Home
Class 11
CHEMISTRY
A solution is separated from pure solven...

A solution is separated from pure solvent by a semipermeable membrane at 298 K. The difference in the height of the solution and the solvent is 0.9 m. If `K_(f)` and freezing point of the solvent are `30 K kg mol^(-1)` and 250.3 K, respectively, the temperature which the solution freezes is :
(Assume density of solution be 1 g/cc)

Promotional Banner

Similar Questions

Explore conceptually related problems

The boiling point of an aqueous solution is 100.18 .^(@)C . Find the freezing point of the solution . (Given : K_(b) = 0.52 K kg mol^(-1) ,K_(f) = 1.86 K kg mol^(-1))

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

Molal depression constant for a solvent is 4.0 kg mol^(-1) . The depression in the freezing point of the solvent for 0.03 mol kg^(-1) solution of K_(2)SO_(4) is : (Assume complete dissociation of the electrolyte)

Molal depression constant for a solvent is 4.0 K kg mol^(-1) . The depression in the freezing point of the solvent for 0.5 mol kg^(-1) solution of KI (Assume complete dissociation of the electrolyte) is _______ .

Molal depression constant for a solvent is 4.0 K kg mol^(-1) . The depression in the freezing point of the solvent for 0.5 mol kg^(-1) solution of KI (Assume complete dissociation of the electrolyte) is _______ .