Given below are two statements:
Statement (I) : Molal depression constant K_f is given by (M₁ RT_f)/(Δ S_(fus)), where symbols have their usual meaning.
Statement (II) : K_f for benzene is less than the K_f for water.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Statement I is the familiar K_f formula in disguise; for Statement II, recall two numbers.

1. Standard expression. The molal depression constant depends only on the solvent. Here M₁ is the solvent's molar mass in kg mol⁻¹ and T_f is its freezing point.

   K_f = (M₁ R T_f²)/(Δ H_(fus)) = (M₁ R T_f)/(Δ H_(fus)/T_f) = (M₁ R T_f)/(Δ S_(fus))

2. Check Statement I. At the freezing point, solid and liquid are in equilibrium, so Δ S_(fus) = Δ H_(fus)/T_f. Substituting this gives exactly the expression in Statement I, so Statement I is correct.
3. Check Statement II. K_f for benzene is 5.12 K kg mol⁻¹ and K_f for water is 1.86 K kg mol⁻¹. Benzene's value is larger, so Statement II is incorrect.
4. So. Statement I is correct and Statement II is incorrect, which is option (a).

The idea to keep. Remember K_f = M₁ R T_f²/Δ H_(fus) and that Δ S_(fus) = Δ H_(fus)/T_f at the freezing point. Benzene's K_f is bigger than water's.

Answer from the NTA final answer key. Solution by PrepWiser.