A chemist has a supply of supersaturated CuSO4 solution that contains 230.0grams per litre of water. The chemist also has another supply of copper sulphate solution with a concentration of 150.0 grams per litre. What quantities of each copper sulphate solution should be mixed together to make 2L of a 200.0grams per litre solution of Copper Sulphate?
This can be done??
If you use 2L of the weaker solution of 150g/L and zero L of the more concentrated solution, then you have 300g/2L and that is too concentrated.
I see a problem below that has been answered by Mathmate that isn't the same problem as this one. This one doesn't allow water to be added and both of the solutions available are more concentrated than the final desired one. I don't think this one is possible.
To find the quantities of each copper sulphate solution that should be mixed, we need to set up an equation based on the principles of concentration and mass conservation. Let's denote the following variables:
- Let x represent the quantity (in liters) of the supersaturated CuSO4 solution to be used.
- Let y represent the quantity (in liters) of the copper sulphate solution with a concentration of 150.0g/L to be used.
Now let's set up the equation:
Total volume equation: x + y = 2L
Mass conservation equation: (mass of CuSO4 in x) + (mass of CuSO4 in y) = (total mass of CuSO4 in the final mixture)
We can calculate the mass of CuSO4 in each solution by multiplying the concentration by the volume:
Mass of CuSO4 in the supersaturated solution = 230.0 g/L * x
Mass of CuSO4 in the 150.0 g/L solution = 150.0 g/L * y
Since we want a final mixture with a concentration of 200.0 g/L, we can write the total mass equation:
(230.0 g/L * x) + (150.0 g/L * y) = (200.0 g/L * 2L)
Now we have a system of equations:
x + y = 2
(230.0 * x) + (150.0 * y) = (200.0 * 2)
We can solve this system of equations using substitution or elimination method. I will use the substitution method here.
From the first equation, we can express x in terms of y as x = 2 - y.
Substituting this into the second equation, we get:
(230.0 * (2 - y)) + (150.0 * y) = (200.0 * 2)
Simplifying the equation:
460.0 - 230.0y + 150.0y = 400.0
Combining like terms:
-80.0y = -60.0
Dividing both sides by -80.0:
y = 0.75
Now substitute the value of y back into x = 2 - y to find x:
x = 2 - 0.75
x = 1.25
So, to make a 2L solution of 200.0 g/L copper sulphate, the chemist should mix 1.25 liters of the supersaturated CuSO4 solution with 0.75 liters of the 150.0 g/L solution.