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An unknown sample required 21.05ml of 0.02047 M KmnO4 to reach the endpoint. How many moles of KmnO4 reacted?(M=mol over liter) A.0.43089 MOL B.0.04309 MOL C.0.004309 MOL D.0.0004309 MOL
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@Legends
use the formula for molarity and rearrange it to solve for moles. \(\sf \large Molarity=\dfrac{moles}{L}\)
um. . . i got 9.72. @aaronq
did you convert mL to L?
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