Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 1.99 m AgNO3(aq). Constants may be found here.

Answer 1

[AgNO3]= 1.99 m   , we know m is molality which is equal to = mol of solute / Mass of solute in kg.

Since the dissociation is 100 % so the i ( number of particles that are formed in the solution ) = 2

Solution:

We use following formulae.

Delta Tf = i x m x kf

Here I = 2 , m is molality, kf is the freezing point constant of water.

Delta Tf = freezing pint of the solvent – freezing point of the solution = 0⁰ C – (freezing point of the solution)

Kf of water = -1.86 ⁰C/ m

Lets plug the given values.

-(freezing point of the solution) = 2 x 1.99 m x ( -1.86 ⁰C/m)

Freezing point of the solution =- 7.40 ⁰C