$1.1 g C o C l_{x} \cdot y N H_{3}$ (Molar mass 267.5) was dissolved in 100 g water. The solution shows freezing point $- {0.306}^{o} C$ . How many solute particles exists in solution for each mole of solute introduced if 100% ionization of complex is noticed? $K_{f}$ for $H_{2} O = 1.86 K m o l^{- 1} k g$ . Also, find the value of $x$ and $y$ .

3, x = 4, y = 5

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4, x = 3, y = 6

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3, x = 5, y = 6

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4, x = 5, y = 6

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Solution

The correct option is B $4, x = 3, y = 6$
The depression in the freezing point:

$Δ T_{f} = {0.306}^{o} C$

$Δ T_{f} = i K_{f} m$

$0.306 = i \times 1.86 \times \frac{1.1}{267.5 \times 0.1}$
The vant Hoff factor $i = 4$
Thus the dissociation of 1 solute molecule gives four particles. The complex is $[C o (N H)_{6}] C l_{3} ⇌ [C o (N H)_{6}]^{3 +} + 3 C l^{-}$
The complex can also be written as $C o C l_{3} \cdot 6 N H_{3}$
Hence, $x = 3, y = 6$

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C o C l_{3} \cdot 6 N H_{3}

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