Zinc iodide is dissolved in 2.5 L of water to make a solution. 1.00 L of this solution is then diluted to make 12.0 L of a 0.50 M solution. What mass of zinc was added to make the original solution?
mols ZnI2 in final solution is
mols ZnI2 = M x L = 0.50 x 12.0L = 6.0 mols. That also = mols in the 1.00 L taken for dilution to the 12.0 L.
This 6.0 mols in the 1.00 L taken for dilution came from 2.5 L of the original; therefore, the original solution must have had 6.0 x 2.5/1.0 = 15 mols ZnI2 to start. g ZnI2 = mols ZnI2 x molar mass ZnI2 = ?g in the original.
Check: grams ZnI2/molar mass ZnI2 = 15.0 mols. Place that in 2.5 L so you have 15.0 mols/2.5L = 6 molar solution.
You take 1.00 L of this and dilute to 12.0 L. The final concentration is
6.0 M x (1.0 L/12.0) = 0.50 M and that's the final concentration in the problem.
Well, that's an interesting question. Let me calculate it for you, but let's hope I don't end up with a zincy situation!
To find the mass of zinc added, we first need to determine the number of moles of zinc iodide in the solution. We can use the equation:
Molarity = moles of solute / volume of solution
Rearranging the equation, we get:
moles of solute = Molarity × volume of solution
So, the moles of zinc iodide in the original solution are:
moles = 0.50 M × 1.00 L = 0.50 mol
Now we can calculate the molar mass of zinc iodide, which is:
Zinc (Zn): 1 atom × 65.38 g/mol = 65.38 g/mol
Iodine (I): 2 atoms × 126.90 g/mol = 253.80 g/mol
Adding those two masses together, we get:
65.38 g/mol + 253.80 g/mol = 319.18 g/mol
Multiplying the moles of zinc iodide by the molar mass, we find the mass of zinc added:
mass = moles × molar mass
mass = 0.50 mol × 319.18 g/mol
mass = 159.59 g
So, approximately 159.59 grams of zinc were added to make the original solution. That's a pretty hefty amount. I hope the zinc was up to the task!
To find the mass of zinc added to make the original solution, we need to use the given information.
Let's start by determining the number of moles of zinc iodide in the final 12.0 L solution.
Molarity (M) is defined as moles of solute per liter of solution. Therefore, we can calculate the number of moles of zinc iodide in the 12.0 L solution using the equation:
moles of solute = molarity × volume of solution
moles of solute = 0.50 M × 12.0 L = 6.00 moles
Since zinc iodide is a 1:1 stoichiometric compound, the number of moles of zinc in the solution is also equal to 6.00 moles.
Now, let's consider the dilution process. We know that 1.00 L of the original solution was diluted to make 12.0 L of the final solution. By the principle of dilution, the number of moles of zinc in both solutions remains the same.
Therefore, the moles of zinc iodide in the original solution are also 6.00 moles.
Next, we need to calculate the mass of zinc added to make the original solution. We can use the atomic mass of zinc to convert the moles of zinc into grams.
The atomic mass of zinc (Zn) is approximately 65.38 g/mol.
mass of zinc = moles of zinc × atomic mass of zinc
mass of zinc = 6.00 moles × 65.38 g/mol = 392.3 grams
Therefore, approximately 392.3 grams of zinc was added to make the original solution.
To find the mass of zinc added to make the original solution, we need to use the information given and perform some calculations.
Let's break down the problem step by step:
Step 1: Find the number of moles of zinc iodide in the 1.00 L solution.
The given concentration of the 1.00 L solution is 0.50 M, which means there are 0.50 moles of zinc iodide for every 1 L of solution.
Therefore, the number of moles in the 1.00 L solution is 0.50 moles.
Step 2: Calculate the number of moles in the original solution.
Since we were given that 1.00 L of the original solution was diluted to make 12.0 L, we can think of it as a ten-fold dilution.
So, the number of moles in the original solution is 10 times the number of moles in the 1.00 L solution:
Number of moles in the original solution = 10 * 0.50 moles = 5.00 moles.
Step 3: Find the molar mass of zinc iodide.
Zinc iodide is a compound consisting of one zinc atom (Zn) and two iodine atoms (I). To find the molar mass, we need to add up the atomic masses of these atoms.
The atomic mass of zinc (Zn) is 65.38 g/mol, and the atomic mass of iodine (I) is 126.90 g/mol. So, the molar mass of zinc iodide (ZnI2) is:
Molar mass of ZnI2 = (1 * 65.38 g/mol) + (2 * 126.90 g/mol) = 318.18 g/mol.
Step 4: Calculate the mass of zinc in the original solution.
To find the mass of zinc, we need to multiply the number of moles of zinc iodide by the molar mass of zinc iodide:
Mass of zinc = Number of moles * Molar mass = 5.00 moles * 318.18 g/mol = 1590.90 g.
So, the mass of zinc added to make the original solution is 1590.90 grams.