Question

Vessel-1 contains w₂ g of a non-volatile solute X dissolved in w₁ g of water. Vessel-2 contains w₂ g of another non-volatile solute Y dissolved in w₁ g of water. Both the vessels are at the same temperature and pressure. The molar mass of X is 80% of that of Y. The van’t Hoff factor for X is 1.2 times of that of Y for their respective concentrations.
The elevation of boiling point for solution in Vessel-1 is _____ % of the solution in Vessel-2.

Answer 1

Correct Answer: 150

To solve the problem, we need to determine the ratio of the boiling point elevation for two solutions in Vessel-1 (solute X) and Vessel-2 (solute Y), expressed as a percentage.

1. Define Variables:
- \(w_1\): Mass of water (solvent) in both vessels
- \(w_2\): Mass of solute in both vessels
- \(M_x\): Molar mass of solute X
- \(M_y\): Molar mass of solute Y
- \(i_x\): van't Hoff factor for solute X
- \(i_y\): van't Hoff factor for solute Y
- \(\Delta T_{b1}\): Boiling point elevation for Vessel-1
- \(\Delta T_{b2}\): Boiling point elevation for Vessel-2

2. Given Information:
\[ M_x = 0.8 M_y, \quad i_x = 1.2 i_y \]

3. Boiling Point Elevation Formula:
The elevation of boiling point is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \] where \(i\) is the van't Hoff factor, \(K_b\) is the ebullioscopic constant (same for both vessels since the solvent is the same), and \(m\) is the molality (moles of solute per kg of solvent).

4. Calculate Molality:
- Molality for Vessel-1 (solute X):
\[ m_x = \frac{\text{moles of X}}{\text{kg of solvent}} = \frac{w_2 / M_x}{w_1 / 1000} = \frac{1000 w_2}{M_x w_1} \] - Molality for Vessel-2 (solute Y):
\[ m_y = \frac{w_2 / M_y}{w_1 / 1000} = \frac{1000 w_2}{M_y w_1} \]

5. Calculate Boiling Point Elevations:
- For Vessel-1:
\[ \Delta T_{b1} = i_x \cdot K_b \cdot m_x = i_x \cdot K_b \cdot \frac{1000 w_2}{M_x w_1} \] - For Vessel-2:
\[ \Delta T_{b2} = i_y \cdot K_b \cdot m_y = i_y \cdot K_b \cdot \frac{1000 w_2}{M_y w_1} \]

6. Find the Ratio \(\Delta T_{b1} / \Delta T_{b2}\):
\[ \frac{\Delta T_{b1}}{\Delta T_{b2}} = \frac{i_x \cdot K_b \cdot \frac{1000 w_2}{M_x w_1}}{i_y \cdot K_b \cdot \frac{1000 w_2}{M_y w_1}} = \frac{i_x}{i_y} \cdot \frac{M_y}{M_x} \] The \(K_b\), \(w_1\), \(w_2\), and 1000 terms cancel out.

7. Substitute Given Relationships:
\[ i_x = 1.2 i_y, \quad M_x = 0.8 M_y \] \[ \frac{i_x}{i_y} = 1.2, \quad \frac{M_y}{M_x} = \frac{M_y}{0.8 M_y} = \frac{1}{0.8} = 1.25 \] \[ \frac{\Delta T_{b1}}{\Delta T_{b2}} = 1.2 \cdot 1.25 = 1.5 \]

8. Express as a Percentage:
\[ \frac{\Delta T_{b1}}{\Delta T_{b2}} = 1.5 \implies \Delta T_{b1} = 1.5 \Delta T_{b2} \] As a percentage:
\[ 1.5 \times 100\% = 150\% \]

Final Answer:
The elevation of boiling point for the solution in Vessel-1 is \(150\%\) of that in Vessel-2.

Answer 2

To solve the problem, we need to find the percentage ratio of elevation in boiling points of solutions in Vessel-1 and Vessel-2.

1. Given data:
- Mass of solute in both vessels: \( w_2 \) g
- Mass of solvent (water) in both vessels: \( w_1 \) g
- Molar mass of solute \( X = 0.8 \times \) molar mass of solute \( Y \)
- Van’t Hoff factor \( i_X = 1.2 \times i_Y \)
- Temperature and pressure are the same for both solutions.

2. Elevation in boiling point formula:
\[ \Delta T_b = i K_b m \] where
\( i \) = van't Hoff factor,
\( K_b \) = ebullioscopic constant (same for both as solvent and conditions same),
\( m \) = molality of solution.

3. Calculate molality for both solutions:
\[ m_X = \frac{\text{moles of } X}{\text{kg of solvent}} = \frac{\frac{w_2}{M_X}}{\frac{w_1}{1000}} = \frac{1000 w_2}{M_X w_1} \] \[ m_Y = \frac{\frac{w_2}{M_Y}}{\frac{w_1}{1000}} = \frac{1000 w_2}{M_Y w_1} \]

4. Ratio of elevation in boiling points:
\[ \frac{\Delta T_{b,X}}{\Delta T_{b,Y}} = \frac{i_X m_X}{i_Y m_Y} = \frac{(1.2 i_Y) \times \frac{1000 w_2}{M_X w_1}}{i_Y \times \frac{1000 w_2}{M_Y w_1}} = 1.2 \times \frac{M_Y}{M_X} \]

5. Substitute \( M_X = 0.8 M_Y \):
\[ \frac{\Delta T_{b,X}}{\Delta T_{b,Y}} = 1.2 \times \frac{M_Y}{0.8 M_Y} = 1.2 \times \frac{1}{0.8} = 1.5 \]

6. Convert ratio to percentage:
\[ \text{Elevation of boiling point for Vessel-1} = 150\% \text{ of that in Vessel-2} \]

Final Answer:
The elevation of boiling point for solution in Vessel-1 is \(\boxed{150\%}\) of that of the solution in Vessel-2.