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: Identify the property.
Adding a non-volatile solute lowers the freezing point - this is depression of freezing point.
Step 2: Write the formula.
$\Delta T_f = K_f \times m$, where $m$ is molality and $K_f$ the cryoscopic constant.
Step 3: List the data.
$m = 1$ molal, $K_f = 1.86$ K kg mol$^{-1}$, and pure water freezes at $0^\circ$C.
Step 4: Compute the depression.
$\Delta T_f = 1.86 \times 1 = 1.86$ K, equivalent to $1.86^\circ$C of lowering.
Step 5: Subtract from the pure value.
$T_f = T_f^0 - \Delta T_f = 0 - 1.86 = -1.86^\circ$C.
Step 6: Match the option.
The new freezing point is $-1.86^\circ$C, which is option (4).
\[ \boxed{T_f = -1.86^\circ\text{C}} \]