Question

atch the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.14 m MnBr2 A. Lowest freezing point 2. 0.12 m CuCl2 B. Second lowest freezing point 3. 0.10 m AlCl3 C. Third lowest freezing point 4. 0.34 m Urea(nonelectrolyte) D. Highest freezing poin

Answer 1

The Lowest freezing point= 0.34 m urea

the highest freezing point= 0.14m MnBr2

second lowest freezing point= 0.10m AlCl3

third lowest freezing point= 0.12m CuCl2

Step-by-step explanation:

The freezing point depression can be calculated from the formula:

T f = i*K *fm ( ΔTf is the freezing point depression, i is van't Hoff factor, Kf is the freezing point depression constant for the solvent

Van't Hoff factor is no of ions involved in reaction, for non electrolytes it is one

T of water = 1.86 degrees kg/ mole

so, 0.14 MnBr2

T= 1.86*3*0.14

= 0.7812

T of 0.12 CuCl2

T= 1.86*3* 0.12

= 0.6696

T of 0.10 Alcl3

T = 1.86*4*0.10

T= 0.744

• For 0.34 mole urea

T = 1.86*1*0.34

T = 0.6324