The solubility of barium nitrate at 100°C is 34 g/100 g of H2O and at 0°C is 5.0 g/100g of H2O. If you start with 100 g of Ba(NO3)2 and make a saturated solution in water at 100°C, how much water is required? If the saturated solution is cooled to 0°C, how much Ba(NO3)2 is precipitated out of solution? The precipitated crystals carry along with them on their surface 4 g of H2O per 100 g of crystals.
To dissolve 100 g Ba(NO3)2 @ 100 C we need
100 g H2O x (100 g/34 g) = ?? g H2O.
When the solution cools to 0 C we get this much Ba(NO3)2 as a ppt.
5 g Ba(NO3)2 x (?? g from above/100) = xx g Ba(NO3)2. I get something like 15 g for this but you need to do it more accurately than that. Since the crystals carry extra H2O molecules on their surface (4 g/100 g crystals), the 15 g crystals will have an extra
4 g x (15/100) = 0.6 so the total mass is 15.6 g. Check my thinking.
Well, it seems like you've stumbled upon a chemistry question. Don't worry, I'm here to clown around and help you out!
To determine the amount of water required to make a saturated solution of barium nitrate at 100°C, we need to use its solubility data. At 100°C, the solubility of barium nitrate is 34 g/100 g of water. Since you have 100 g of barium nitrate and you want to saturate the solution, you'll need to figure out how much water is needed to dissolve that.
Let's set up a proportion:
34 g barium nitrate / 100 g water = 100 g barium nitrate / x g water
Cross-multiplying and solving for x, we find that approximately 294 g of water is required to dissolve 100 g of barium nitrate at 100°C. So, now you know how much water you need to make your solution!
Now, let's move on to the second part of the question. You have 100 g of barium nitrate dissolved in 294 g of water at 100°C. If we cool this solution to 0°C, we know that the solubility of barium nitrate drops to 5.0 g/100 g of water.
To find out how much barium nitrate precipitates out of the solution, we need to calculate the amount that exceeds its solubility at 0°C. The excess amount will crystallize and form solid barium nitrate.
First, let's find out how much barium nitrate can dissolve at 0°C using the new solubility:
5.0 g barium nitrate / 100 g water = x g barium nitrate / 294 g water (the original amount of water)
You can solve for x and find that approximately 14.7 g of barium nitrate can dissolve at 0°C. Since you started with 100 g, you can subtract 14.7 g from that to find the amount of barium nitrate precipitated out of the solution.
So, approximately 85.3 g of barium nitrate will precipitate out of the solution as it is cooled down.
Now, remember that the precipitated crystals carry along with them 4 g of water per 100 g of crystals. If we assume that all the barium nitrate precipitates out of the solution, we can calculate the amount of water carried with the crystals.
Using a proportion:
4 g water / 100 g crystals = y g water / 85.3 g crystals
Solving for y, we find that approximately 3.45 g of water will be carried along with the precipitated barium nitrate crystals.
So, to recap: you'll need approximately 294 g of water to make a saturated solution of 100 g of barium nitrate at 100°C. When the solution is cooled to 0°C, approximately 85.3 g of barium nitrate will precipitate out, carrying along with it approximately 3.45 g of water.
I hope I was able to help you with your chemistry conundrum! If you have any more questions or need further clown-tastic assistance, feel free to ask!
To find out how much water is required to make a saturated solution of barium nitrate at 100°C, we need to calculate the amount of water that can dissolve 100 g of Ba(NO3)2 at that temperature.
Given:
Solubility of barium nitrate at 100°C = 34 g/100 g of H2O
Mass of barium nitrate (Ba(NO3)2) = 100 g
To find the amount of water required, we can use the following equation:
Mass of solute (Ba(NO3)2) / Mass of solvent (H2O) = Solubility
Plugging in the values, we have:
100 g / Mass of water required = 34 g/100 g
Cross-multiplying, we get:
Mass of water required = (100 g * 100 g) / 34 g
Mass of water required = 294.12 g (approximately)
Therefore, approximately 294.12 g of water is required to make a saturated solution at 100°C.
To calculate the amount of Ba(NO3)2 that precipitates out of the solution when it is cooled to 0°C, we need to find the difference between the initial amount dissolved and the solubility at 0°C.
Given:
Solubility of barium nitrate at 0°C = 5.0 g/100 g of H2O
Initially, we started with 100 g of Ba(NO3)2. To find the amount of Ba(NO3)2 that remains dissolved, we subtract the solubility at 0°C from the initial amount dissolved:
100 g - 5.0 g = 95.0 g
Therefore, 95.0 g of Ba(NO3)2 remains dissolved in the solution at 0°C. The remaining 5.0 g of Ba(NO3)2 will precipitate out of the solution.
However, it is mentioned that the precipitated crystals carry along with them 4 g of water per 100 g of crystals. So, to calculate the total mass of precipitated crystals, we need to account for this water content.
Mass of precipitated crystals = Mass of Ba(NO3)2 precipitated + Water carried by the crystals
Given:
Mass of Ba(NO3)2 precipitated = 5.0 g
We can calculate the water carried by the crystals using the ratio provided:
Water carried by the crystals = (Mass of Ba(NO3)2 precipitated) * (4 g/100 g)
Plugging in the values, we have:
Water carried by the crystals = 5.0 g * (4 g/100 g)
Water carried by the crystals = 0.2 g
Therefore, the total mass of the precipitated crystals, including the water carried by them, is:
Mass of precipitated crystals = 5.0 g (Ba(NO3)2) + 0.2 g (water carried by the crystals)
Mass of precipitated crystals = 5.2 g
Thus, approximately 5.2 g of Ba(NO3)2 will precipitate out of the solution when cooled to 0°C.
To find out how much water is required to make a saturated solution of barium nitrate at 100°C, we can use the given solubility value of 34 g/100 g of water. The solubility value tells us that at that temperature, 34 grams of barium nitrate can dissolve in 100 grams of water.
Since you start with 100 grams of barium nitrate, you'll need to use the solubility value to determine the amount of water required. We can set up a proportion:
(34 g barium nitrate / 100 g water) = (100 g barium nitrate / x g water)
Simplifying the proportion:
34 / 100 = 100 / x
Cross multiplying:
34x = 100 * 100
x = (100 * 100) / 34
x ≈ 294 g
Therefore, approximately 294 grams of water is required to make a saturated solution of barium nitrate at 100°C.
Now, let's calculate how much barium nitrate will precipitate out of solution when the saturated solution is cooled to 0°C. We can subtract the solubility at 0°C (5.0 g/100 g water) from the solubility at 100°C (34 g/100 g water).
The difference in solubility is 34 g/100 g water - 5.0 g/100 g water = 29 g/100 g water.
Knowing this, we can find the amount of precipitated barium nitrate using a similar proportion:
(29 g barium nitrate / 100 g water) = (x g barium nitrate / 294 g water)
Simplifying the proportion:
29 / 100 = x / 294
Cross multiplying:
29 * 294 = 100 * x
x ≈ 85.2 g
Therefore, approximately 85.2 grams of barium nitrate will precipitate out of solution when the saturated solution is cooled to 0°C.
Additionally, we are told that the precipitated crystals carry along with them 4 g of water per 100 g of crystals. So, to find the total amount of water carried by the precipitated barium nitrate, we can set up another proportion:
(4 g water / 100 g crystals) = (x g water / 85.2 g crystals)
Simplifying the proportion:
4 / 100 = x / 85.2
Cross multiplying:
4 * 85.2 = 100 * x
x ≈ 3.408 g
Therefore, approximately 3.408 grams of water will be carried along with the precipitated barium nitrate crystals.