Relative Elevation in Boiling Point of Non Volatile Solutions

To find the elevation of boiling point for the solution in Vessel-1 as a percentage of that in Vessel-2, we start with the formula for the elevation of boiling point ($\Delta T_b$):

$\Delta T_b = i \cdot K_b \cdot m$

where:

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

The molality $m$ is given by: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{w_2 / M}{w_1 / 1000}$

where $w_2$ is the mass of the solute in grams, $M$ is the molar mass of the solute in $\text{g mol}^{-1}$, and $w_1$ is the mass of water in grams.

For Vessel-1 (containing solute X):

$\Delta T_{b,1} = i_X \cdot K_b \cdot \frac{w_2}{M_X \cdot (w_1 / 1000)}$

For Vessel-2 (containing solute Y):

$\Delta T_{b,2} = i_Y \cdot K_b \cdot \frac{w_2}{M_Y \cdot (w_1 / 1000)}$

Taking the ratio of $\Delta T_{b,1}$ to $\Delta T_{b,2}$:

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

Given from the problem statement:

1. $M_X = 80\% \text{ of } M_Y = 0.80 M_Y \implies \frac{M_Y}{M_X} = \frac{1}{0.80} = 1.25$
2. $i_X = 1.2 i_Y \implies \frac{i_X}{i_Y} = 1.2$

Substitute these ratios into the equation: $\frac{\Delta T_{b,1}}{\Delta T_{b,2}} = 1.2 \times 1.25 = 1.5$

To express $\Delta T_{b,1}$ as a percentage of $\Delta T_{b,2}$: $\text{Percentage} = \left( \frac{\Delta T_{b,1}}{\Delta T_{b,2}} \right) \times 100\% = 1.5 \times 100\% = 150\%$

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