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Chemistry 21 Online
OpenStudy (kkutie7):

How much KMnO4 is needed to prepare 250mL of 2.5*10^-3 M KMnO4 solution?

OpenStudy (abmon98):

Number of moles=Concentration(mol/dm^3)*Volume(dm^3) convert volume units from cubic decimeter to milliliter 1 dm3 = 1000 ml x dm^3=250 ml Cross multiplication to find x (volume) but in dm^3 instead of ml 250*1/1000*2.5*10^-3=Number of moles of KMnO4

OpenStudy (kkutie7):

Why dm^3 instead of L?

OpenStudy (abmon98):

its the same

OpenStudy (kkutie7):

Let me do the work and see if I got it right. \[250mL*\frac{ 1L}{ 1000mL } = .25L\] \[.25L*\frac{ 2.5*10^{-3}mols }{ L }= 6.25*10^{-4}\] \[6.25*10^{-4}mols*\frac{ 158.032g KMnO4 }{ 1 mol KMnO4 }= .09877g\] Is this right?

OpenStudy (abmon98):

good job :)

OpenStudy (kkutie7):

Thanks for the help

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