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

Options

A. 125mL
B. 25mL
C. 12.5mL
D. 37.5mL

Answer

C

Explanation

The balanced equation is
`2KMnO_(4) + 3H_(2)SO_(4) + 5H_(2)C_(2)O_(4) to K_(2)SO_(4) + 2MnSO_(4) + 8H_(2)O + 10CO_(2)`
`because ` molarity `= (n_(B))/(V_(ml)) xx 1000`
`therefore 0.025 =(n_(B))/(20) xx 1000`
`therefore n_(B) = (0.025)/(50) = 5 xx 10^(-4)` mol of `KMnO_(4)`
since 2 moles of `KMnO_(4)` completely oxidises 5 moles of oxalic acid
` = ( 5)/(2) xx 5 xx 10^(-4)` mol of oxalic acid
` = 1.25 xx 10^(-3)` mol of oxalic acid
`because` molarity `= ( n_(B))/(V_(ml)) xx 1000`
`therefore V_(ml) = (n_(B))/("molarity") xx 1000 = (1.25 xx 10^(-3))/(0.1) xx 1000`
(`because` molarity of oxalic acid = 0.1M)
` = 12.5 mL` or `12.5 cm^(3)`