Question

A solution of 0.3 g of benzoic acid in 20 g benzene freeze at 0.317°C below the freezing point of the solvent calculate da degree of association of the acid assuming it to exist in the dimeric state apprant molar mass of the acid kf =5.1 kkg mol-¹.​

Answer 1

Answer:

Answer

Correct option is

A

99.2 %

Given,

W  

B

​

=2g,K  

f

​

=4.9KKgmol  

−1

 

W  

A

​

=25g,ΔT  

f

​

=1.62K

Now,

ΔT  

f

​

=K  

f

​

×  

M  

B

​

 

W  

B

​

 

​

×  

W  

A

​

 

1000

​

 

M  

B

​

=  

1.62×25

4.9×2×1000

​

=241.98g/mol

2C  

6

​

H  

5

​

COOH⇌(C  

6

​

H  

5

​

COOH)  

2

​

 

If x is the degree of association, (1−x) mole of benzoic acid left undissociated & corresponding x/2 as associated moles of C  

6

​

H  

5

​

COOH at equilibrium.

∴ Total number of moles of particles at eqm,

⇒1−x+  

2

x

​

=1−  

2

x

​

=i

⇒i=  

Abnormalmolecularmass

Normalmolecularmass

​

 

⇒1−  

2

x

​

=  

241.98

122

​

 

⇒  

2

x

​

=1−  

241.98

122

​

⇒x=0.992=99.2%

∴ Degree of association of benzene =99.2%

Explanation: