Step-by-Step Solution

1. Identify Given Information

Let the properties of the solution in Vessel-1 be denoted by subscript 1 and those in Vessel-2 by subscript 2. The solutes are X and Y respectively.

• Mass of solute X, $w_{X} = w_2 \mathrm{~g}$

• Mass of solvent (water), $w_{\text{solvent}} = w_1 \mathrm{~g}$

• Molar mass of solute X = $M_X$

• van't Hoff factor for solute X = $i_X$

• Mass of solute Y, $w_{Y} = w_2 \mathrm{~g}$

• Mass of solvent (water), $w_{\text{solvent}} = w_1 \mathrm{~g}$

• Molar mass of solute Y = $M_Y$

• van't Hoff factor for solute Y = $i_Y$

We are given the following relationships:

• Molar mass of X is 80% of Y: $M_X = 0.80 \times M_Y$
• van't Hoff factor for X is 1.2 times that of Y: $i_X = 1.2 \times i_Y$

2. Recall the Formula for Elevation of Boiling Point

The elevation of boiling point ($\Delta T_b$) is a colligative property given by the formula: $\Delta T_b = i \times K_b \times m$ where:

• $i$ is the van't Hoff factor
• $K_b$ is the molal boiling point elevation constant (ebullioscopic constant) of the solvent.
• $m$ is the molality of the solution.

3. Calculate the Molality of Each Solution

Molality ($m$) is defined as moles of solute per kilogram of solvent. $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{w_{\text{solute}} / M_{\text{solute}}}{w_{\text{solvent}} / 1000}$

• For the solution in Vessel-1 (solute X): $m_1 = \frac{w_2 / M_X}{w_1 / 1000} = \frac{1000 \cdot w_2}{w_1 \cdot M_X}$

• For the solution in Vessel-2 (solute Y): $m_2 = \frac{w_2 / M_Y}{w_1 / 1000} = \frac{1000 \cdot w_2}{w_1 \cdot M_Y}$

4. Write Expressions for the Elevation of Boiling Point for Both Solutions

• For Vessel-1: $(\Delta T_b)_1 = i_X \times K_b \times m_1 = i_X \times K_b \times \left(\frac{1000 \cdot w_2}{w_1 \cdot M_X}\right)$

• For Vessel-2: $(\Delta T_b)_2 = i_Y \times K_b \times m_2 = i_Y \times K_b \times \left(\frac{1000 \cdot w_2}{w_1 \cdot M_Y}\right)$ Since the solvent is water in both vessels, the value of $K_b$ is the same for both.

5. Determine the Ratio of the Elevations of Boiling Point

We need to find the ratio of the elevation of boiling point of the solution in Vessel-1 to that in Vessel-2. $\frac{(\Delta T_b)_1}{(\Delta T_b)_2} = \frac{i_X \times K_b \times \left(\frac{1000 \cdot w_2}{w_1 \cdot M_X}\right)}{i_Y \times K_b \times \left(\frac{1000 \cdot w_2}{w_1 \cdot M_Y}\right)}$

Canceling the common terms ($K_b$, $1000$, $w_1$, $w_2$): $\frac{(\Delta T_b)_1}{(\Delta T_b)_2} = \frac{i_X / M_X}{i_Y / M_Y} = \left(\frac{i_X}{i_Y}\right) \times \left(\frac{M_Y}{M_X}\right)$

6. Substitute the Given Relationships

From the problem statement, we have:

• $\frac{i_X}{i_Y} = 1.2$
• $M_X = 0.80 \times M_Y \implies \frac{M_Y}{M_X} = \frac{1}{0.80}$

Substituting these values into the ratio expression: $\frac{(\Delta T_b)_1}{(\Delta T_b)_2} = (1.2) \times \left(\frac{1}{0.80}\right) = \frac{1.2}{0.8} = \frac{12}{8} = \frac{3}{2} = 1.5$

7. Express the Result as a Percentage

The question asks for the elevation of boiling point for the solution in Vessel-1 as a percentage of the solution in Vessel-2. $\text{Percentage} = \frac{(\Delta T_b)_1}{(\Delta T_b)_2} \times 100 \%$ $\text{Percentage} = 1.5 \times 100 \% = 150 \%$

Thus, the elevation of boiling point for the solution in Vessel-1 is 150% of the solution in Vessel-2.