What is the required concentration (in percent by mass) for an aqueous ethylene glycol (C2H6O2) solution to have a boiling point of 104.0 °C?

Verified step by step guidance

1. <insert step 1: Understand that the boiling point elevation formula is \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the boiling point elevation, \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant of the solvent, and \( m \) is the molality of the solution.>
2. <insert step 2: Calculate \( \Delta T_b \) by subtracting the normal boiling point of water (100.0 °C) from the given boiling point (104.0 °C).>
3. <insert step 3: Use the van't Hoff factor \( i = 1 \) for ethylene glycol, as it does not dissociate in solution.>
4. <insert step 4: Rearrange the boiling point elevation formula to solve for molality \( m \): \( m = \frac{\Delta T_b}{i \cdot K_b} \). Use the known \( K_b \) value for water, which is 0.512 °C kg/mol.>
5. <insert step 5: Convert the molality to percent by mass. Assume a certain mass of water (e.g., 1 kg) to find the mass of ethylene glycol using its molality, then calculate the percent by mass using the formula: \( \text{percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%.\)>