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.
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.
The elevation of boiling point for solution in Vessel-1 is _____ % of the solution in Vessel-2.
Updated On: Mar 9, 2025
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Correct Answer: 150
Solution and Explanation
Boiling Point Elevation Calculation
1. Expression for Elevation of Boiling Point in Vessel-1 (\( (\Delta T_b)_I \)):
\[ (\Delta T_b)_I = \frac{i_X \times w_2}{M_X} \times 1000 \times \frac{w_1}{K_B} \]
2. Expression for Elevation of Boiling Point in Vessel-2 (\( (\Delta T_b)_{II} \)):
\[ (\Delta T_b)_{II} = \frac{i_Y \times w_2}{M_Y} \times 1000 \]
3. Ratio of Elevations:
\[ \frac{(\Delta T_b)_I}{(\Delta T_b)_{II}} = \frac{i_X}{w_1} \times K_B \times \frac{M_Y}{i_Y \times M_X} \]
4. Substituting \( M_X = 0.8M_Y \) and \( i_X = 1.2i_Y \):
\[ \frac{(\Delta T_b)_I}{(\Delta T_b)_{II}} = \frac{1.2}{1} \times \frac{1}{0.8} = 1.5 \]
5. Percentage Increase:
Quick Tip:
\[ (\Delta T_b)_I = 150\% \, \text{of} \, (\Delta T_b)_{II} \]
Final Answer:
The elevation of boiling point in Vessel-1 is 150% of the elevation in Vessel-2.
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