$\text{Vessel-1 contains}{\mathrm{ w}}_2\text{ g of a non-volatile solute}\mathrm{ X }\text{dissolved in}{\mathrm{ w}}_1\text{ g of water.}\text{Vessel-2 contains}{\mathrm{ w}}_2\text{ g of}$

$\text{another non-volatile solute}\mathrm{ Y }\text{dissolved in }{w}_1\text{g of water.}\text{Both the vessels are at the same temperature}$

$\text{and pressure.}\text{The molar mass of}\mathrm{ X}\text{ is}\mathrm{ 80}\%\text{of that of}\mathrm{ Y}.\text{The van’t Hoff factor for}\mathrm{ X }\text{is }1.2\text{ times that}$

$\text{of}\mathrm{ Y }\text{for their respective concentrations.}$
$\text{The elevation of boiling point for solution in Vessel-1 is}\text{\_\_\_\_\_\_}\%\text{of the solution in Vessel-2.}$

The correct answer is 150.

Step 1: Understand the formula for elevation of boiling point
The elevation of boiling point ($\Delta T_b$) of a solution 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 ebullioscopic constant of the solvent (water)
• $m$ is the molality of the solution, which is given by $m=\frac{\text{mass of solute in grams}}{\text{molar mass of solute}\times \text{mass of solvent in kg}}$

Step 2: Express elevation of boiling point for both vessels
For Vessel-1 (containing solute X):
Mass of solute = $w_2$ g, Mass of solvent = $w_1$ g, Molar mass = $M_X$, and van ’t Hoff factor = $i_X$.

$\Delta T_{b,1}=i_X\times K_b\times \frac{w_2\times 1000}{M_X\times w_1}$

For Vessel-2 (containing solute Y):
Mass of solute = $w_2$ g, Mass of solvent = $w_1$ g, Molar mass = $M_Y$, and van ’t Hoff factor = $i_Y$.

$\Delta T_{b,2}=i_Y\times K_b\times \frac{w_2\times 1000}{M_Y\times w_1}$

Step 3: Ratio of elevation of boiling points
Dividing $\Delta T_{b,1}$ by $\Delta T_{b,2}$ (noting that $w_1$, $w_2$, and $K_b$ cancel out):

$\frac{\Delta T_{b,1}}{\Delta T_{b,2}}=\left(\frac{i_X}{i_Y}\right)\times \left(\frac{M_Y}{M_X}\right)$

Step 4: Substitute the given relationships
1. The molar mass of X is 80% of that of Y:

$M_X=0.80\times M_Y\Rightarrow \frac{M_Y}{M_X}=\frac{1}{0.80}=1.25$

2. The van ’t Hoff factor for X is 1.2 times that of Y:

$i_X=1.2\times i_Y\Rightarrow \frac{i_X}{i_Y}=1.2$

Substituting these ratios into the equation:

$\frac{\Delta T_{b,1}}{\Delta T_{b,2}}=1.2\times 1.25=1.5$

Step 5: Calculate the percentage
To express the elevation of boiling point for Vessel-1 as a percentage of Vessel-2:

$\text{Percentage}=\frac{\Delta T_{b,1}}{\Delta T_{b,2}}\times 100=1.5\times 100=150\%$

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