$\begin{array}{1 1}(a)\;200& (b)\;225 \\(c)\;400& (d)\;446\end{array} $

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Number of moles of $Na^+=2$

$2NaCl\equiv 2Na$

$Na+Hg\rightarrow Na(Hg)$

By electrolysis we can get a maximum of 2 moles of sodium which can combine with exactly 2 moles of mercury to give amalgam.

$\therefore$ Maximum weight of Na amalgam (assuming equimolar Na and Hg)=46+400=446g

Hence (d) is the correct answer.

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