If $I_2$  liberated in the first reaction is completely consumed in the second reaction then which of the following is correct 
$C{l}_2+NaI\to NaCl+I_2\mathrm{.......}\left(1\right);I_2+NaOH\to NaI+NaI{O}_3+H_{2}O\mathrm{......}\left(2\right)$ 

equivalents of $I_2$ in the first reaction = equivalents of $I_2$ in the second reaction 

moles of $I_2$ in the first reaction = moles of $I_2$ in the second reaction 

equivalent weight of $I_2$ in the first reaction = equivalent weight of $I_2$ in the second reaction

n-factor of $I_2$  in the first reaction = n-factor of  $I_2$ in the second reaction