\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\\ n_{Cl_2}=0,05\left(mol\right)\\Tacó: n_{NaOH}=2n_{Cl_2}=0,1\left(mol\right)\\ \Rightarrow V_{NaOH}=\dfrac{0,1}{1}=0,1\left(l\right)\\ Tacó:n_{NaCl}=n_{NaClO}=n_{Cl_2}=0,05\left(mol\right)\\\Rightarrow CM_{NaCl}=CM_{NaClO}=\dfrac{0,05}{0,1}=0,5M\)