Question

The molal freezing point constant for water is \(1.86^\circ C\). The freezing point of \(0.1\,m\;NaCl\) solution is expected to be:

• A. \(-1.86^\circ C\)
• B. \(-0.372^\circ C\)
• C. \(-0.186^\circ C\)
• D. \(0.372^\circ C\)

Hint:

Electrolytes produce more particles in solution, so freezing point depression becomes larger. For \(NaCl\), always remember: \[ i \approx 2 \] because it dissociates into two ions.

Answer 1

The Correct Option is B

Concept: Depression in freezing point is a colligative property and is given by: \[ \Delta T_f = iK_fm \] where:
• \( \Delta T_f \) = depression in freezing point
• \( i \) = van't Hoff factor
• \( K_f \) = molal freezing point constant
• \( m \) = molality The freezing point of solution is: \[ T_f = T_f^\circ - \Delta T_f \] where \(T_f^\circ\) is the freezing point of pure solvent.

Step 1: Finding van't Hoff factor for \(NaCl\).
Sodium chloride dissociates as: \[ NaCl \rightarrow Na^+ + Cl^- \] Thus, total ions formed \(=2\). Therefore, \[ i=2 \]

Step 2: Calculating depression in freezing point.
Given: \[ K_f = 1.86^\circ C \] \[ m=0.1 \] \[ i=2 \] Using formula: \[ \Delta T_f = iK_fm \] Substituting values: \[ \Delta T_f = 2 \times 1.86 \times 0.1 \] \[ \Delta T_f = 0.372^\circ C \]

Step 3: Calculating freezing point of solution.
Freezing point of pure water: \[ 0^\circ C \] Hence, \[ T_f = 0 - 0.372 \] \[ T_f = -0.372^\circ C \] Therefore, the freezing point of the solution is: \[ -0.372^\circ C \]