What volume of 1.53 M KMnO4 , in liters, contains 164g of solute? Volume=___ L

Answer 1

First, find the molar mass of the solute, which is KMnO₄:

On the periodic table, we can see that the molar mass of K is 39.10 g/mol, the molar mass of Mn is 54.94 g/mol, and the molar mass of O is 16.00 g/mol. There are 1 K, 1 Mn, and 4 O's in KMnO₄ so 39.10 + 54.94 + 4(16.00) = 158.04 g/mol of KMnO₄ which is the same as 158.04 g /1 mol of KMnO₄.

Since we're dealing with molarity (M), we need moles. We just found g/mol and we're given 164 g in the problem, so we need to multiply 164 g by mol/g in order to cancel the grams and get moles. That tells us to flip our g/mol fraction from above upside-down to get mol/g. If we didn't flip it, the grams wouldn't cancel.

164 g of KMnO₄ * 1 mol of KMnO₄ / 158.04 g of KMnO₄ = 1.04 mol of KMnO₄

Next, use the fact that M = mol/L to solve for L (the volume in liters):

1.53 M = 1.04 mol/L --> multiply both sides by L and divide by 1.53 M to isolate L --> L = 0.680

So, the volume is 0.680 L.