Question:
31 g of ethylene glycol (C$_2$H$_6$O$_2$) is mixed with 500 g of solvent ($K_f = 2$ K kg mol$^{-1}$). What is the freezing point of the solution in K? (Freezing point of solvent = 273 K)
31 g of ethylene glycol (C$_2$H$_6$O$_2$) is mixed with 500 g of solvent ($K_f = 2$ K kg mol$^{-1}$). What is the freezing point of the solution in K? (Freezing point of solvent = 273 K)
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Non-electrolytes: van’t Hoff factor = 1.
Updated On: May 2, 2026
- 272
- 271
- 270
- 274
- 275
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The Correct Option is B
Solution and Explanation
Concept: Freezing point depression
\[ \Delta T_f = K_f \times m \] ---
Step 1: Molar mass
\[ C_2H_6O_2 = 2(12) + 6(1) + 2(16) = 62 \] ---
Step 2: Moles
\[ \frac{31}{62} = 0.5 \text{ mol} \] ---
Step 3: Molality
\[ m = \frac{0.5}{0.5} = 1 \] ---
Step 4: Depression
\[ \Delta T_f = 2 \times 1 = 2 \text{ K} \] ---
Step 5: Final freezing point
\[ 273 - 2 = 271 \text{ K} \] --- Final Answer: \[ \boxed{271 \text{ K}} \]
\[ \Delta T_f = K_f \times m \] ---
Step 1: Molar mass
\[ C_2H_6O_2 = 2(12) + 6(1) + 2(16) = 62 \] ---
Step 2: Moles
\[ \frac{31}{62} = 0.5 \text{ mol} \] ---
Step 3: Molality
\[ m = \frac{0.5}{0.5} = 1 \] ---
Step 4: Depression
\[ \Delta T_f = 2 \times 1 = 2 \text{ K} \] ---
Step 5: Final freezing point
\[ 273 - 2 = 271 \text{ K} \] --- Final Answer: \[ \boxed{271 \text{ K}} \]
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