When #"18 g"# of ethylene glycol (#C_2H_6O_2#) is dissolved in #"150 g"# of pure water, what is the freezing point of the solution? (The freezing point depression constant for water is #1.86^@ "C"cdot"kg/mol"#.)
Approx.
Ethylene glycol is a molecular species, and we need to calculate the molality of its water solution.....
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To find the freezing point of the solution, we can use the formula for freezing point depression:
ΔTf = i * Kf * m
Where: ΔTf = Freezing point depression i = Van't Hoff factor (for ethylene glycol, which is a nonelectrolyte, i = 1) Kf = Freezing point depression constant for water (given as 1.86°C⋅kg/mol) m = Molality of the solution (moles of solute per kilogram of solvent)
First, we need to calculate the molality of the solution:

Calculate the moles of ethylene glycol: moles = mass / molar mass moles = 18 g / 62.07 g/mol (molar mass of ethylene glycol) moles ≈ 0.290 mol

Calculate the total mass of the solution: mass of solution = mass of ethylene glycol + mass of water mass of solution = 18 g + 150 g = 168 g

Calculate the molality: molality = moles of solute / mass of solvent (in kg) molality = 0.290 mol / 0.150 kg molality ≈ 1.933 mol/kg
Now, we can plug the values into the freezing point depression formula:
ΔTf = (1) * (1.86°C⋅kg/mol) * (1.933 mol/kg) ΔTf ≈ 3.61°C
Since the freezing point depression is the difference between the freezing point of the pure solvent (water) and the freezing point of the solution, we subtract ΔTf from the freezing point of water:
Freezing point of water  ΔTf = 0°C  3.61°C = 3.61°C
Therefore, the freezing point of the solution is approximately 3.61°C.
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