Question

Observe the statements and choose the correct option: Statement-I: Boiling point of water is more than that of ethanol. Statement-II: Ebullioscopic constant of water is more than that of ethanol.

• A. Statement-l and statement-Il both are correct
• B. Statement-l and statement-ll both are incorrect
• C. Statement-I is correct but statement-ll is incorrect
• D. Statement-I is incorrect but statement-ll is correct

Answer 1

Answer: (C)

C

Answer 2

Statement-I: Boiling point of water is more than that of ethanol.

• The boiling point of pure water at 1 atm pressure is 100 °C.
• The boiling point of pure ethanol at 1 atm pressure is approximately 78.37 °C.
• Comparing these values, 100 °C > 78.37 °C.
• Therefore, Statement-I is correct. Water has a higher boiling point than ethanol primarily due to stronger and more extensive hydrogen bonding networks in water molecules compared to ethanol.

Statement-II: Ebullioscopic constant of water is more than that of ethanol.

The ebullioscopic constant ($K_b$), also known as the molal elevation constant, is a characteristic property of a solvent. It is used in the colligative property of boiling point elevation. The formula for $K_b$ is:

$K_b = \frac{R T_b^2 M_{solvent}}{1000 \Delta H_{vap}}$

Where:

• $R$ is the ideal gas constant (8.314 J/mol·K)
• $T_b$ is the boiling point of the pure solvent in Kelvin
• $M_{solvent}$ is the molar mass of the solvent in g/mol
• $\Delta H_{vap}$ is the molar enthalpy of vaporization of the solvent in J/mol

Let's calculate $K_b$ for water and ethanol:

For Water:

• $T_b = 100 \text{ °C} = 373.15 \text{ K}$
• $M_{solvent} = 18.015 \text{ g/mol}$
• $\Delta H_{vap} = 40.65 \text{ kJ/mol} = 40650 \text{ J/mol}$

$K_b (\text{water}) = \frac{(8.314 \text{ J/mol·K}) \cdot (373.15 \text{ K})^2 \cdot (18.015 \text{ g/mol})}{1000 \cdot (40650 \text{ J/mol})}$

$K_b (\text{water}) = \frac{8.314 \cdot 139242.22 \cdot 18.015}{40650000} \approx 0.513 \text{ K·kg/mol}$

For Ethanol:

• $T_b = 78.37 \text{ °C} = 351.52 \text{ K}$
• $M_{solvent} = 46.07 \text{ g/mol}$
• $\Delta H_{vap} = 38.56 \text{ kJ/mol} = 38560 \text{ J/mol}$

$K_b (\text{ethanol}) = \frac{(8.314 \text{ J/mol·K}) \cdot (351.52 \text{ K})^2 \cdot (46.07 \text{ g/mol})}{1000 \cdot (38560 \text{ J/mol})}$

$K_b (\text{ethanol}) = \frac{8.314 \cdot 123566.6 \cdot 46.07}{38560000} \approx 1.229 \text{ K·kg/mol}$

Comparing the $K_b$ values:

$K_b (\text{water}) \approx 0.513 \text{ K·kg/mol}$ $K_b (\text{ethanol}) \approx 1.229 \text{ K·kg/mol}$

It is clear that $K_b (\text{ethanol}) > K_b (\text{water})$. Therefore, Statement-II, which claims that the ebullioscopic constant of water is more than that of ethanol, is incorrect.

Conclusion:

Statement-I is correct. Statement-II is incorrect.

The correct option is C.