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Byju's Answer

Standard XII

Chemistry

Nucleophilic Substitution in Aryl Halides

Acetic acid f...

Question

Acetic acid forms a dimer in non polar solvent benzene through hydrogen bonding the observed molecular mass of acetic acid assuming $25 %$ dimerisation in benzene is $x$ . The value of $0.875 x$ is

60.0

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60.00

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Solution

$M_{obs} = \frac{M_{actual}}{i}$

$α = 0.25$
$∴ i = 1 - \frac{α}{2} = 1 - \frac{0.25}{2}$
$\Rightarrow i = 0.875$
$∴ M_{obs} (x) = \frac{60}{0.875}$
The value of $0.875 x$ is $0.875 \times \frac{60}{0.875} = 60$

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Similar questions

Q. 1 mol

C H_{3} C O O H

is added in 250 g benzene. Acetic acid dimerises in benzene due to hydrogen bond.

K_{b}

of benzene is 2 K

k g m o l^{- 1}

. The boiling point has increased by 6.4 K. % dimerisation of acetic acid is :

Q. The freezing point of a solution of acetic acid (mole fraction =

0.02

) in benzene is

277.4 K

. Acetic acid exists partly as a dimer

2 A ⇌ A_{2} .

Determine the equilibrium constant for dimerisation. Freezing point of benzene is

278.4 K

and

K_{f}

for benzene is

5

Acetic acid undergoes dimerisation, when dissolved in benzene (as shown in figure)
Molecular mass of acetic acid is found to be

120

Which among the following relation is correct?

(where,

α

= degree of association,

d

= observed vapour density,

D

= Theoretical vapour density).

Q. The freezing point of

0.02

mole fraction of acetic acid in benzene is

277.4

K. Acetic acid exists partly as dimer. Calculate the equilibrium constant for dimerisation. Freezing point of benzene is

278.4

K and heat of fusion of benzene is

10.042

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Q. The freezing point of a solution containing

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[Assume acetic acid dimerizes in benzene and

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= 5.12

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]

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Acetic acid forms a dimer in non polar solvent benzene through hydrogen bonding the observed molecular mass of acetic acid assuming 25% dimerisation in benzene is x. The value of 0.875x is

Mobs=Mactuali α=0.25 ∴i=1−α2=1−0.252 ⇒i=0.875 ∴Mobs(x)=600.875 The value of 0.875x is 0.875×600.875=60

Acetic acid forms a dimer in non polar solvent benzene through hydrogen bonding the observed molecular mass of acetic acid assuming $25 %$ dimerisation in benzene is $x$ . The value of $0.875 x$ is

$M_{obs} = \frac{M_{actual}}{i}$

$α = 0.25$
$∴ i = 1 - \frac{α}{2} = 1 - \frac{0.25}{2}$
$\Rightarrow i = 0.875$
$∴ M_{obs} (x) = \frac{60}{0.875}$
The value of $0.875 x$ is $0.875 \times \frac{60}{0.875} = 60$