Lets say you have 25 ml of a 3.25M copper(II) chlorate solution. How many moles of copper (II) chlorate would be in the solution and how many moles of the copper(II) ion would be in the solution?
Copper(II) chlorate has the formula of Cu(ClO3)2.
mols Cu(ClO3)2 = M x L = ?
mols Cu(II) ion = mols Cu(ClO3)2 since there is 1 mol Cu ions in 1 mol Cu(ClO3)2.
The problem doesn't ask but how many mols chlorate (ClO3)^- do you have. You will have twice as many as there are mols Cu(ClO3)2 since there are two chlorataes for each copper chlorate.
wewwo
To determine the number of moles of copper(II) chlorate in the solution, we will use the equation:
moles = concentration (M) × volume (L)
First, let's convert the volume from milliliters (ml) to liters (L):
25 ml ÷ 1000 ml/L = 0.025 L
Now we can calculate the number of moles of copper(II) chlorate using the given concentration:
moles of copper(II) chlorate = 3.25 M × 0.025 L = 0.08125 moles
So, there are approximately 0.08125 moles of copper(II) chlorate in the solution.
Next, we need to determine the number of moles of the copper(II) ion in the solution. Copper(II) chlorate dissociates into one copper(II) ion (Cu2+) and two chlorate ions (ClO3-).
From the formula of copper(II) chlorate, Cu(ClO3)2, we can see that there are 2 moles of Cu2+ ions for every 1 mole of copper(II) chlorate.
Therefore, the number of moles of the copper(II) ion in the solution is:
moles of Cu2+ = 2 × 0.08125 moles = 0.1625 moles
So, there are approximately 0.1625 moles of the copper(II) ion in the solution.