Question #16097

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Aug 28, 2016

#"Molar mass of "KMnO_4#
#=39+55+4*16=158g/"mol"#

#0.02(M) KMnO_4#
#=0.02"mol"/Lxx158g/"mol"=3.16g/L" "KMnO_4#

So to make 1L 0.02M solution(secondary standard) approximately 3.16g of #KMnO_4# is taken in a 1L volumetric flask and then dissolved in water to make volume 1L.

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Question #16097

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