An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. Required:
a. Determine the freezing point of the solution. Express you answer in degrees Celsius. (Assume a density of 1.00 g/mL for water.)
b. Compute the boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

a. TFinal = -6.57°C

b. Tfinal = 101.80°C

Explanation:

When a solute is added to a solvent producing an ideal solution, the freezing point of the solution decreases with regard to pure solvent. Also, boiling point increases with regard to pure solvent.

The formulas are:

Freezing point:

ΔT = Kf×m×i

Where Kf is freezeing point depression constant of water (1.86°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).

Boiling point:

ΔT = Kb×m×i

Where K is freezeing point depression constant of water (0.51°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).

Moles of 21.4g of ethylene glycol (Molar mass: 62.07g/mol) are:

21.4g C₂H₆O₂ ₓ (1mol / 62.07g) = 0.345 moles

And kg of 97.6mL of water = 97.6g are 0.0976kg. Molality of the solution is:

0.345mol / 0.0976kg = 3.5325m

Replacing in the formulas:

a. Freezing point:

ΔT = 1.86C/m×3.5325m×1

ΔT = 6.57°C

0°C - Tfinal = 6.57°C

b. Boiling point:

ΔT = 0.51°C/m×3.5325m×1

ΔT = 1.80°C

Tfinal - 100°C = 1.80°C

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the answer is c i took the test