$\begin{array}{1 1}(a)\;6\\(b)\;4\\(c)\;8\\(d)\;2\end{array}$

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- Let $T_f^0$ be the freezing point of pure solvent and $T_f$ be its freezing point when non-volatile solute is dissolved in it. The decrease in freezing point $\Delta T_f = T_f^0 - T_f$
- Depression of freezing point ($\Delta T_f$) for dilute solution (ideal solution) is directly proportional to molality. Thus, $\Delta T_f=K_f \times m$
- Since, $m= \large \frac{w_2 / M_2}{w / 1000}$. Substituting the value of molality we obtain, $\Delta T_f = \large \frac{K_f \times w_2 \times 1000}{M_2 \times w_1}$

Decrease in freezing point of $CS_2,$ $\Delta T_f=K_f \times m$

$\Rightarrow K_f= \large\frac{\text{weight of sulphur}}{\text{molecular weight of sulphur}\times 0.1\times n}$

Atomicity of sulphur in $CS_2$ (n) is $\large\frac{K_f\times 2.56}{32\times 0.1\times 0.010}$ $= \large\frac{0.1\times 2.56}{32\times 0.1\times 0.01}$$= \large\frac{256}{32} = 8$

$\therefore$ Atomicity of sulphur(n)=8

Hence (c) is the correct answer.

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