Question
Calculate the boiling point and freezing point of the following. 3m of sugar water, 3m of Lithium nitrate, 3m of lead (II) chlorate and 3m of ammonium phsphate.
Answer:
When a solute is dissolved in a solvent, the freezing point of a solution is lowered according to the equation:
^Tf = i•Kf•m
and the boiling point of a solution is elevated according to the equation:
^T = i•Kb•m
where;
^Tb = freezing point depression or boiling point elevation(°C)
i = the van’t Hoff factor
Kf = the molal freezing point constant (1.86°C/m)
Kb = the molal boiling point constant (0.51°C/m)
m = the molality, which is the moles of solute per kilograms of solvent
FREEZING POINTS
3m sugar (sucrose) C12H22O11 (molecular dissolving)
C12H22O11(aq) ——-> C12H22O11-(aq)
3m ………………………………….… 3m (i = 1)
^T = i•Kf•m
^T = (1)•(1.86°C/m)•(3m)
^T = 5.58°C
T(normal) – T(depressed) = 5.58°C
0°C – T(depressed) = 5.58°C
T(depressed) = – 5.58°C
3m LiNO3 (ionic dissolving)
LiNO3(aq) ——–> Li^+(aq) + NO3^-(aq)
3m …………………….. 3m ………… 3m (i = 2)
^T = i•Kf•m
^T = (2)•(1.86°C/m)•(3m)
^T = 11.16°C
T(normal) – T(depressed) = 11.16°C
0°C – T(depressed) = 11.16°C
T(depressed) = – 11.16°C
3m Pb(ClO3)2 (ionic dissolving):
Pb(ClO3)2 ——–> Pb^2+(aq) + 2ClO3^-(aq)
3m …………………….. 3m ……………. 6m (i = 3)
^T = i•Kf•m
^T = (3)•(1.86°C/m)•(3m)
^T = 16.74°C
T(normal) – T(depressed) = 16.74°C
0°C – T(depressed) = 16.74°C
T(depressed) = – 16.74°C
3m (NH4)3PO4 (ionic dissolving)
(NH4)3PO4(aq) ——> 3NH4^+(aq) + PO4^3-(aq)
3m …………………………….. 9m ……………. 3m (i = 4)
^T = i•Kf•m
^T = (4)•(1.86°C/m)•(3m)
^T = 22.32°C
T(normal) – T(depressed) = 22.32°C
0°C – T(depressed) = 22.32°C
T(depressed) = – 22.32°C
BOILING POINTS
3m sugar (sucrose) C12H22O11 (molecular dissolving)
C12H22O11(aq) ——-> C12H22O11-(aq)
3m ………………………………….… 3m (i = 1)
^T = i•Kb•m
^T = (1)•(0.51°C/m)•(3m)
^T = 1.53°C
T(elevated) – T(normal) = 1.53°C
T(elevated) = 100°C + 1.53°C
T(elevated) = 101.53°C
Similarly for
3m LiNO3 , T(elevated) = 103.06°C
3m Pb(ClO3)2, T(elevated) = 104.59°C
3m (NH4)3PO4, T(elevated) = 106.12°C
