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 find out how much water is required to make a saturated solution of barium nitrate at 100°C, we can use the given solubility information.
The solubility of barium nitrate at 100°C is 34 g/100 g of water. This means that 100 g of water can dissolve 34 g of barium nitrate.
If we start with 100 g of barium nitrate, we need to determine the amount of water required to dissolve it completely. We can set up a proportion using the solubility information:
(34 g of Ba(NO3)2) / (100 g of H2O) = (100 g of Ba(NO3)2) / (x g of H2O)
Cross-multiplying, we get:
34g * x = 100 g * 100 g
x = (100 g * 100 g) / 34 g
x ≈ 294 g of H2O
Therefore, to make a saturated solution of barium nitrate at 100°C, approximately 294 g of water is required.
Now, let's determine how much barium nitrate precipitates out of solution when the saturated solution is cooled to 0°C.
The solubility of barium nitrate at 0°C is 5.0 g/100 g of water. This means that 100 g of water can dissolve 5.0 g of barium nitrate.
The amount of water in the saturated solution at 0°C will still be 294 g, as water does not evaporate during the cooling process.
To find out how much barium nitrate precipitates, we can subtract the amount of barium nitrate that remains dissolved at 0°C from the initial amount added.
The initial amount of barium nitrate added was 100 g.
We can set up a proportion using the solubility information at 0°C:
(5.0 g of Ba(NO3)2) / (100 g of H2O) = (x g of Ba(NO3)2) / (294 g of H2O)
Cross-multiplying, we get:
5.0 g * 294 g = 100 g * x
1470 g = 100 g * x
x = 1470 g / 100 g
x ≈ 14.7 g of Ba(NO3)2
Therefore, when the saturated solution is cooled to 0°C, approximately 14.7 g of barium nitrate will precipitate out of solution.
Additionally, since the precipitated crystals carry along with them 4 g of water per 100 g of crystals, we can calculate the amount of water carried by the precipitated barium nitrate.
The amount of barium nitrate precipitated is approximately 14.7 g.
(4 g H2O / 100 g Ba(NO3)2) x (14.7 g Ba(NO3)2) ≈ 0.588 g of H2O.
Therefore, approximately 0.588 g of water will be carried along with the precipitated barium nitrate crystals.
To determine how much water is required to make a saturated solution of barium nitrate at 100°C, we need to find the amount of barium nitrate that can dissolve in the given amount of water at that temperature.
The solubility of barium nitrate at 100°C is given as 34 g/100 g of H2O. This means that 100 g of water can dissolve 34 g of barium nitrate.
Since we have 100 g of barium nitrate, we can use the solubility to calculate the amount of water needed.
Amount of water required = (mass of barium nitrate) / (solubility of barium nitrate at 100°C)
= 100 g / (34 g/100 g)
= 294.12 g
Therefore, to make a saturated solution of barium nitrate at 100°C with 100 g of Ba(NO3)2, you would need approximately 294.12 g of water.
Now, let's move on to the second part of the question. We have a saturated solution of barium nitrate at 100°C, and we cool it down to 0°C. We need to determine how much Barium Nitrate (Ba(NO3)2) will precipitate out of the solution.
To find this, we need to calculate the difference between the initial amount of barium nitrate dissolved at 100°C and the solubility of barium nitrate at 0°C.
The solubility of barium nitrate at 0°C is given as 5.0 g/100 g of H2O. This means that 100 g of water can dissolve 5.0 g of barium nitrate at 0°C.
The initial amount of barium nitrate dissolved at 100°C is 100 g.
Amount of barium nitrate precipitated = (initial amount of barium nitrate) - (solubility of barium nitrate at 0°C)
= 100 g - (5.0 g/100 g)
= 100 g - 5 g
= 95 g
Therefore, when the saturated solution of barium nitrate at 100°C is cooled down to 0°C, approximately 95 g of barium nitrate will precipitate out of the solution.
Lastly, we are given that the precipitated crystals carry along with them on their surface 4 g of water per 100 g of crystals. Since we have 95 g of barium nitrate precipitated, we can calculate the amount of water carried along with it.
Amount of water carried along = (amount of barium nitrate precipitated) * (amount of water per 100 g of crystals)
= 95 g * (4 g/100 g)
= 3.8 g
Therefore, approximately 3.8 g of water will be carried along with the 95 g of precipitated barium nitrate crystals.