Phenol associates in benzene to a certain extent to form dimer. A solution containing 2.0 × 10-2 kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is (Kf for benzene = 5.12 K kg mol-1)

Question

Phenol associates in benzene to a certain extent to form dimer. A solution containing 2.0 × 10-2 kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is (Kf for benzene = 5.12 K kg mol-1) (A) 73.4% (B) 50.1% (C) 42.3% (D) 25.1%

Tardigrade Admin · Accepted Answer

M2(obs) = (Kf⋅ w⋅1000/W⋅Δ Tf) = (5.12×2.0× 10-1× 1000/1.0× 0.69) = 148.4 Calculated molecular mass of phenol = 94 i = (M2(cal)/M2(obs)) = (94/148.4) = 0.633 beginmatrix2C6H5OH&leftharpoons&(C6H5OH)2 1-α&&α/2 endmatrix Total species =(1-α) + (α/2) = 1- (α/2) i = (1-α /2/1) or (α/2) = 1-i or α = 2(1- i) = 2(1- 0.633) = 0.734 = 73.4 %

Q. Phenol associates in benzene to a certain extent to form dimer. A solution containing 2.0×10−2kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is (Kf​ for benzene =5.12Kkgmol−1)

73.4%

50.1%

42.3%

25.1%

Q. Phenol associates in benzene to a certain extent to form dimer. A solution containing $2.0 \times 1 0^{- 2} k g$ of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is ( $K_{f}$ for benzene $= 5.12 K k g m o l^{- 1}$ )