Question

 

finishing this table.

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 Δt = iKfm for m. Note: Kf = 1.86 oC•kg/mol; Δt = iKfm, therefore m = Δt/iKf;

Table 1: Freezing Data Point
Solution	Freezing Point (OC)	Δt (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

Answer 1

Ans 1) Let us first calculate the molality and vant hoff factor(i) for glycerol

We all know Vant Hoff factor (i) is the ratio of observed number of particles ( after dissociation / association ) to the original number of particles (before dissociation /association) only in case of electrolyte.

i.e. i= obserbed number of particles / original number of particle

Now we know glycerol is a non-electrolyte so there is no chance of association or dissociation of particles and in case of non electrolyte ,vant hoff factor (i) = 1

therefore for glycerol , i= 1

Now , we know

where m is the molality

Therefore

where K_f = molal depression constant

Given

K_f = 1.86 ⁰C Kg/mol

therefore molality

      = 3.1 / 1x1.86

       = 1.666 m

Ans 2) Let us calculate the molality and vant hoff factor for NaCl

We Know, NaCl is a electrolyte which dissociates

i.e NaCl -> Na⁺ + Cl^-

Here we can see after dissociation,

observed number of particle = 1+1 = 2

and before dissociation

original number of particle = 1

Therefore , i=2/1

                 = 2

Now, we know

Given ,

Therefore

      = 7.3 / 2 x 1.86

      = 1.962 m

Ans 3) Let us calculate the molality and vant hoff factor for CaCl₂

We know , CaCl₂ is a electrolyte which dissociates as

CaCl₂ -> Ca²⁺ + 2Cl^-

Here we can see after dissociation

observed number of particle = 1+2=3

and before dissociation

original number of particle = 1

therefore i= 3/1

              = 3

we Know

Given ,

      = 4.3 / 3 x 1.86

      = 0.770 m

Now , Together we can conclude our answer as

For glycerol ,

i= 1 , molality m = 1.666

For NaCl

i=2 ,molality ,m=1.962

For CaCl₂

i=3 , molality , m= 0.770 m