Question

The volume of 0.1 Moxalic acid that can be completely
oxidized by 20 mL of 0.025 M KMnO4 solution is
(a) 125 mL
(b) 25 mL
(c) 12.5 mL
(d) 37.5 mL
​

Answer 1

${\red {\huge {\underline {\underline {\mathtt {Answer:-}}}}}}$

$\mathtt{2KMnO_4+5H_2C_2O_4+6H^{+}→2K +^{+} }$

$\mathtt{2Mn^{2+}+10CO_2+8H_2O}$

$\mathtt{\frac{M_{KMnO_4 }V _{KMnO_4}}{2} = \frac{M_{KMnO_4 }V _{KMnO_4}}{2}}$

$\mathtt{\frac{ 0.025M×20ml}{2} = \frac{0 .1M\times V_{H_2C_2O_4}}{2}} </p><p></p><p>$

$\mathtt{V_{H_2 C_2O_4}=12.5ml}$

$\rm{The \:volume\:of \:0.1 M\:oxalic \:acid \: that } \\ \rm {can \: be \: completely \: oxidized \: by \: 20 ml \: of }\\ \rm{ 0.025M \: KMnO_4 \: solution \: is \: 12.5ml}$

12.5ml

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