Question

The value of x is ______.

Answer 1

Correct Answer: 100.1

Step 1: Understanding the given data
The molality of the AgNO₃ solution is given as 0.1 molal (solution-A). This means that for every 1 kg of solvent, there are 0.1 moles of AgNO₃ dissolved.
The equation for the boiling point elevation (\( \Delta T_b \)) is:
ΔT_b = i k_b m
Where:
- \( \Delta T_b \) is the elevation in boiling point.
- \( i \) is the van't Hoff factor, which accounts for the number of particles formed in solution (for AgNO₃, \( i = 2 \) because it dissociates into Ag⁺ and NO₃^–).
- \( k_b \) is the ebullioscopic constant (which is a property of the solvent, and here \( k_b = 0.5 \) is assumed).
- \( m \) is the molality of the solution, which is 0.1 mol/kg.
Step 2: Applying the formula
The formula for the change in boiling point is \( \Delta T_b = i k_b m \). Substituting the given values:
\[ \Delta T_b = 2 \times 0.5 \times 0.1 \] Simplifying this: \[ \Delta T_b = 0.1 \, \text{°C} \] Step 3: Using the boiling point elevation
The boiling point of the pure solvent (water) is 100ºC. The new boiling point of the solution is given by: \[ T_{\text{s}} - T^{\circ} = \Delta T_b \] Substituting the value of \( \Delta T_b \): \[ T_{\text{s}} - 100 = 0.1 \] Solving for \( T_{\text{s}} \): \[ T_{\text{s}} = 100 + 0.1 = 100.1 \, \text{°C} \] Step 4: Conclusion
Therefore, the boiling point of solution-A (AgNO₃ solution) is \( T_{\text{s}} = 100.1 \, \text{°C} \). The value of \( x \) is 100.1.

Answer 2

Step 1: Understanding the given data
We are given two solutions:
- Solution A: 0.1 molal silver nitrate (AgNO₃) solution.
- Solution B: A mixture of 0.1 molal AgNO₃ solution and 0.1 molal barium chloride (BaCl₂) solution.
The densities of the solutions A and B are assumed to be the same as that of water, and the salts dissociate completely.
The ebullioscopic constant (\( K_b \)) is given as 0.5 K·kg·mol^-1, and the boiling point of pure water is 100°C.
We are tasked with calculating the difference in the boiling points of water in solutions A and B, and expressing it in the form of

\( y \times 10^{-2} \, \text{°C} \)

Step 2: Boiling Point Elevation Formula
The formula for boiling point elevation is:
\[ \Delta T_b = i \cdot K_b \cdot m \] where:
- \( \Delta T_b \) is the elevation in boiling point.
- \( i \) is the van't Hoff factor, which represents the number of particles the solute dissociates into.
- \( K_b \) is the ebullioscopic constant (given as 0.5 K·kg·mol^-1).
- \( m \) is the molality of the solution.
Step 3: Calculate the boiling point elevation for solution A
In solution A (AgNO₃):
- The van’t Hoff factor for AgNO₃ is \( i = 2 \) because it dissociates into 2 ions: Ag⁺ and NO₃^-.
- The molality of solution A is 0.1 mol/kg.
Using the boiling point elevation formula:
\[ \Delta T_{bA} = 2 \cdot 0.5 \cdot 0.1 = 0.1^\circ C \]
So, the boiling point of solution A is 100 + 0.1 = 100.1°C.
Step 4: Calculate the boiling point elevation for solution B
In solution B (mixture of AgNO₃ and BaCl₂):
- AgNO₃ dissociates into 2 ions (Ag⁺ and NO₃^-) as in solution A, so \( i = 2 \) for AgNO₃.
- BaCl₂ dissociates into 3 ions (Ba²⁺ and 2 Cl^-), so \( i = 3 \) for BaCl₂.
The molality of both solutions is 0.1 mol/kg, and they are mixed in equal volumes. Therefore, the effective molality of solution B is:
\[ m_B = \frac{0.1 + 0.1}{2} = 0.1 \, \text{mol/kg} \] Now, for solution B, the total van’t Hoff factor is the sum of the individual factors for AgNO₃ and BaCl₂:
\[ i_B = 2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2} = 2.5 \] Using the boiling point elevation formula:
\[ \Delta T_{bB} = 2.5 \cdot 0.5 \cdot 0.1 = 0.125^\circ C \]
So, the boiling point of solution B is 100 + 0.125 = 100.125°C.
Step 5: Calculate the difference in boiling points
The difference in boiling points between solution A and solution B is:
\[ \Delta T_b = 100.125 - 100.1 = 0.025^\circ C \] Expressing this difference in terms of \( y \times 10^{-2} \, ^\circ C \), we get:
\[ y = 2.5 \] Step 6: Conclusion
Therefore, the value of \( |y| \) is 2.5.