$(a)\;2 : 1\qquad(b)\;4 : 1\qquad(c)\;3 : 2\qquad(d)\;1 : 2$

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Step 1:

Let a mole of HCOOH and b mole of $H_2C_2O_4$ are present in original mixture.

Mole of $CO$ formed =a+b

$a\Rightarrow$ From HCOOH

$b\Rightarrow$ From $H_2C_2O_4$

Mole of $CO_2$ formed =b

Total mole of gases =$a+b+b=a+2b$

Step 2:

$\therefore CO_2$ is observed by KOH and volume reduces by $\large\frac{1}{6}$

Mole of $CO_2=\large\frac{1}{6}$$(a+2b)$

$b=\large\frac{1}{6}$$(a+2b)$

$\large\frac{a}{5}$$=4$

$a : b=4 : 1$

Hence (b) is the correct option.

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