A non-volatile solute is dissolved in water. The \(\Delta T_f\) of the resultant solution is 0.052 K. What is the freezing point of the solution (in K)?
(Given: \( K_b \) of water = 0.52 K kg mol\(^{-1}\), \( K_f \) of water = 1.86 K kg mol\(^{-1}\), Freezing point of water = 273 K)
(Given: \( K_b \) of water = 0.52 K kg mol\(^{-1}\), \( K_f \) of water = 1.86 K kg mol\(^{-1}\), Freezing point of water = 273 K)
Show Hint
- \( 272.628 \)
- \( 273.186 \)
- \( 273.000 \)
- \( 272.814 \)
The Correct Option is D
Solution and Explanation
Step 1: Use the freezing point depression formula \[ \Delta T_f = i \times K_f \times m \] where: - \(\Delta T_f\) is the depression in freezing point, - \(K_f\) is the cryoscopic constant, - \(m\) is the molality of the solute.
Step 2: Calculate the new freezing point \[ T_f = T_0 - \Delta T_f \] where: - \( T_0 \) is the normal freezing point of water (273 K), - \( \Delta T_f \) is given as 0.052 K. \[ T_f = 273 - 0.052 \] Step 3: Compute the final answer \[ T_f = 272.814 { K} \] Thus, the correct answer is \(\mathbf{272.814 \, K}\).
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