An ethylene glycol solution contains 28.3 g of ethylene glycol, C₂H₆O₂ in 97.2 mL of water. Calculate the freezing point of the solution. The density of water 1.00 g/mL.

• A. –8.72°C
• B. –0.848°C
• C. –0.541°C
• D. –17.4°C

Answer: (A)

Verified step by step guidance

1. Determine the molar mass of ethylene glycol (C₂H₆O₂) by adding the atomic masses of its constituent elements: carbon (C), hydrogen (H), and oxygen (O). Use the periodic table to find these values.
2. Calculate the number of moles of ethylene glycol using the formula: \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \). Substitute the given mass of ethylene glycol and the molar mass you calculated.
3. Find the mass of water in grams using the density formula: \( \text{mass} = \text{density} \times \text{volume} \). Given the density of water is 1.00 g/mL and the volume is 97.2 mL, calculate the mass of water.
4. Calculate the molality of the solution using the formula: \( \text{molality} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). Convert the mass of water from grams to kilograms and use the moles of ethylene glycol calculated earlier.
5. Use the freezing point depression formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor (which is 1 for ethylene glycol), \( K_f \) is the freezing point depression constant for water (1.86 °C/m), and \( m \) is the molality. Calculate the change in freezing point and subtract it from the normal freezing point of water (0°C) to find the freezing point of the solution.