The complex \(\text{CrCl}_3 \cdot x \text{NH}_3\) can exist as a coordination compound. A 0.1 molal aqueous solution of this complex shows a depression in the freezing point of 0.558°C. Assuming 100% ionization of this complex and coordination number of Cr is 6, what is the formula of the complex?
(Given \(K_f = 1.86 \, \text{K kg mol}^{-1}\))
1. | \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3\) | 2. | \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} \) |
3. | \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \) | 4. | \(\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]\) |