Question

when phenol is dissolved in benzene its 2 particle are associated to form 1larger molecule when 2green of powder is dissolved in 100gm of benzene then there is a decrease of 0.69 C in freezing point find the degree of association of phenol​

Answer 1

Answer:2C  

6

​  

H  

5

​  

OH⇌(C  

6

​  

H  

5

​  

OH)  

2

​  

 

    1 mole                      0

   1−αα/2

Total=1−α+  

2

α

​  

=1−  

2

α

​  

 

i=  

1

1−(α/2)

​  

=1−  

2

α

​  

 

ΔT  

f

​  

=iK  

f

​  

W  

2

​  

×1000Mw  

2

​  

×W  

1

​  

 

0.69=  

94×1.0kg

i×5.12×20×10  

3

 

​  

kg

i=1−  

2

α

​  

,α=0.733=73.3%

Explanation:

Answer 2

Given:

w = 2 gm

W = 100 gm

ΔTf = 0.69 C

2 molecules of phenol associate to form 1 molecule in benzene.

To Find:

The degree of association of phenol​.

Calculation:

- For benzene, Kf = 5.12 K/mol

- For phenol, M.wt = 94 gm

- We know that:

ΔTf = i × Kf × m

⇒ ΔTf = i × Kf × (w × 1000) / (M.wt × W)

⇒ 0.69 = i × 5.12 × (2 × 1000) / (94 × 100)

⇒ i = (0.69 × 94) / (5.12 × 20)

⇒ i = 64.86 / 102.4

⇒ i = 0.633

- The degree of association can be given as:

α = n (1 - i) / (n -1)

⇒ α = 2 × (1 - 0.633) / (2 -1)

⇒ α = 2 × 0.367

⇒ α = 0.734

- So, the degree of association of phenol​ is 0.734.