Answer:
KMnO4
Add 79.0 grams of KMnO4 to a flask and then add 500 ml to make a 0.5L of a 1.0M KMnO4 solution.
Step-by-step explanation:
The correct way to write potassium permanganate is to use the proper periodic table symbols, which include the initial capital followed by any lower case characters, if needed.
KMnO4
We are asked to prepare 0.500 liters of a 1 M KmNO4 solution. First, lets find the molar mass of KMnO4 by adding the atomic masses for each atom in the molecule. I arrive at 158.0, and we can replace ATM with grams/mole. 1 mole of KMnO4 amounts to 158.0 grams of the substance.
Next, remember the definition of M (Molar). I Molar is defined as 1 mole/liter. #M would mean a solution that has 3 moles for every liter (or 1000ml).
Lets write 1.0M as 1.0 moles/liter.
We would need to dissolve 158.0 grams (1 mole of KMnO4) in 1 liter to make a 1M solution of KMnO4.
But we are only asked for 0.500 liter, which is 1/2 the amount we just used. Instead of 1 liter, 500 ml would only require (1/2) the moles of KMnO4, or 79.0 grams KMnO4. Add 79.0 grams of KMnO4 to a flask and then add 500 ml to make a 0.5L of a 1.0M KMnO4 solution.
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Another approach is to write an equation based on the definition of Molar (mole/liter). It is not easy to follow since the units tend to add too much clutter, but all the conversion factors and relationships can be expressed in one equation, where X is the grams KMNo4 needed to make 0.5L of a 1.0M KMnO4 solution.
(X grams KMnO4)/(158.0 g/mole)/(0.5L) = (1.0 moles KmNO4)/liter
X/158g/mole/0.5L = 1 mole/L
X/158g/mole= (1 mole/L)*(0.5L )
X/158g/mole= (1 mole)*(0.5 )
X= (1 mole)*(0.5 )*(158g/mole)
X = (1)*(0.5 )*(158g)
X = 79 grams KMnO4 The same answer as before.