Freezing-point depression fundamentals: How does the depression in freezing point (ΔTf) of a solution depend on the amounts of solute and solvent, assuming a non-electrolyte and dilute solution?

Difficulty: Easy

Correct Answer: Both (a) and (b).

Explanation:


Introduction / Context:
Cryoscopic (freezing-point) measurements are widely used to estimate molar mass and colligative properties. Understanding how ΔTf scales with solute amount and solvent quantity is essential for lab calculations and industrial formulations (antifreeze, brines, food science).


Given Data / Assumptions:

  • Dilute, ideal solution of a non-electrolyte (i = 1).
  • Solvent properties captured by cryoscopic constant Kf.
  • Temperature range small enough for linear approximation.


Concept / Approach:
The relation is ΔTf = Kf * m, where m is molality. By definition, m = moles of solute / kilograms of solvent. Therefore, ΔTf increases linearly with moles of solute for a fixed solvent mass and decreases if the mass of solvent increases while solute moles stay constant.


Step-by-Step Solution:

Step 1: Write ΔTf = Kf * (n_solute / kg_solvent).Step 2: For constant kg_solvent, ΔTf ∝ n_solute (direct proportionality).Step 3: For constant n_solute, ΔTf ∝ 1 / kg_solvent (inverse proportionality).Step 4: Conclude both statements (a) and (b) are correct for the specified conditions.


Verification / Alternative check:
Units: Kf has units of °C·kg/mol; multiplying by mol/kg yields °C, confirming consistency.


Why Other Options Are Wrong:

  • Only (a) or only (b): Incomplete; both relationships hold.
  • Neither: Contradicted by the standard cryoscopic equation.


Common Pitfalls:
For electrolytes, van’t Hoff factor i > 1 increases ΔTf. Also, confusing molality with molarity; molality uses mass of solvent, not volume of solution, so temperature changes do not affect it directly.


Final Answer:
Both (a) and (b).

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