7 posts • Page 1 of 1
You would need to convert the amount of KMnO4 to moles and the ml into liters. Once you find that, you know the initial molarity and the initial volume then you would use M1V1=M2V2 since you also know the final volume.
You would need to use to molar mass in order to find the number of moles of KMnO4. You would find the molar mass of this molecule using the given values, and use the formula (n=g/MM) to find the number of moles of KMnO4 present. This will help you find the initial molarity and complete the problem.
Convert the 5 grams of KMnO4 into moles by dividing it by its molar mass. Then, since you’ll want to be able to the M1V1=M2V2 formula, divide the number of moles by the volume of the original solution (0.15000 L) - this will be M1. Since they’re only using 20.00 mL of the original solution, V1 will be 0.02000 L. V2 will be the 0.25000 L. Plug the three variables in to solve for M2. :)
This problem demonstrates a little confusion because we have basically 2 problems that take place which causes 3 volume to be there. We have the first problem that is a simple M=n/v to give us the molarity for the second problem which is a MV=MV. Therefore, we see slight hardships. But first convert to moles from grams using 158.04g/mol and then Find molarity with first volume give with the moles found. Then use molarity with .02 equals new molarity times .250 shown in work.
- (13.64 MiB) Downloaded 1 time
Who is online
Users browsing this forum: No registered users and 1 guest