When Hg is used as cathode, Na+will be reduced to Na instead of H2O. Thus formed Na will form sodium amalgam, Na-Hg.
At cathode:Na++ 1e-+ Hg -------> Na-Hg
2.0 moles of NaCl present in the solution contain 2 moles of Na+. Hence two moles of Na-Hg is formed.
The mass of 2 moles of Na-Hg = 2 (23 + 200) = 446 g.