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Read the following paragraph and answer the question given below:
Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02 mol fraction of acetic acid in benzene is 277.4 K.
The molal cryoscopic constant of benzene in `K "molality"^(-1)` is

A

`4.0`

B

5.6

C

4.5

D

`5.0`

Text Solution

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The correct Answer is:
D
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The freezing point of 0.02 mole fraction acetic acid in benzene is 277.4 K . Acetic acid exists partly as dimer. Calculate the equilibrium constant for dimerization. The freezing point of benzene is 278.4 K and the heat the fusion of benzene is 10.042 kJ mol^(-1) . Assume molarity and molality same.

Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02mol fraction of acetic acid in benzene is 277.4K. The degree of dimerisation of acetic acid in benzene is

Knowledge Check

  • Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02mol fraction of acetic acid in benzene is 277.4K. The molal cryoscopic constant of benzene in K "molality"^(-1) is

    A
    `4.0`
    B
    5.6
    C
    4.5
    D
    `5.0`
  • Read the following paragraph and answer the question given below: Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02 mol fraction of acetic acid in benzene is 277.4 K. The degree of dimerisation of acetic acid in benzene is

    A
    `0.24`
    B
    `0.72`
    C
    `0.48`
    D
    `0.64`
  • Read the following paragraph and answer the question given below: Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02 mol fraction of acetic acid in benzene is 277.4 K. The equilibrium constant for dimerisation of acetic in benzene is

    A
    2.4
    B
    3.39
    C
    2.42
    D
    2.56
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    Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02mol fraction of acetic acid in benzene is 277.4K. The equilibrium constant for dimerisation of acetic acid in benzene is

    The depression in freezing point of benzene is 0.45^(@)C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, the percentage association of acetic acid in benzene will be ( K_(f) for benzene =5.12K kg mol^(-1) )

    Read the paragraph carefully and answer the following questions: Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K. Freezing point of benezene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer than answer the following based upon above data K_(1) for benzene is

    Read the paragraph carefully and answer the following questions: Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K. Freezing point of benezene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer than answer the following based upon above data Calculate the equilibrium constant for dimerisation of acetic acid

    Read the paragraph carefully and answer the following questions: Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K. Freezing point of benezene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer than answer the following based upon above data What should be the molality of the solution formed