Home
Class 12
CHEMISTRY
A very small amount of a non-volatile so...

A very small amount of a non-volatile solute (that does not dissociate) is dissolved in `56.8 cm^(3)` of benzene (density `0.889 g cm^(3))`. At room temperature, vapour pressure of this solution is `98.88 mm Hg` while that of benzene is `100 mm Hg` . Find the molality of this solution. If the freezing temperature of this solution is `0.73` degree lower than that of benzene, what is the value of molal the freezing point depression constant of benzene?

Text Solution

Verified by Experts

We know,
`(P^(@)-P_(S))/(P_(S)) = (w)/(m) xx (M)/(W)`
`:. (100 - 98.88)/(98.88) = (w xx 78 xx 1000)/(m xx W xx 1000)`
or molality `((w xx 1000)/(mxxW)) = (1.12 xx 1000)/(78 xx 98.88) = 0.1452`
Also `Delta T = K'_(f) xx "molality"`
`0.73 = K'_(f) xx 0.1452`
`:. K'_(f) = 5.028 K "molality"^(-1)`
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 very small amount of non-volatile solute, non-electrolyte is dissolved in 50 cm^(3) of benzene (d=0.9 g cm^(-3)) . At room temperature, vapour prssure of this solution is 99 mm Hg while that of benzene is 100 Hg. If DeltaT_(f) is the 0.73 What is value of K_(f)?

A small amount of a non-volatile solute is dissilved in 64.5 cm^(3) of acetone ( density 0.791 g//cm^(3) ) the vapour pressure of this solution at room temperature is 260 mm Hg , while that of acetone is 285 mm Hg . What is the molality of the solution ?

The vapour pressure of pure benzene at 25^@C is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution of

If at certain temperature, the vapour pressure of pure water is 25 mm Hg and that of a very dilute aqueous urea solution is 24.5 mm Hg, the molality of the solution is

At 25^(@)C , the vapour pressure of pure water is 25.0 mm Hg . And that of an aqueous dilute solution of urea is 20 mm Hg . Calculate the molality of the solution.