Question

Acetic acid in benzene exhibits a molar mass of 80 g/mol. Its degree of dimerisation is​

Answer 1

Answer:

Acetic acid undergoes dimerisation, when dissolved in benzene.

Molecular mass of acetic acid is found 120.

The correct relation is α=2(

d

d−D

)

where d= Observed vapour density

D= Theoretical vapour density

α= degree of association

n= number of molecules that undergo association.

For association of molecule,

α=

d(1−

n

1

)

d−D

Substitute n=2,

α=

d(1−

2

1

)

d−D

α=

d(

2

1

)

d−D

α=2(

d

d−D

)

Hence, option (C) is the correct answer.

Answer 2

Answer:  The molar mass of $80 g/mol$ of Acetic acid in Benzene is $160 g/mol$.

Explanation:

• What is Molar Mass ?

The Molar Mass/ Molecular Weight can be defined as the sum of total mass in grams of atoms present to make up a molecule per mole.

• What is Degree of Dimerisation ?

The degree of Dimersation can be defined as the addition reaction in which two molecules of same compound reacts with each other to give an adduct.

• What will be degree of dimerisation of Acetic Acid in Benzene exhibiting a molar mass of $80 g/mol$ ?

Acetic acid dimerises in Benzene due to the presence of Hydrogen  bond between its molecules, its molar mass becomes double the actual amount.

Hence, the molar mass becomes:

$80 g/mol$ × $2$ $= 160 g/mol$

• If α is the degree of dimerisation of Acetic acid in Benzene ,

Then van't Hoff factor ' i ' is

$2(CH_{3} COOH$⇄ $(CH_{2}COOH )_{2}$

2 mole of Acetic acid associates to form 1 mole of dimerisation.

For accociate,

$i = 1 - (1-\frac{1}{n} )\alpha$

For association in the absence of dissociation, the van't Hoff factor is :

$i \leq 1$

Hence,

$i = 1 - (1 - \frac{1}{n} )$$\alpha$

$i = 1 - (1 - \frac{1}{2} )\alpha$

$i = 1 - \frac{\alpha }{2}$

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