Home
Class 12
CHEMISTRY
A solution containing 0.1 mol of naphtha...

A solution containing 0.1 mol of naphthalene and 0.9 mol of benzene is cooled out until some benzene freezes out. The solution is then decanted off from the solid and warmed upto `353 K` where its vapour pressure was found to be `670 mm`. The freezing point and boiling point of benzene are `278.5 K` and `353 K` respectively, and its enthalpy of fusion is `10.67 KJ "mol"^(-1)`. Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume ideal behaviour.

Text Solution

Verified by Experts

The correct Answer is:
`270.39 K, 12.14g`;
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

A solution containing 0.1 mole of naphthalenen and 0.9 mole of benzenen is cooled out until some benzene freezes out. The solution is then decanted off from the solid and warmed upto 353K where its vapour pressure was found to be 670 torr. The freezing point and and boiling point of benzene are 278.5K and 353K respectively and its enthalpy of fusion is 10.67 Lk "mol"^(-1) . Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume ideal behavious.

What is the freezing of 0.4 molal solution of acetic acid in benzene in which it dimerises to the extent of Freezing point of benzene is 278.K and its molar heat of fusion is 10-042KJ mol^(-1)

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.

The freezing point depression of 0.1 molal solution of acetic acid in benzene is 0.256 K , K_(f) for benzene is 5.12 K Kg mol^(-1) . What conclusion can you draw about the molecular state of acetic acid in benzence ?

The freezing point depression constant of benzene is 5.12 K kg mol^(-1) . The freezing point depression for the solution of molality 0.078m containing a non-electrolyte solute in benzene is

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.

The freezing point of 0.02 mol fraction solution of acetic acid (A) in benzene (B) is 277.4K . Acetic acid exists partly as a dimer 2A = A_(2) . Calculate equilibrium constant for the dimerisation. Freezing point of benzene is 278.4K and its heat of fusion DeltaH_(f) is 10.042 kJ mol^(-1) .