Question

Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.13 m CoI2 A. Lowest freezing point 2. 0.10 m Cr(CH3COO)3 B. Second lowest freezing point 3. 0.16 m Mn(CH3COO)2 C. Third lowest freezing point 4. 0.44 m Glucose(nonelectrolyte) D. Highest freezing point

Answer 1

1. CoI₂  --> Co²⁺ + 2 I^-

for strong electrolyte van't hoff factor

i = number of ions

i = 1 + 2 or i = 3

Given molality = 0.13 m

Delta (T_f) = i *K_f *m

Put the value

Delta (T_f) = 3*K_f*0.13

Delta (T_f) = 0.39* K_f   ------(1)

2.

Cr(CH₃COO)₃ ---> Cr³⁺  + 3 CH₃COO^-

i = 1+3 or i = 4

molality = 0.10 m

Delta (T_f) = i *K_f *m

Delta (T_f) = 4*K_f *0.10

Delta (T_f) = 0.40*K_f ------(2)

3. Mn(CH₃COO)₂ ----> Mn²⁺ + 2 CH₃COO^-

i = 1+2 or i = 3

molality = 0.16 m

Delta (T_f) = i *K_f *m

Delta (T_f) = 3*K_f *0.16

Delta (T_f) = 0.48*K_f ----(3)

4.

Glucose will not dissolve into ions hence

i = 1

molality = 0.44 m

Delta (T_f) = i *K_f *m

Delta (T_f) = 1*K_f *0.44

Delta (T_f) = 0.44*K_f ---(4)

**** if delta( T_f ) will be high then freezing point will also be maximum( in negative).

From the above four equation

1. CoI₂ A) lowest freezing point

2. Cr(CH₃COO)₃    B) second lowest freezing point

3. Mn(CH₃COO)₂   D) Highest freezing point

4. Glucose. C) third lowest freezing point