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