please help chem question

Question

Assuming 100% dissociation, calculate the freezing point ( 𝑇f ) and boiling point ( 𝑇b ) of 1.75 𝑚 AgNO3(aq) .

Answer 1

The equation to be used for the freezing point depression and boiling point elevation is

∆T = imK

∆T = change in freezing or boilling point = ?

i = van't Hoff factor = 2 for AgNO3 since it dissociates into 2 particle (Ag+ and NO3-)

m = molality = 1.75 m

K = freezing or boiling constant for water (assuming the solution is aqueous)

K_freezing = 1.86º/m

K_boiling = 0.512º/m

Freezing point calculation

∆T = (2)(1.75)(1.86) = 6.5º

New freezing point = -6.5ºC since water normally freezes at 0º

Boiling point calculation

∆T = (2)(1.75)(0.512) = 1.8º

New boiling point = 101.8º since water normally boils at 100ºC