1.

The volume of 0.1 M oxalic acid that can be completely oxidized by 20 mL of 0.025 M KMnO_(4) solution is

Answer»

125 mL
25 mL
12.5 mL
37.5 mL

Solution :The BALANCED equation is
`2 KMnO_(4)+3 H_(2)SO_(4)+5 H_(2)C_(2)O_(4) rarr K_(2)SO_(4)+2 MnSO_(4)+8 H_(2)O+10 CO_(2)`
`:'` Molarity `=n_(B)/V_(ml)xx1000`
`:. 0.025=n_(B)/20xx1000`
`:. n_(B)=0.025/50=5xx10^(-4)` mol of `KMnO_(4)`
SINCE 2 moles of `KMnO_(4)` completely oxidises 5 moles of oxalic ACID
`:. 5xx10^(-4)` mol of `KMnO_(4)` will completely oxidize
`=5/2xx5xx10^(-4)` mol of oxalic acid
`=1.25xx10^(-3)` mol of oxalic acid
`:'` Molarity `=n_(B)/V_(ml)xx1000`
`:. V_(ml)=n_(B)/("molarity")xx1000=(1.25xx10^(-3))/0.1 xx1000`
`( :' " Molarity of oxalic acid"=0.1 M)`
`=12.5 ml or 12.5 CM^(3)`.


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