I need help finishing this table. I just need the i and molality for glycerol, N

Question

I need help finishing this table. I just need the i and molality for glycerol, NaCl, and CaCl2. (I don't need any of the other values for water) I guess I am most confused about the i value. In the hint below, it states the "values", but are these the actual values that I am supposed to use for i- or am I missing something? Thanks for your help.

Hint: The column “i” applies to electrolyte (ionic) solutions and represents the ratio of moles of particles in solution to moles of formula units dissolved. In this case, glycerol has a value of 1 (it does not dissociate into separate units), NaCl has a value of 2 (it separates into 2 units—1 Na+ and 1 Cl-), and CaCl2 has a value of 3 (it separates into 3 units—1 Ca2+ and 2 Cl-). The molality column is experimentally determined by solving the equation t = iKfm for m. Note: Kf = 1.86 oC•kg/mol; t = iKfm, therefore m = t/iKf;

Table 1: Freezing Data Point

Solution

Freezing Point (OC)

t (Ti – Tf)

i

Molality (m)

Solute Mass (g)

Solvent Mass (kg)

H2O

-1.3OC

Glycerol

- 4.4OC

3.1

6.3

0.025

NaCl

-8.6OC

7.3

2.0

0.025

CaCl2

-5.6OC

4.3

2.0

0.025

Table 1: Freezing Data Point

Solution

Freezing Point (OC)

t (Ti – Tf)

i

Molality (m)

Solute Mass (g)

Solvent Mass (kg)

H2O

-1.3OC

Glycerol

- 4.4OC

3.1

6.3

0.025

NaCl

-8.6OC

7.3

2.0

0.025

CaCl2

-5.6OC

4.3

2.0

0.025

Explanation / Answer

Answer;-

Table 1: Freezing Data Point

Solution

Freezing Point (OC)

t (Ti – Tf)

i

Molality (m)

Solute Mass (g)

Solvent Mass (kg)

H2O

-1.3OC

Glycerol

- 4.4OC

3.1

0.61

2.74

6.3

0.025

NaCl

-8.6OC

7.3

2.86

1.37

2.0

0.025

CaCl2

-5.6OC

4.3

3.31

0.72

2.0

0.025

Table 1: Freezing Data Point

Solution

Freezing Point (OC)

t (Ti – Tf)

i

Molality (m)

Solute Mass (g)

Solvent Mass (kg)

H2O

-1.3OC

Glycerol

- 4.4OC

3.1

0.61

2.74

6.3

0.025

NaCl

-8.6OC

7.3

2.86

1.37

2.0

0.025

CaCl2

-5.6OC

4.3

3.31

0.72

2.0

0.025

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