How many grams of KMnO4 should be used to prepare 2.00 L of a 0.500M solution?

To calculate the grams of KMnO4 needed to prepare a 0.500M solution, we need to use the equation:

Molarity (M) = (moles of solute) / (liters of solution)

First, let's rearrange the equation to solve for moles of solute:

moles of solute = Molarity (M) × liters of solution

Since we are given the molarity (M) as 0.500M and the liters of solution as 2.00 L, we can substitute these values into the equation.

moles of solute = 0.500M × 2.00 L
moles of solute = 1.00 mol

Now, to find the grams of KMnO4 needed, we need to use the molar mass of KMnO4. The molar mass is the sum of the atomic masses of each element in the compound.

Molar mass of KMnO4 = (1 × atomic mass of K) + (1 × atomic mass of Mn) + (4 × atomic mass of O)

The atomic masses are:
K: 39.10 g/mol
Mn: 54.94 g/mol
O: 16.00 g/mol

Substituting these values into the equation, we can calculate the molar mass of KMnO4:

Molar mass of KMnO4 = (1 × 39.10 g/mol) + (1 × 54.94 g/mol) + (4 × 16.00 g/mol)
Molar mass of KMnO4 = 158.04 g/mol

Finally, to find the grams of KMnO4 needed, we can use the equation:

grams of KMnO4 = moles of solute × molar mass of KMnO4

Substituting the values we have:

grams of KMnO4 = 1.00 mol × 158.04 g/mol
grams of KMnO4 = 158.04 g

Therefore, 158.04 grams of KMnO4 should be used to prepare 2.00 L of a 0.500M solution.

To determine the number of grams of KMnO4 needed to prepare the solution, you need to use the formula:

\( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters (L)}} \)

Rearranging the formula, we get:

\( \text{moles of solute} = \text{Molarity (M)} \times \text{volume of solution in liters (L)} \)

First, convert the given volume of the solution to liters:

2.00 L = 2.00 L

Next, substitute the values into the formula:

\(\text{moles of solute} = 0.500 \, \text{M} \times 2.00 \, \text{L}\)

Simplifying the calculation:

\(\text{moles of solute} = 1.00 \, \text{mol}\)

To find the number of grams, you need to use the molar mass of KMnO4, which is:

\( \text{Molar mass of KMnO4} = 39.10 \, \text{g/mol (K)} + 54.94 \, \text{g/mol (Mn)} + 16.00 \, \text{g/mol (O)} \times 4 \)

Simplifying the calculation:

\( \text{Molar mass of KMnO4} \approx 158.04 \, \text{g/mol} \)

Now, use the molar mass to calculate the grams of KMnO4:

\( \text{grams of KMnO4} = \text{moles of solute} \times \text{molar mass of KMnO4} \)

Substituting the known values:

\( \text{grams of KMnO4} = 1.00 \, \text{mol} \times 158.04 \, \text{g/mol} \)

Simplifying the calculation:

\( \text{grams of KMnO4} = 158.04 \, \text{g} \)

Therefore, you need approximately 158.04 grams of KMnO4 to prepare 2.00 L of a 0.500 M solution.

How many moles do you need? That is M x L = moles.

Then grams = moles/molar mass