1 mol CH 3 COOH is added in 250 g benzene. Acetic acid dimerises in benzene due to hydrogen bond, Kb of benzene is 2 K kgmol -1. The boiling point has increased by 6.4 K . % dimerisation of acetic acid is:

Question

1 mol CH 3 COOH is added in 250 g benzene. Acetic acid dimerises in benzene due to hydrogen bond, Kb of benzene is 2 K kgmol -1. The boiling point has increased by 6.4 K . % dimerisation of acetic acid is: (A) 50 (B) 40 (C) 30 (D) 20

Tardigrade Admin · Accepted Answer

Δ T b = i K b ⋅ m Given molality =(1 × 1000/250)=4 m , 6.4= i × 2 × 4 or i =0.8 For dimerisation i=1-(β/2) ⇒ 0.8-1=(-β/2) or β=0.4 ⇒ 40 %

Q. $1 m o l C H_{3} COO H$ is added in $250 g$ benzene. Acetic acid dimerises in benzene due to hydrogen bond, $K_{b}$ of benzene is $2 K$ kgmol $^{- 1}$ . The boiling point has increased by $6.4 K .%$ dimerisation of acetic acid is:

50

40

30

20

$Δ T_{b} = i K_{b} \cdot m$

Given molality $= \frac{1 \times 1000}{250} = 4 m, 6.4 = i \times 2 \times 4$ or $i = 0.8$

For dimerisation

$i = 1 - \frac{β}{2}$

$\Rightarrow 0.8 - 1 = \frac{- β}{2}$ or

$β = 0.4 \Rightarrow 40%$

Q. 1molCH3​COOH is added in 250g benzene. Acetic acid dimerises in benzene due to hydrogen bond, Kb​ of benzene is 2K kgmol −1. The boiling point has increased by 6.4K.% dimerisation of acetic acid is:

50

40

30

20

ΔTb​=iKb​⋅m Given molality =2501×1000​=4m,6.4=i×2×4 or i=0.8 For dimerisation i=1−2β​ ⇒0.8−1=2−β​ or β=0.4⇒40%

Q. $1 m o l C H_{3} COO H$ is added in $250 g$ benzene. Acetic acid dimerises in benzene due to hydrogen bond, $K_{b}$ of benzene is $2 K$ kgmol $^{- 1}$ . The boiling point has increased by $6.4 K .%$ dimerisation of acetic acid is:

$Δ T_{b} = i K_{b} \cdot m$

Given molality $= \frac{1 \times 1000}{250} = 4 m, 6.4 = i \times 2 \times 4$ or $i = 0.8$

For dimerisation

$i = 1 - \frac{β}{2}$

$\Rightarrow 0.8 - 1 = \frac{- β}{2}$ or

$β = 0.4 \Rightarrow 40%$