Consider the reaction $$2KBrO_3+12H^++10e^-\rightarrow Br_2+6H_2O+2K^+$$. From above reaction the equivalent weight of $$KBrO_3$$ can be calculated as, where $$M$$ is molecular weight of $$KBrO_3$$

A

$$\frac{M}{5}$$

B

$$\frac{M}{10}$$

C

$$\frac{M}{12}$$

D

$$\frac{M}{2}$$