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\(\mathrm{PtCl}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) can exist as a hydrated complex. Its 1 molal aq. solution has depression in freezing point of 3.72. Assume \(100 \%\) ionisation and \(K_{f}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\), then complex is : (a) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{4}\) (b) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \mathrm{Cl} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{4}\right] \cdot 4 \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
The correct complex is (b) \([\mathrm{Pt}(\mathrm{H}_2 \mathrm{O})_{4} \mathrm{Cl}_{2}] \mathrm{Cl}_{2} \cdot 2 \mathrm{H}_2 \mathrm{O}\).

Step by step solution

01

Calculate the Molal Depression Constant (Kf)

Use the molal depression constant (Kf) given for water. The value is given as 1.86°C kg/mol.
02

Calculate the Molality of the Solution

Use the depression in freezing point and the Kf value to calculate molality (m) using the formula \(\Delta T_f = K_f \times m\). The depression in freezing point (\(\Delta T_f\)) is given as 3.72°C.
03

Determine the Van't Hoff Factor (i)

Solve for the van't Hoff factor (i) using the calculated molality and the given freezing point depression. Since the solution is 1 molal, the expected freezing point depression for a non-dissociating compound would be equal to the Kf value. The ratio of the actual freezing point depression to the expected provides the van't Hoff factor. The formula used is \(i = \frac{\Delta T_f}{K_f}\).
04

Identify the Correct Formula of the Complex

Use the van't Hoff factor to determine the number of particles the complex dissociates into when dissolved in water. Match this number with the dissociation pattern of the given options to identify the correct complex formula.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Freezing Point Depression
Understanding the concept of freezing point depression is crucial when working with solutions. It refers to the process where the freezing point of a pure solvent becomes lower when a non-volatile solute is dissolved in it. For instance, when salt is added to ice, the resulting solution has a lower freezing point than the pure ice. This phenomenon can be quantified using the formula ewline \[\Delta T_f = K_f \times m \]ewline where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the cryoscopic constant (also known as the molal freezing-point depression constant), and \(m\) is the molality of the solution. This principle is not only essential in theoretical chemistry but also has practical applications such as in the making of ice cream or the formulation of anti-freeze.
Molality Calculation
Molality is a measure of the concentration of a solute in a solution. Unlike molarity, molality is not affected by changes in temperature or pressure because it is based on the mass of the solvent, not its volume. To calculate the molality (m), you need to know the moles of solute and the mass of the solvent in kilograms. The formula for molality is:ewline \[m = \frac{\text{moles of solute}}{\text{mass of solvent in kilograms}}\]ewline In the context of freezing point depression, molality helps determine the extent to which the freezing point will be lowered. For example, when solving problems that involve the depression in freezing point, like the given textbook problem, being able to calculate molality precisely is a crucial step.
Van't Hoff Factor
The van't Hoff factor (i) plays a pivotal role in colligative properties, including freezing point depression. It represents the number of particles that a compound dissociates into when it dissolves in a solvent. For nonelectrolytes, the van't Hoff factor is 1 since these compounds do not dissociate. Electrolytes, however, break down into ions; thus, their van't Hoff factor is greater than 1.ewline The formula to determine the van't Hoff factor in the context of freezing point depression is:ewline \[i = \frac{\Delta T_f}{K_f \times m}\]ewline This calculation allows you to understand the extent of the compound's dissociation in solution. When complex ions, like those in the given exercise, dissolve, they can form several species in solution, which is why knowledge of the van't Hoff factor is crucial for identifying the correct stoichiometry of the resulting complex.
Hydrated Complex Ions
Hydrated complex ions are a form of chemical species where central metal ions are surrounded by a certain number of water molecules. These complex ions can dissociate in solution, leading to a range of stoichiometries depending on the formula of the complex. In colligative property calculations, the dissociation of these hydrated complexes must be accounted for because each particle contributes to the properties such as freezing point depression. A complex such as \(\mathrm{PtCl}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) can dissociate into different ions, with each possible dissociation pattern potentially showing a unique van't Hoff factor (i). Recognizing the formation and dissociation of these hydrated complex ions is essential when predicting the behavior of a solution upon freezing.

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Most popular questions from this chapter

An aqueous solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) having a concentration of \(34.2\) gram/litre has an osmotic pressure of \(2.38\) atmospheres at \(17^{\circ} \mathrm{C}\). For an aqueous solution of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to be isotonic with this solution, its concentration should be : (a) \(34.2\) gram per litre (b) \(17.1\) gram per litre (c) \(18.0\) gram per litre (d) \(36.0\) gram per litre

Equal weight of a solute are dissolved in equal weight of two solvents \(A\) and \(B\) and formed very dilute solution. The relative lowering of vapour pressure for the solution \(B\) has twice the relative lowering of vapour pressure for the solution \(A\). If \(M_{A}\) and \(M_{B}\) are the molecular weights of solvents \(A\) and \(B\) respectively, then: (a) \(M_{A}=M_{B}\) (b) \(M_{B}=2 \times M_{A}\) (c) \(M_{A}=4 M_{B}\) (d) \(M_{A}=2 M_{B}\)

The boiling point of an azeotropic mixture of water-ethanol is less than that of both water and ethanol. Then: (a) the mixture will show negative deviation from Raoult's law (b) the mixture will show positive deviation from Raoult's law (c) the mixture will show no deviation from Raoult's law (d) this mixture cannot be considered as true solution

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\(1 \mathrm{~kg}\) of water under a nitrogen pressure of 1 atmosphere dissolves \(0.02 \mathrm{gm}\) of nitrogen at 293K. Calculate Henry's law constant : (a) \(7.7 \times 10^{4}\) atm (b) \(7.7 \times 10^{3}\) atm (c) \(2 \times 10^{-5} \mathrm{~atm}\) (d) \(2 \times 10^{-2} \mathrm{~atm}\)

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