List - I ( Solution) List - II ( Property) (P) 0.1 molal aq. urea (1) Δ T b = 0.1 3 ∘ C (Q) 0.1 molar NaCl (at 2 5 ∘ C ) ( α = 1 ) (2) π = 4.89 atm (R) 0.1 mol PhOH added to 1 kg benzene ( ℓ ) (Assu

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List - I ( Solution) List - II ( Property) (P) 0.1 molal aq. urea (1) Δ T b ​ = 0.1 3 ∘ C (Q) 0.1 molar NaCl (at 2 5 ∘ C ) ( α = 1 ) (2) π = 4.89 atm (R) 0.1 mol PhOH added to 1 kg benzene ( ℓ ) (Assume 100% dimerization) (3) Freezing point = 272.8 K (S) 0.1 mol PhOH ( K a ​ = 1 0 − 10 ) added to 1 kg H 2 ​ O (4) Relative lowering of vapour pressure = 0.18% Given ( K b ​ ) H 2 ​ O ​ = 0.5 m o l K − K g ​ ] ( K f ​ ) C 6 ​ H 6 ​ ​ = 5 m o l K − K g ​ ​ ; ( K f ​ ) H 2 ​ O ​ ​ = 2 m o l K − K g ​ ​ ; ( K b ​ ) C 6 ​ H 6 ​ ​ = 2.6 m o l K − K g ​ ​

TestHub · Accepted Answer

(P) Δ T f ​ = 1 × 2 × 0.1 = 0.2 T f ​ ​ = 272.8 K ​ ​ ​ ​ ( Q ) π = i CRT π = 2 × 0.1 × 0.082 × 298 π = 4.89 atm (R) Δ T b ​ = iK . m = 2 1 ​ × 2.6 × 0.1 (S) X solute ​ = 1000/18 0.1 ​ = 1000 1.8 ​ or 0.18%

List - I (Solution)	List - II (Property)
(P) 0.1 molal aq. urea	(1) $Δ T_{b} = 0.1 3^{\circ} C$
(Q) 0.1 molar NaCl (at $2 5^{\circ} C$ ) ( $α = 1$ )	(2) $π = 4.89 atm$
(R) 0.1 mol PhOH added to 1 kg benzene ( $ℓ$ ) (Assume 100% dimerization)	(3) Freezing point $= 272.8 K$
(S) $0.1 mol PhOH (K_{a} = 1 0^{- 10})$ added to $1 kg H_{2} O$	(4) Relative lowering of vapour pressure $= 0.18%$

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