The solubility of copper(ii) tetraoxosulphate vi at 85°c is 60g in 100g and at 15°c is 18.8g in 100g. if 120g of solution saturated at 85°c are cooled to 150°c, what mass of copper(ii) sulphate crystal would be deposted?
Answer
Your question makes no sense. If you start with T = 85 and you COOL it to 150, that isn't cooling it very much. I assume you meant it was cooled to 15 C. When you correct and repost make sure you call the compound by a correct name. One correct name, approved by the International Union of Pure and Applied Chemicaty (IUPAC) is copper (II) sulfate. The name you used is not accepted by IUPAC.
Ok
Tell me question that we come out in 3rd term exam for ss2
Assuming you meant it was cooled to 15 C, then
60 g in 100 @ 85 and 18.8 g om 100 @ 15 C.
So how much CuSO4 is in the saturated solution @ 85 C. That's
60 x 120/100 = 72 grams. How much will be dissolved @ 15 C. That's
18.8 x 120/100 = 22.6 g
Start with 72 g - 22.6 g dissolved = ? grams CuSO4 deposited.
where did you learn to call it copper(ii) tetraoxosulfate (vi)?
To determine the mass of copper(II) sulfate crystal deposited when the saturated solution is cooled from 85°C to 15°C, we need to calculate the difference in solubility between the two temperatures and use that to find the amount of copper(II) sulfate that will precipitate out.
First, let's find the amount of copper(II) sulfate dissolved in the solution at 85°C. The solubility is given as 60g in 100g of solution. Therefore, for every 100g of solution, there are 60g of copper(II) sulfate dissolved.
We are given that there are 120g of the saturated solution. To calculate the amount of copper(II) sulfate dissolved in this solution at 85°C, we can set up a proportion:
(60g / 100g) = (x / 120g)
Cross-multiplying the above equation, we have:
100x = 60 * 120
100x = 7200
x = 7200 / 100
x = 72g
So, at 85°C, there are 72g of copper(II) sulfate dissolved in 120g of the solution.
Next, we need to find the amount of copper(II) sulfate that will remain dissolved when the solution is cooled to 15°C. The solubility at this temperature is given as 18.8g in 100g of solution.
Using a similar proportion, we can calculate the amount of copper(II) sulfate that remains dissolved:
(18.8g / 100g) = (y / 120g)
Cross-multiplying, we get:
100y = 18.8 * 120
100y = 2256
y = 2256 / 100
y = 22.56g
Therefore, at 15°C, there will be approximately 22.56g of copper(II) sulfate dissolved in the 120g of the solution.
Finally, to find the mass of copper(II) sulfate crystal deposited when the saturated solution is cooled, we subtract the amount that remains dissolved at 15°C from the initial amount dissolved at 85°C:
Mass of copper(II) sulfate deposited = Initial mass - Remaining mass
= 72g - 22.56g
= 49.44g
Therefore, approximately 49.44g of copper(II) sulfate crystal would be deposited when the 120g of solution is cooled from 85°C to 15°C.