83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ___________ K. (Nearest integer)
Use: Molal Freezing point depression constant of water = 1.86 K kg mol\(^{-1}\)
Freezing point of water = 273 K
Atomic masses: C: 12.0 u, O: 16.0 u, H: 1.0 u
Show Hint
Remember the definitions of concentration units. Molality (\(m\)) is moles of solute per kg of \textbf{solvent}, while Molarity (M) is moles of solute per liter of \textbf{solution}. Colligative property formulas like freezing point depression and boiling point elevation use molality.
Step 1: Understanding the Question
We need to calculate the freezing point of an aqueous solution of ethylene glycol. This is a problem based on the colligative property of freezing point depression. Step 2: Key Formula or Approach
The depression in freezing point (\(\Delta T_f\)) is given by:
\[ \Delta T_f = K_f \times m \]
where \(K_f\) is the molal freezing point depression constant and \(m\) is the molality of the solution.
The freezing point of the solution (\(T_f\)) is then:
\[ T_f = T_f^0 - \Delta T_f \]
where \(T_f^0\) is the freezing point of the pure solvent. Step 3: Detailed Calculation Calculate the molar mass of ethylene glycol (C\(_2\)H\(_6\)O\(_2\)):
Molar Mass = 2(C) + 6(H) + 2(O) = 2(12.0) + 6(1.0) + 2(16.0) = 24 + 6 + 32 = 62 g/mol. Calculate the moles of ethylene glycol:
Mass of ethylene glycol = 83 g
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{83 \text{ g}}{62 \text{ g/mol}} \approx 1.3387 \text{ mol} \]
Calculate the molality (m) of the solution:
Mass of solvent (water) = 625 g = 0.625 kg
\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{1.3387 \text{ mol}}{0.625 \text{ kg}} \approx 2.142 \text{ mol/kg} \]
Calculate the depression in freezing point (\(\Delta T_f\)):
\(K_f\) for water = 1.86 K kg mol\(^{-1}\)
\[ \Delta T_f = 1.86 \text{ K kg mol}^{-1} \times 2.142 \text{ mol kg}^{-1} \approx 3.984 \text{ K} \]
Calculate the freezing point of the solution (\(T_f\)):
\(T_f^0\) (water) = 273 K
\[ T_f = 273 \text{ K} - 3.984 \text{ K} = 269.016 \text{ K} \]
Step 4: Final Answer
The freezing point of the solution to the nearest integer is 269 K.
