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Which one of the following aqueous solution will have the lowest freezing point?
- a)0.1 m urea
- b)0.1 m sucrose
- c)0.1 m CH₃COOH
- d)0.1 m CaCl₂
Correct answer is option 'D'. Can you explain this answer?
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Which one of the following aqueous solution will have the lowest freez...
Solution:
The freezing point of a solution is lowered due to the presence of solute particles. The more the solute particles, the greater the lowering of the freezing point. This phenomenon is known as depression in freezing point.
Calculation of Van’t Hoff factor (i):
i = (number of particles in solution after dissociation) / (number of particles in solution before dissociation)
a) Urea (CH4N2O)
Urea does not dissociate in water. So, i = 1
b) Sucrose (C12H22O11)
Sucrose does not dissociate in water. So, i = 1
c) Acetic acid (CH3COOH)
Acetic acid partially dissociates in water to form acetate ions (CH3COO-) and hydrogen ions (H+). So, i = 2
CH3COOH ⇌ CH3COO- + H+
d) Calcium chloride (CaCl2)
Calcium chloride dissociates completely in water to form calcium ions (Ca2+) and chloride ions (2Cl-). So, i = 3
CaCl2 ⇌ Ca2+ + 2Cl-
Calculation of depression in freezing point (ΔTf):
ΔTf = Kf x i x molality
where Kf is the freezing point depression constant and molality is the number of moles of solute per kilogram of solvent.
Kf for water = 1.86 °C/m
a) Urea (CH4N2O)
ΔTf = Kf x i x molality = 1.86 x 1 x 0.1 = 0.186 °C
b) Sucrose (C12H22O11)
ΔTf = Kf x i x molality = 1.86 x 1 x 0.1 = 0.186 °C
c) Acetic acid (CH3COOH)
ΔTf = Kf x i x molality = 1.86 x 2 x 0.1 = 0.372 °C
d) Calcium chloride (CaCl2)
ΔTf = Kf x i x molality = 1.86 x 3 x 0.1 = 0.558 °C
Conclusion:
The aqueous solution of calcium chloride (CaCl2) will have the lowest freezing point because it has the highest value of depression in freezing point (0.558 °C).
The freezing point of a solution is lowered due to the presence of solute particles. The more the solute particles, the greater the lowering of the freezing point. This phenomenon is known as depression in freezing point.
Calculation of Van’t Hoff factor (i):
i = (number of particles in solution after dissociation) / (number of particles in solution before dissociation)
a) Urea (CH4N2O)
Urea does not dissociate in water. So, i = 1
b) Sucrose (C12H22O11)
Sucrose does not dissociate in water. So, i = 1
c) Acetic acid (CH3COOH)
Acetic acid partially dissociates in water to form acetate ions (CH3COO-) and hydrogen ions (H+). So, i = 2
CH3COOH ⇌ CH3COO- + H+
d) Calcium chloride (CaCl2)
Calcium chloride dissociates completely in water to form calcium ions (Ca2+) and chloride ions (2Cl-). So, i = 3
CaCl2 ⇌ Ca2+ + 2Cl-
Calculation of depression in freezing point (ΔTf):
ΔTf = Kf x i x molality
where Kf is the freezing point depression constant and molality is the number of moles of solute per kilogram of solvent.
Kf for water = 1.86 °C/m
a) Urea (CH4N2O)
ΔTf = Kf x i x molality = 1.86 x 1 x 0.1 = 0.186 °C
b) Sucrose (C12H22O11)
ΔTf = Kf x i x molality = 1.86 x 1 x 0.1 = 0.186 °C
c) Acetic acid (CH3COOH)
ΔTf = Kf x i x molality = 1.86 x 2 x 0.1 = 0.372 °C
d) Calcium chloride (CaCl2)
ΔTf = Kf x i x molality = 1.86 x 3 x 0.1 = 0.558 °C
Conclusion:
The aqueous solution of calcium chloride (CaCl2) will have the lowest freezing point because it has the highest value of depression in freezing point (0.558 °C).
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Which one of the following aqueous solution will have the lowest freezing point?a)0.1 m ureab)0.1 m sucrosec)0.1 m CH₃COOHd)0.1 m CaCl₂Correct answer is option 'D'. Can you explain this answer?
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