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}.