Potassium permanganate (KMnO4) solutions are used for the determination of iron in samples of unknown concentration. As a laboratory assistant, you are supposed to prepare 200. mL of a 0.300 M KMnO4 solution. What mass of KMnO4, in grams, do you need?
I got 60, but that's wrong. I know you have to get moles by using the molarity formula and then change it into grams using molar mass, but I can't get the right answer/ Any help is appreciated!
the molar mass of KMnO4 is
39.098 + 54.938 + 15.999*4 = 158.032
200 ml = 200 mL / 1000 mL/L = 0.2 L
0.2 L * 0.3 moles/L = 0.06 Moles KMnO4
= 0.06 * 158.032 g/mol = 9.48 g KMnO4
You must have changed your method because you're no longer reporting 60. You're right, 60 is millimoles KMnO4 you need which is 0.06 mols. Then g = mols x molar mass = 0.06 x 158.034 = 9.48 g KMnO4.
That looks good to me.
It says that the molarity is.3M so you would do .3 mol/1 L*.2L. .6mol*157g=94.2. That's my new answer.
nope. 0.3 x 0.2 is right but that answer is 0.06 and not 0.6 and that x 158.34 (not 157) = 9.48g.
Well, it seems like you've got the right approach! Let's break it down step by step.
First, we need to find the moles of KMnO4 required. We can use the formula:
moles = molarity x volume (in liters)
Given:
Molarity (M) = 0.300 M
Volume (L) = 0.200 L (since 200 mL = 0.200 L)
moles = 0.300 M x 0.200 L
moles = 0.060 mol
Now, we need to find the molar mass of KMnO4. Adding up the atomic masses of potassium (K), manganese (Mn), and oxygen (O):
Molar mass of KMnO4 = 39.1 g/mol + 54.9 g/mol + (16.0 g/mol x 4) = 158.0 g/mol
Finally, we can find the mass of KMnO4 needed using the equation:
mass = moles x molar mass
mass = 0.060 mol x 158.0 g/mol
mass = 9.48 g (rounded to two decimal places)
So, the mass of KMnO4 needed is approximately 9.48 grams.
To find the mass of KMnO4 needed to prepare a 0.300 M solution, you can follow these steps:
1. Start by determining the number of moles of KMnO4 needed using the molarity formula:
Molarity (M) = moles (mol) / volume (L)
Rearrange the formula to solve for moles:
Moles (mol) = Molarity (M) x Volume (L)
In this case, the molarity is given as 0.300 M and the volume is given as 0.200 L (200 mL converted to L):
Moles (mol) = 0.300 M x 0.200 L = 0.060 mol
2. Next, use the molar mass of KMnO4 to convert moles to grams. The molar mass of KMnO4 is calculated by summing the atomic masses of its constituent elements:
Molar mass of K = 39.10 g/mol
Molar mass of Mn = 54.94 g/mol
Molar mass of O = 16.00 g/mol (Note: There are four oxygen atoms in KMnO4, so we multiply the molar mass by four)
Molar mass of KMnO4 = (1 x Molar mass of K) + (1 x Molar mass of Mn) + (4 x Molar mass of O)
= (1 x 39.10 g/mol) + (1 x 54.94 g/mol) + (4 x 16.00 g/mol)
= 39.10 g/mol + 54.94 g/mol + 64.00 g/mol
= 158.04 g/mol
Finally, convert moles to grams using the molar mass:
Mass (g) = Moles (mol) x Molar mass (g/mol)
= 0.060 mol x 158.04 g/mol
≈ 9.48 g
Therefore, you would need approximately 9.48 grams of KMnO4 to prepare 200 mL of a 0.300 M solution.