III) sulfate dissolved in the 100 grams of water, we need to look up the solubility of cerium (III) sulfate at 50 degrees Celsius.
Let's assume that at 50 degrees Celsius, the solubility of cerium (III) sulfate is x grams per 100 grams of water.
So, at 50 degrees Celsius, the solution contains x grams of cerium (III) sulfate dissolved in 100 grams of water.
Now, we need to find the solubility of cerium (III) sulfate at 0 degrees Celsius.
Let's assume that at 0 degrees Celsius, the solubility of cerium (III) sulfate is y grams per 100 grams of water.
Since we want to saturate the solution at 0 degrees Celsius, we need to determine how many grams of cerium (III) sulfate are needed to reach this solubility.
Using the relationship between the solubities of cerium (III) sulfate at 50 degrees Celsius and 0 degrees Celsius, we can set up the following proportion:
x grams / 100 grams = y grams / 100 grams
Cross-multiply to solve for y:
100 * x = 100 * y
y = x
So, the solubility of cerium (III) sulfate at 0 degrees Celsius is the same as the solubility at 50 degrees Celsius, which is x grams per 100 grams of water.
Thus, to saturate the solution with cerium (III) sulfate at zero degrees Celsius, we need to add x grams of cerium (III) sulfate to 100 grams of water, assuming the same solubility as at 50 degrees Celsius.
To determine the grams of cerium (III) sulfate needed to saturate the solution at zero degrees Celsius, we need to understand the concept of solubility and how it is affected by temperature.
Solubility refers to the maximum amount of solute that can dissolve in a given amount of solvent at a particular temperature. Typically, as the temperature decreases, the solubility of most substances decreases as well.
In this case, we know that at 50 degrees Celsius, 100 grams of water is saturated with cerium (III) sulfate. This means that 100 grams of water can dissolve the maximum amount of cerium (III) sulfate at that temperature.
To find out how many grams of cerium (III) sulfate are needed to saturate the solution at zero degrees Celsius, we need to find the solubility value for cerium (III) sulfate at both temperatures (50 degrees Celsius and zero degrees Celsius) and compare them.
The next step would be to search for the solubility data for cerium (III) sulfate at different temperatures. This information can be found through reliable sources such as textbooks, scientific literature, or online databases. Let's assume we find the following solubility values:
- At 50 degrees Celsius: 40 grams of cerium (III) sulfate dissolved in 100 grams of water.
- At 0 degrees Celsius: 20 grams of cerium (III) sulfate dissolved in 100 grams of water.
Now we can calculate the grams of cerium (III) sulfate needed at zero degrees Celsius. Since we want to saturate 100 grams of water, and at 0 degrees Celsius, the solubility is 20 grams of cerium (III) sulfate per 100 grams of water, we can set up a proportion:
(20 grams of cerium (III) sulfate / 100 grams of water) = (x grams of cerium (III) sulfate / 100 grams of water)
Cross-multiplying the proportion:
20 grams of cerium (III) sulfate * 100 grams of water = 100 grams of water * x grams of cerium (III) sulfate
2000 = 100x
Dividing both sides by 100:
20 = x
Therefore, to saturate 100 grams of water at zero degrees Celsius, 20 grams of cerium (III) sulfate must be added.
Please note that the given solubility values in this explanation are hypothetical. In practice, it is necessary to refer to reliable sources to find accurate solubility data.