1.

The molarity (m) of `KMnO_(4)` in the acidic medium is (densityh of `KMnO_(4)` solution`=1.58gmL^(-1)Mw(KMnO_(4))=158gmol^(-1))`A. 0.025B. 0.25C. 0.12D. 0.012

Answer» Correct Answer - D
`d_(sol)=M((Mw_(2))/(1000)+(1)/(m))`
`M=0.02`
`1.58=0.02((158)/(1000)+(1)/(m))`
Solve for m:
`m=0.012`


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