Question

Colligative properties (boiling point increase)
a. Order the following solutions in ascending order of their boiling point: 0.0825 m NaCl, 0.050 m KCl, 0.050 M Na2SO4, 0.055m Ca (NO3) 2, 0.050 m Na3PO4.

If the boiling point of a 0.100 m NaCl aqueous solution is 100.5 ◦C, answer…
b. How much will the boiling point of a 0.0825 m KCl solution be?
c. How much will be the boiling point of a solution 0.0825 m Ca (NO3) 2?
d. Determine the osmotic pressure of the solutions in part a.

Answer 1

(a). Greater is the value of ΔTb (Elevation in boiling point) greater will be the boiling point of the solution.

ΔTb = i.kb,m ............(1)

# For 0.0825 m NaCl, ΔTb =2 x 0.0825 kb = 0.165 kb

# For 0.050 m KCl, ΔTb =2 x 0.050 kb = 0.100 kb

# For 0.050 M Na2SO4, ΔTb =3 x 0.050 kb = 0.150 kb

# For 0.055m Ca (NO3)2, ΔTb =3 x 0.055 kb = 0.165 kb

# For 0.050 m Na3PO4, ΔTb =4 x 0.050 kb = 0.200 kb

Therefore, ascending order of their boiling point is :

0.050 m KCl < 0.050 M Na2SO4 < 0.0825 m NaCl = 0.055m Ca (NO3)2< 0.050 m Na3PO4  

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(b). Tb - Tb0 = i.kb.m

Tb - 100 ºC = (2).(0.512 ºC/m).(0.0825 m)

Tb - 100 ºC = 0.08448 ºC

Tb = 0.08448 ºC + 100 ºC

Tb = 100.08448 ºC

Therefore, boiling point of KCl solution = 100.08448 ºC

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(c). Tb - Tb0 = i.kb.m

Tb - 100 ºC = (3).(0.512 ºC/m).(0.0825 m)

Tb - 100 ºC = 0.12672 ºC

Tb = 0.12672 ºC + 100 ºC

Tb = 100.12672 ºC

Therefore, boiling point of KCl solution = 100.12672 ºC

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(d).To determine the osmotic pressure, Temperature should be given

Osmotic pressure (π) = iCRT

Osmotic pressure of NaCl = (2).(0.0825),(0.0821 L atm K-1mol-1)(298K) = 4.037 atm

Osmotic pressure of KCl = (2).(0.050),(0.0821 L atm K-1mol-1)(298K) = 2.45 atm

Osmotic pressure of Na2SO4 = (3).(0.050),(0.0821 L atm K-1mol-1)(298K) = 3.67 atm

Osmotic pressure of Ca(NO3)2 = (3).(0.055),(0.0821 L atm K-1mol-1)(298K) = 4.037 atm

Osmotic pressure of Na3PO4 = (4).(0.050),(0.0821 L atm K-1mol-1)(298K) = 4.89 atm