Respuesta :
Answer:
1)- 0.11 m Fe(NO₃)₃ ⇒ A- Lowest freezing point
2)- 0.18 m NaOH ⇒ D- Highest freezing point
3)- 0.21 m FeSO₄ ⇒ B- Second lowest freezing point
4)- 0.38 m Glucose ⇒ C- Third lowest freezing point
Explanation:
Freezing point depression is given by the following equation:
ΔTf= Kf x m x i
As it is a depression point (final temperature is lower than initial temperature), as higher is ΔTf, lower is the freezing point. Kf is the cryoscopic constant. For water, Kf= 1.853 K·kg/mol. At high m (molality of solute) and i (Van't Hoff factor, dissociated species), the freezing point will be low.
Kf is the same for all solution, so we can simply calculate m x i and order the solutions from high m x i (lowest freezing point) to low m x i (highest freezing point):
- Fe(NO₃)₃⇒ Fe³⁺ + 3 NO₃⁻ -------> i= 1 + 3= 4
m x i = 0.11 x 4 = 0.44
A) Lowest freezing point
- NaOH ⇒ Na⁺ + OH⁻ -------------> i= 1+1= 2
m x i = 0.18 x 2= 0.36
D) Highest freezing point
- FeSO₄ ⇒ Fe²⁺ + SO₄⁻ -------------> i= 1+1= 2
m x i = 0.21 x 1= 0.42
B) Second lowest freezing point
- Glucose -------------> non electrolyte : i=1
m x i = 0.38 x 1 = 0.38
C) Third lowest freezing point
A fall in the hotness and coldness at which the matter freezes is called freezing point depression.
It can be calculated using:
[tex]\Delta \text{T}_{\text{f}} &= \text{K}_{\text{f}} \times \text{m} \times \text{ i}[/tex]
- The initial temperature is higher than the final temperature as it is a depression point.
- Higher the [tex]\Delta \text{T}_{\text{f}}[/tex] lower will be the freezing point.
- [tex]\text{k}_{\text{f}}[/tex] is the constant for cryoscopic.
- When m (molality) and the i (Van't Hoff factor) are high the freezing point would be low.
For all the solution [tex]\text{k}_{\text{f}}[/tex] value is constant hence, m × i can be calculated to know the order of lowest and highest freezing points.
The correct matches are:
1) 0.11 m Fe(NO₃)₃ ⇒ Option A. Lowest freezing point
- Fe(NO₃)₃⇒ Fe³⁺ + 3 NO₃⁻ ⇒ i = 1 + 3 = 4
- m x i = 0.11 x 4 = 0.44
- Option A. Lowest freezing point
2) 0.18 m NaOH ⇒ Option D. Highest freezing point
- NaOH ⇒ Na⁺ + OH⁻ ⇒ i = 1+1 = 2
- m x i = 0.18 x 2= 0.36
- Option D. Highest freezing point
3) 0.21 m FeSO₄ ⇒ Option B. Second lowest freezing point
- FeSO₄ ⇒ Fe²⁺ + SO₄⁻ ⇒ i = 1+1 = 2
- m x i = 0.21 x 1= 0.42
- Option B. Second lowest freezing point
4) 0.38 m Glucose ⇒ Option C. Third lowest freezing point
- Glucose ⇒ non electrolyte : i = 1
- m x i = 0.38 x 1 = 0.38
- Option C. Third lowest freezing point
To learn more about freezing point depression refer to the link:
https://brainly.com/question/25165904
