Which flask has the solution with the lowest freezing point
1 answer:
Answer:
Here's what I get
Explanation:
Assume you have 0. 1 mol·kg⁻¹ solutions of A) KCl, B) CH₃OH, C) Ba(OH)₂, and D) CH₃COOH.
The formula for freezing point depression is
Where
b = the molal concentration
K_f = the freezing point depression constant
i = the van't Hoff i factor
The i-factor is the number of solute particles produced by one formula unit of the substance.
The only difference in the solutions is the i-factor of the solutes.
A) KCl
KCl(aq) ⟶ K⁺(aq) +Cl⁻(aq); i = 2
B) CH₃OH
Methanol is a nonelectrolyte.
CH₃OH(aq) ⟶ CH₃OH(aq); i = 1
C) Ba(OH)₂
Ba(OH)₂(aq) ⟶ Ba²⁺(aq) + 2OH⁻(aq); i = 3
D) CH₃COOH
CH₃COOH(aq) ⇌ CH₃COO⁻(aq)+ H⁺(aq); i ⪆ 1
The Ba(OH)₂ solution has the greatest i-value. It therefore has the greatest freezing point depression and the lowest freezing point.
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Answer:
The ratio of f at the higher temperature to f at the lower temperature is 5.356
Explanation:
Given;
activation energy, Ea = 185 kJ/mol = 185,000 J/mol
final temperature, T₂ = 525 K
initial temperature, T₁ = 505 k
Apply Arrhenius equation;
Where;
is the ratio of f at the higher temperature to f at the lower temperature
R is gas constant = 8.314 J/mole.K
Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356