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The 2.56gm of sulphur in 100gm of $CS_2$ has depression in freezing point of $0.010^{\large\circ}$C. $K_f=0.1^{\large\circ}C$/molal. Hence atomicity of sulphur in $CS_2$

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

1 Answer

  • 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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