Vessel-1 contains w 2 g 𝑤 2 𝑔 of a non-volatile solute X 𝑋 dissolved in w 1 g 𝑤 1 𝑔 of water. Vessel-2 contains w 2 g 𝑤 2 𝑔 of another non-volatile solute Y 𝑌 dissolved in w 1 g 𝑤 1 𝑔 of water. Both the vessels are at the same temperature and pressure. The molar mass of X 𝑋 is 80% of that of Y 𝑌 . The van’t Hoff factor for X 𝑋 is 1.2 times of that of Y 𝑌 for their respective concentrations. The elevation of boiling point for solution in Vessel-1 is _____ % of the solution in Vessel-2.

 

To determine the percentage by which the boiling point elevation in Vessel-1 (with solute $X$) is greater than that in Vessel-2 (with solute $Y$), we'll follow these steps:

Given:

1. Masses:

   • Mass of solute $X$ in Vessel-1 = $w_2$ g

   • Mass of solute $Y$ in Vessel-2 = $w_2$ g

   • Mass of water in both vessels = $w_1$ g

2. Molar Masses:

   • Molar mass of $X$ ($M_X$) = 80% of $M_Y$, so:

     $$M_X=0.8M_Y$$

3. Van't Hoff Factors:

   • $i_X$ (for $X$) = 1.2 × $i_Y$ (for $Y$):

     $$i_X=1.2i_Y$$

4. Same temperature and pressure in both vessels.

Boiling Point Elevation Formula:

The boiling point elevation ($\mathrm{\Delta }{T}_b$) is given by:

$$\mathrm{\Delta }{T}_b=i\cdot K_b\cdot m$$

where:

• $i$ = van't Hoff factor

• $K_b$ = ebullioscopic constant (same for both solutions)

• $m$ = molality of the solution

Molality ($m$):

$$m=\frac{\text{moles of solute}}{\text{kg of solvent}}=\frac{w_2\mathrm{/}M}{{\text{w}}_1\mathrm{/}1000}$$$$m=\frac{w_2}{M}\times \frac{1000}{w_1}$$

For Vessel-1 (Solute $X$):

$$\mathrm{\Delta }{T}_{b,X}=i_X\cdot K_b\cdot (\frac{w_2}{M_X}\times \frac{1000}{w_1})$$

Substitute $M_X=0.8M_Y$ and $i_X=1.2i_Y$:

$$\mathrm{\Delta }{T}_{b,X}=1.2i_Y\cdot K_b\cdot (\frac{w_2}{0.8M_Y}\times \frac{1000}{w_1})$$$$\mathrm{\Delta }{T}_{b,X}=1.2\times \frac{1}{0.8}\cdot i_Y{K}_b(\frac{w_2}{M_Y}\times \frac{1000}{w_1})$$$$\mathrm{\Delta }{T}_{b,X}=1.5\cdot i_Y{K}_b{m}_Y$$

For Vessel-2 (Solute $Y$):

$$\mathrm{\Delta }{T}_{b,Y}=i_Y\cdot K_b\cdot (\frac{w_2}{M_Y}\times \frac{1000}{w_1})$$$$\mathrm{\Delta }{T}_{b,Y}=i_Y{K}_b{m}_Y$$

Ratio of Boiling Point Elevations:

$$\frac{\mathrm{\Delta }{T}_{b,X}}{\mathrm{\Delta }{T}_{b,Y}}=\frac{1.5\cdot i_Y{K}_b{m}_Y}{i_Y{K}_b{m}_Y}=1.5$$

Percentage Increase:

$$\text{Percentage}=\left(\frac{\mathrm{\Delta }{T}_{b,X}-\mathrm{\Delta }{T}_{b,Y}}{\mathrm{\Delta }{T}_{b,Y}}\right)\times 100\mathrm{\%}$$$$\text{Percentage}=(1.5-1)\times 100\mathrm{\%}=50\mathrm{\%}$$

Final Answer:

The boiling point elevation in Vessel-1 is 50% higher than in Vessel-2.

$$\boxed{{\displaystyle 50}}$$