$ {$\underset{\mathrm{Cold, dilute}}{ {Cl_2 + 2NaOH}}$ ->$\underset{\mathrm{sod. hypochlorite }}{ { NaCl + NaCIO + H_2O }}$ }$
$ {$\underset{\mathrm{ hot cone }}{ { 3C12 + 6NaOH}}$-> $\underset{\mathrm{ sod. chlorate}}{ {5NaCl + NaCIO3 + 3H2O }}$ }$
So sodium chloride is formed in both cases.