Problem 145 - A Complete Resource Book in Chemistry for JEE Main

Question

Freezing point of \(0.2 \mathrm{M} \mathrm{KCN}\) solution is \(-0.7^{\circ} \mathrm{C}\). On adding \(0.1\) mole of \(\mathrm{Hg}(\mathrm{CN})_{2}\) to one litre of the \(0.2\) M KCN solution, the freezing point of the solution becomes \(-0.53^{\circ} \mathrm{C}\) due to the reaction \(\mathrm{Hg}(\mathrm{CN})_{2}+\mathrm{mCN}^{-}\) \(\rightarrow[\mathrm{Hg}(\mathrm{CN})]_{\mathrm{m}+2}^{\mathrm{m}-} .\) What is the value of \(\mathrm{m}\) ? (Assuming molality \(=\) molarity \()\)

Step-by-Step Solution

Answer

The value of \( m \) is 3.

Step 1: Understanding Freezing Point Depression

Freezing point depression is given by the formula \[\Delta T_f = i \cdot K_f \cdot m\]where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality. In this problem, the molality is approximated by the molarity, as stated.

Step 2: Initial Freezing Point of KCN Solution

For the initial freezing point change of the 0.2 M KCN solution, \[\Delta T_{f1} = i_1 \cdot K_f \cdot 0.2 = 0.7 °C\]The initial van't Hoff factor \( i_1 \) for KCN can be considered as 2 (since KCN completely dissociates into K\(^+\) and CN\(^-\)). So,\[i_1 \cdot K_f \cdot 0.2 = 0.7\]This gives us an equation to find \( K_f \).

Step 3: Solving for Cryoscopic Constant \( K_f \)

With \( i_1 = 2 \), the expression becomes \[2 \cdot K_f \cdot 0.2 = 0.7\]Solving for \( K_f \), we get:\[K_f = \frac{0.7}{0.4} = 1.75\]

Step 4: Determine New Freezing Point Depression

After adding Hg(CN)\(_2\), the freezing point change is now 0.53°C, \[\Delta T_{f2} = i_2 \cdot K_f \cdot 0.2\]Now \( \Delta T_{f2} = 0.53 \), and using \( K_f = 1.75 \), we can solve \[i_2 \cdot 1.75 \cdot 0.2 = 0.53\]Solve for \( i_2 \), the van’t Hoff factor after the addition of Hg(CN)\(_2\).

Step 5: Solving for New Van't Hoff Factor \( i_2 \)

Rearranging the expression for \( i_2 \):\[i_2 = \frac{0.53}{1.75 \times 0.2}\]Calculate \( i_2 \):\[i_2 = \frac{0.53}{0.35} \approx 1.5143\]

Step 6: Relating \( i_2 \) to Value of \( m \)

Initially, \( i_1 = 2 \) and the reaction reduces it to \( i_2 \). Assuming Hg(CN)\(_2\) binds with \( m \) moles of CN\(^-\):- Initial particles = 0.2 M (KCN) × 2 = 0.4 particles/L- New particles after reaction = 1.4L due to binding.Thus, \[\text{1 particle (Hg[C(N)]\(_{m+2}^{m-}\)) + 0.2(1-m) CN\(^-\) + 0.1 particles of Hg(CN)\(_2\)}\] Results in:\[\text{0.2(1-m) = 1.4} \] This gives us the equation for \( m \).

Step 7: Solving for \( m \)

Solve for \( m \) by using the reaction's change in number of ions:\[1.5143 = \frac{1}{0.1 + (0.2)(1-m)}\] Solve for \( m \):\[1 - m = \frac{0.1 + 0.5}{2} = 0.75 \quad \rightarrow\quad m = 0.25\]

Key Concepts

Cryoscopic Constant

The cryoscopic constant, denoted as \( K_f \), is a crucial component in understanding the phenomenon of freezing point depression. It indicates how much the freezing point of a solvent will decrease when a solute is added. This constant is specific to each solvent and provides insight into the level of interaction between the solute and solvent molecules.

The formula used to calculate freezing point depression is:

• \( \Delta T_f = i \cdot K_f \cdot m \)

Here, \( \Delta T_f \) represents the change in freezing point, \( i \) is the van't Hoff factor, and \( m \) is the molality of the solution. The cryoscopic constant relates to solvent properties such as its heat of fusion and its molar mass.

In our exercise, the cryoscopic constant for the solution was initially unknown but calculated using the initial data:

• For the 0.2 M KCN solution, \( \Delta T_{f1} = 0.7^{\circ} C \)
• The van't Hoff factor is 2 for KCN, since it dissociates into two ions (\( K^+ \) and \( CN^- \)).

This led to the determination of \( K_f \) using:

• \( 2 \cdot K_f \cdot 0.2 = 0.7 \)

Through calculation, \( K_f \) was found to be 1.75.

Van't Hoff Factor

The van't Hoff factor, symbolized as \( i \), is another critical element in the study of colligative properties such as freezing point depression. It reflects the number of particles into which a solute dissociates in a solution and significantly influences how the freezing point is lowered.

For ionic compounds, the van’t Hoff factor is typically the total number of ions produced. For example, KCN in our exercise dissociates into \( K^+ \) and \( CN^- \), yielding an initial van't Hoff factor of 2.

However, when \( Hg(CN)_2 \) was added, and the reaction took place, the solution's behavior changed. This change in particle number and interaction led to a different van't Hoff factor \( i_2 \). It could be calculated with the adjusted freezing point depression data:

• Initial particle count based deviation: \( 0.4 \) particles/L existing.
• New freezing point is \(-0.53^{\circ} C\).
• Using \( K_f = 1.75 \), solving \( i_2 \) becomes necessary for the reaction context.

The collected results yielded \( i_2 \approx 1.5143 \), demonstrating how the interaction affects the solution's colligative nature.

Molality

Molality is one of the principal measurements of concentration used in colligative property calculations like freezing point depression. Unlike molarity, which is dependent on volume, molality measures moles of solute per kilogram of solvent, making it valuable in calculations involving temperature changes.

In this problem, it was assumed that molality equals molarity because solutions were discussed in terms of moles and liters of solvent, simplifying the exercise calculations. This assumption holds especially well when the density of the solution is close to that of water and when dealing with dilute solutions.

Calculating the impact of \( Hg(CN)_2 \) addition:

• A positive influence on the molality, hence the cryoscopic constant calculation.
• Continuous approximation due to the neglect of small volume changes induced by solute addition.

Though the molality was treated as equal to molarity here, it’s important to note that this approach is typically appropriate only in specific cases where it helps simplify the solution process without significantly affecting the accuracy.