Home
Class 12
CHEMISTRY
Pure benzene freezes at 5.45^(@)C at a c...

Pure benzene freezes at `5.45^(@)C` at a certain place but a 0.374m solution of tetrachloroethane in benzene freezes at `3.55^(@)C`. The `K_(f)` for benzene is

A

`5.08 K kg mol^(-1)`

B

`508 K kg mol^(-1)`

C

`0.508 K kg mol^(-1)`

D

`50.8^(@)C kg mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaT_(f) = (5.45 - 3.55) = K_(f) xx m`
or `1.90 = k_(f) xx 0.374`
or `k_(f) = (1.90)/(0.374) = 5.08`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    DINESH PUBLICATION|Exercise REVISION QUESTION|186 Videos

  • SOLUTIONS

    DINESH PUBLICATION|Exercise OBJECTIVE TYPE MCQs|47 Videos

  • SOLID STATE

    DINESH PUBLICATION|Exercise Brain storming|10 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    DINESH PUBLICATION|Exercise ULTIMATE PREPARATORY PACKAGE|20 Videos

Similar Questions

Explore conceptually related problems

Pure benzene freeze at 5.45^(@)"C" . A solution containing 7.24g of "C"_(2)"H"_(2)"Cl_(4) in 115.3 g of benzene was observed to freeze at 3.55^(@)"C". What is the molal freezing point constant of benzene?

Pure benzene freezes t 5.3^(@)C . A solution of 0.223 g of phenylacetic acid (C_(6)H_(5)CH_(2)COOH) in 4.4 g of benzene (K_(f) = 5.12 Kkg mol^(-1)) freezes at 4.47^(@)C . From the observation one can conclude that :

The freezing point of a solution contaning 0.3 g of acetic acid in 43 g of benzene reduces by 0.3^(@) . Calculate the Van's Hoff factor "( K_(f) for benzene = 5.12 K kg mol^(-1) )"