Question

What is the freezing point of 1 molal aqueous solution of a non volatile solute? ($K_f = 1.86$ K kg mol$^{-1}$) ($T_f^0$ for water = 0$^\circ$C)

• A. $-0.93^\circ$C
• B. $-2.43^\circ$C
• C. $-3.72^\circ$C
• D. $-1.86^\circ$C

Hint:

By definition, the cryoscopic constant $K_f$ is the exact amount by which the freezing point drops when exactly 1 mole of solute is dissolved in 1 kg of solvent. Since the solution is 1 molal, the drop is exactly equal to the value of $K_f$, pulling the freezing point down from 0$^\circ$C to $-K_f^\circ$C.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:

We need to find the new freezing point of an aqueous solution whose molality is exactly 1 molal, given the cryoscopic constant ($K_f$) of water and the freezing point of pure water.


Step 2: Key Formula or Approach:

The depression in freezing point ($\Delta T_f$) for a solution containing a non-volatile, non-electrolyte solute is given by the formula: $$\Delta T_f = K_f \times m$$ where $K_f$ is the freezing point depression constant and $m$ is the molality of the solution. The freezing point of the solution ($T_f$) is then determined by: $$T_f = T_f^0 - \Delta T_f$$

Step 3: Detailed Explanation:
Given: Molality ($m$) = 1 molal = 1 mol kg$^{-1}$ Cryoscopic constant ($K_f$) = 1.86 K kg mol$^{-1}$ Freezing point of pure water ($T_f^0$) = 0$^\circ$C Calculate the depression of freezing point ($\Delta T_f$): $$\Delta T_f = 1.86 \text{ K kg mol}^{-1} \times 1 \text{ mol kg}^{-1} = 1.86 \text{ K}$$ A temperature difference of 1.86 K is equivalent to a difference of 1.86$^\circ$C. Now, determine the final freezing point ($T_f$): $$T_f = 0^\circ\text{C} - 1.86^\circ\text{C} = -1.86^\circ\text{C}$$

Step 4: Final Answer:
The freezing point of the 1 molal solution is $-1.86^\circ$C, which corresponds to option (D).