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Chapter 4: Problem 70

A solution of \(0.2\) mole \(\mathrm{KI}(\alpha=100 \%)\) in 1000 \(\mathrm{g}\) water freezes at \(\mathrm{T}_{1}{ }^{\circ} \mathrm{C}\). Now to this solution \(0.1\) mole \(\mathrm{HgI}_{2}\) is added and the resulting solution freezes at \(\mathrm{T}_{2}{ }^{\circ} \mathrm{C}\). Which of the following is correct: (1) \(T_{1}=T_{2}\) (2) \(\mathrm{T}_{1}>\mathrm{T}_{2}\) (3) \(\mathrm{T}_{1}<\mathrm{T}_{2}\) (4) Can't be prediced

Short Answer

Expert verified
T1 ≈ T2. The freezing points are nearly the same.

Step by step solution

01

Molality Calculation of Initial Solution

First, calculate the molality of the initial KI solution. Molality (m) is given by the moles of solute per kilogram of solvent. Here, the moles of KI are 0.2 and the mass of the solvent (water) is 1000 g or 1 kg. Therefore, the initial molality is: \( m = \frac{0.2 \text{ moles}}{1 \text{ kg}} = 0.2 \text{ m} \)
02

Freezing Point Depression of Initial Solution

Next, use the formula for freezing point depression, which is \( \Delta T_f = i \times K_f \times m \). For KI, which dissociates completely (\(\alpha=100\%\)) into two ions (K+ and I-), the van't Hoff factor (i) is 2. Assuming the cryoscopic constant (K_f) for water is about 1.86 °C/m, the freezing point depression is: \( \Delta T_{f} = 2 \times 1.86 \ ^\circ\text{C/m} \times 0.2 \ m = 0.744 \ ^\circ\text{C} \)Since the normal freezing point of water is 0 °C, the solution's freezing point is: \( T_1 = 0^\text{°C} - 0.744^\text{°C} = -0.744^\text{°C} \)
03

Effect of Adding HgI2

When \(0.1\) moles of HgI2 are added, consider any interaction effects that may occur. HgI2 has a low solubility in water and does not dissociate significantly. Therefore, the main contributing species for freezing point depression due to added HgI2 will still mainly be KI. This implies that the freezing point of the resulting solution should not significantly change. Therefore, \( T_2 \approx T_1 \).
04

Conclusion

Based on the above steps, addition of a small amount of HgI2 does not significantly alter the freezing point because it does not contribute additional ions to the solution. Thus, \( T_1 \approx T_2 \).

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Key Concepts

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

molality calculation
Molality is a measure of the concentration of a solution expressed in terms of the amount of solute per unit mass of solvent. It is an important concept in chemistry, especially when dealing with colligative properties like freezing point depression.
To calculate molality, use the formula: \( m = \frac{moles \ of \ solute}{kilogram \ of \ solvent} \).
In the exercise, we have 0.2 moles of KI dissolved in 1000 grams of water (which is 1 kilogram). So, the molality \( m \) is: \( m = \frac{0.2 \ moles}{1 \ kg} = 0.2 \ m \).
It is important to note that molality is temperature-independent since it is based on the mass of the solvent, not its volume.
van't Hoff factor
The van't Hoff factor (\(i\)) is a measure of the effect of a solute on the colligative properties of a solution. It represents the number of particles into which a solute dissociates in solution. For ionic compounds, this is usually the total number of ions formed.
In the example given, KI dissociates completely into K+ and I- ions. This gives us a van't Hoff factor (\(i\)) of 2.
The formula to calculate the change in freezing point (freezing point depression) for a solution is: \( \Delta T_f = i \times K_f \times m \).
Here, \( K_f \) is the cryoscopic constant specific to the solvent, and molality \( m \) is as previously calculated.
cryoscopic constant
The cryoscopic constant (\( K_f \)) is a proportionality factor that relates the freezing point depression to the molality of the solute in a solution. Each solvent has its own unique cryoscopic constant.
For water, \( K_f \) is approximately 1.86 °C/m. This means that for every molal (m) of solute added to 1 kilogram of water, the freezing point of the water decreases by 1.86°C.
Using the van't Hoff factor calculated earlier (which was 2 for KI), and the molality (which was 0.2 m), we get the freezing point depression: \( \Delta T_f = 2 \times 1.86 °C/m \times 0.2 m = 0.744 °C \).
The normal freezing point of pure water is 0°C, so the solution with KI freezes at \( -0.744 °C \).

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