Home
Class 12
CHEMISTRY
1.0 g of a non-electrolyte solute( mol. ...

`1.0 g` of a non-electrolyte solute( mol. Mass `250.0 g mol^(-1)`) was dissolved in `5.12 g` benzene. If the freezing point depression constant, `K_(f)` of benzene is `5.12 K kg mol^(-1)`, the freezing point of benzene will be lowered by:

A

`0.5 K`

B

`0.2 K`

C

`0.4 K`

D

`0.3 K`

Text Solution

Verified by Experts

The correct Answer is:
C

`Delta T_(f) = (1000xx K_(f) xx w)/(m xx W)`
`Delta T_(f) = (1000 xx 5.12 xx 1)/(250 xx 51.2) = 0.4 K`
Promotional Banner

Topper's Solved these Questions

  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 3B Objectice Problems|1 Videos
  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 9 Advanced Numerical|12 Videos
  • CHEMICAL KINETICS

    P BAHADUR|Exercise Exercise 9|13 Videos
  • ELECTROCHEMISTRY

    P BAHADUR|Exercise Exercise (9) ADVANCED NUMERICAL PROBLEMS|36 Videos

Similar Questions

Explore conceptually related problems

1.00 g of a non-electrolyte solute (molar mass 250g mol^(–1) ) was dissolved in 51.2 g of benzene. If the freezing point depression constant K_(f) of benzene is 5.12 K kg mol^(–1) , the freezing point of benzene will be lowered by:-

1.0 g of a non-electrolyte solute (molar mass 250gmol^(-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is 5.12K kgmol^(-1) the lowering in freezing point will be

1.0 g of non-electrolyte solute dissolved in 50.0 g of benzene lowered the freezing point of benzene by 0.40 K . The freezing point depression constant of benzene is 5.12 kg mol^(-1) . Find the molecular mass of the solute.