A 3.4g sample of h2o2 solution containing x% H2O2 by mass requires x ml of a KMnO4 solution for complete oxidation under acidic condition. the molarity of kmno4 solution is: a)1 b)0.5 c)0.4 d)0.2

Answer 1

I think I know what the author of the problem expects; however, I don't think this is a viable problem BECAUSE you have two unknowns. That is you have x mL KMnO4 as well as y% H2O2 in the 3.4 g sample. That author assumes the x%(which is in grams) and x mL is the same and they can't be. But I think the following is what the author expects so here is the solution.
2KMnO4 + 5H2O2 + 3H2SO4 =>2MnSO4 + K2SO4 + 8H2O + 5O2

Assuming the x values CAN be the same let's try the 4 answers and see if they fit. For 1 M, we have 1M(xmL) = 1 millimils KMnO4. That will equal (5x/2) mmols H2O2 and (5x/2)*0.034(the millimolar mass H2O2) = 0.085x g H2O2. Then
[(0.085x g)/3.4]*100 = x%. Check it out and the left side doesn't equal the right side.
Do the same for 0.5M, 0.4M and 0.2M
One of them matches. Post your work if you get stuck.

Answer 2

assumes the x%(which is in grams) and x mL is the same and they can't be. But I think the following is what the author expects so here is the solution.
2KMnO4 + 5H2O2 + 3H2SO4 =>2MnSO4 + K2SO4 + 8H2O + 5O2

Assuming the x values CAN be the same let's try the 4 answers and see if they fit. For 1 M, we have 1M(xmL) = 1 millimils KMnO4. That will equal (5x/2) mmols H2O2 and (5x/2)*0.034(the millimolar mass H2O2) = 0.085x g H2O2. Then
[(0.085x g)/3.4]*100 = x%. Check it out and the left side doesn't equal the right side.
Do the same for 0.5M, 0.4M and 0.2M
One of them matches. Post your work if you get stuck.