A solution containing 25.6gm of sulphur dissolved in 1000gm of naphthalene gave a freezing point lowering of 0.680, then molecular formula of sulphur is

$\left({\mathrm{K}}_{\mathrm{F}}\right.$ for naphthalene =6.8 K kgmol } $- 1$)

$\Delta {\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}·\mathrm{m}$

Whereas,

${\mathrm{K}}_{\mathrm{f}}$ = molal depression constant
m= molality of solution $=\frac{\text{ mass of solute }}{\text{ molecular mass of solute }}\times \frac{1000}{\text{ mass of solvent (in gm) }}$
$$\begin{gathered} Mass of dissolved sulphur \left(solute\right) =2.56g \\ Mass of solvent \left(Naphthalene\right) =100g \end{gathered}$$

$Molecular mass of solute =M \left(Assumed\right)$

$\therefore 0.68=6.8\times \left(\frac{2.56}{\mathrm{M}}\times \frac{1000}{100}\right)$

$\Rightarrow \mathrm{M}=256$
Hence, the correct option (D).  ${\mathrm{S}}_8$