Answer :
Answer:To find the freezing point of a solution of 35 grams of
CaCl
2
CaCl
2
in 1.5 liters of water, we can follow these steps:
Explanation:Calculate the molality of the solution:
First, determine the molar mass of
CaCl
2
CaCl
2
:
Molar mass of
CaCl
2
=
40.08
(
Ca
)
+
2
×
35.45
(
Cl
)
=
110.98
g/mol
Molar mass of CaCl
2
=40.08(Ca)+2×35.45(Cl)=110.98g/mol
Next, calculate the number of moles of
CaCl
2
CaCl
2
in 35 grams:
Moles of
CaCl
2
=
35
g
110.98
g/mol
≈
0.315
mol
Moles of CaCl
2
=
110.98g/mol
35g
≈0.315mol
Then, calculate the molality of the solution. Molality (
m) is defined as the number of moles of solute per kilogram of solvent. Given that the density of water is approximately
1
kg/L
1kg/L:
Mass of water
=
1.5
L
×
1
kg/L
=
1.5
kg
Mass of water=1.5L×1kg/L=1.5kg
Molality
=
0.315
mol
1.5
kg
≈
0.21
m
Molality=
1.5kg
0.315mol
≈0.21m
Determine the van't Hoff factor (
i) for
CaCl
2
CaCl
2
:
CaCl
2
CaCl
2
dissociates into 3 ions in water:
CaCl
2
→
Ca
2
+
+
2
Cl
−
CaCl
2
→Ca
2+
+2Cl
−
Therefore, the van't Hoff factor
=
3
i=3.
Calculate the freezing point depression (
Δ
ΔT
f
):
The formula for freezing point depression is:
Δ
=
⋅
⋅
ΔT
f
=i⋅K
f
⋅m
Given that
K
f
for water is 1.86 \, ^\circ\text{C}/\text{m}:
\Delta T_f = 3 \times 1.86 \, ^\circ\text{C}/\text{m} \times 0.21 \, \text{m} \approx 1.17 \, ^\circ\text{C}
Determine the new freezing point of the solution:
The normal freezing point of water is 0 \, ^\circ\text{C}.
The freezing point of the solution is:
\text{Freezing point} = 0 \, ^\circ\text{C} - 1.17 \, ^\circ\text{C} = -1.17 \, ^\circ\text{C}
Therefore, the freezing point of the solution of 35 grams of
CaCl
2
CaCl
2
in 1.5 liters of water is approximately -1.17 \, ^\circ\text{C}.
