What mass of ethylene glycol (C2H6O2) in grams, must be added to 1.0 kg of H2O to that produces a solution that boils at 105.0 Celsius? The molar mass of ethylene glycol is 62.07 g/mol and the kb for H2O is 0.512 C/m.
The correct answer and explanation is :
To solve this problem, we will use the concept of boiling point elevation, which is a colligative property. The equation for boiling point elevation is:
$$
\Delta T_b = i \cdot K_b \cdot m
$$
Where:
- $\Delta T_b$ is the boiling point elevation (in °C),
- $i$ is the van’t Hoff factor (which represents the number of particles the solute breaks into in solution),
- $K_b$ is the ebullioscopic constant (boiling point elevation constant) for the solvent,
- $m$ is the molality of the solution (in mol/kg).
Step 1: Identify the Known Information
- Boiling point of water before the solute is added = 100.0°C.
- Final boiling point of the solution = 105.0°C.
- The boiling point elevation, $\Delta T_b = 105.0°C – 100.0°C = 5.0°C$.
- The ebullioscopic constant for water, $K_b = 0.512 \, \text{°C/m}$.
- The molar mass of ethylene glycol (C₂H₆O₂) is 62.07 g/mol.
- The mass of water = 1.0 kg.
Step 2: Calculate the Molality of the Solution
The molality ($m$) is defined as:
$$
m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}
$$
Using the boiling point elevation equation, solve for molality:
$$
\Delta T_b = K_b \cdot m
$$
$$
5.0 = 0.512 \cdot m
$$
$$
m = \frac{5.0}{0.512} = 9.77 \, \text{mol/kg}
$$
Step 3: Calculate the Moles of Ethylene Glycol Needed
Now, we know that the molality is 9.77 mol/kg, and we have 1.0 kg of water. Therefore, the moles of ethylene glycol required are:
$$
\text{moles of ethylene glycol} = m \cdot \text{mass of solvent}
$$
$$
\text{moles of ethylene glycol} = 9.77 \, \text{mol/kg} \times 1.0 \, \text{kg} = 9.77 \, \text{mol}
$$
Step 4: Calculate the Mass of Ethylene Glycol
To find the mass of ethylene glycol, we use the relationship:
$$
\text{mass} = \text{moles} \times \text{molar mass}
$$
$$
\text{mass} = 9.77 \, \text{mol} \times 62.07 \, \text{g/mol} = 606.7 \, \text{g}
$$
Final Answer:
The mass of ethylene glycol that must be added to 1.0 kg of water to produce a solution that boils at 105.0°C is 606.7 grams.
Explanation:
The process involved using the concept of boiling point elevation to relate the change in boiling point to the amount of solute. We calculated the required molality using the known boiling point elevation, then used that molality to determine the number of moles of ethylene glycol needed. Finally, we converted moles to grams using the molar mass of ethylene glycol.