Assuming 100% dissociation, calculate the freezing point and boiling point of 1.

Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 1.48 m Na2SO4(aq). Constants may be found here.

(°C/m)

Normal freezing

point (°C)

(°C/m)

Normal boiling

point (°C)

Tf_____________ degrees celcius

Tb_____________ degrees celcius

Solvent Formula Kf value*

(°C/m)

Normal freezing

point (°C)

Kb value

(°C/m)

Normal boiling

point (°C)

water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3 1.22 78.4 carbon
tetrachloride   CCl4 29.8 –22.9 5.03 76.8 camphor   C10H16O 37.8 176
*When using positive Kf values, assume that Tf is the absolute value of the change in temperature. If you would prefer to define Tf as "final minus initial" temperature, then Tf will be negative and so you must use negative Kf values. Either way, the freezing point of the solution should be lower than that of the pure solvent.

Tf_____________ degrees celcius

Tb_____________ degrees celcius

Explanation / Answer

Apply

dTf = -Kf*m*i

for Na2SO4 = 2Na+ and SO4-2 are in solutoin, so total of 2+1 = 3 ions

i = 3

dTf = -1.86*1.48*3 = -8.2584

Tf = 0-8.2584 = -8.2584

For

boiling

dTb = Kb*mï = 0.512*1.48*3 =2.27328 C

Tb = 100+2.27328 = 102.27328° C

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