Respuesta :

Answer:

0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 >  0.20 m MgCl2 > 0.35 m NaCl

Explanation:

Step 1:

0.10m Na3PO4

Na3PO4 → 3Na+ + PO4^-3-

⇒ van't Hoff factor is 3 + 1 = 4

0.10 m * 4 = 0.40

0.35m NaCl

NaCl → Na+ + Cl-

⇒ van't Hoff factor = 1+1 = 2

0.35 * 2 = 0.70

0.20m MgCl2

MgCl2 → Mg^2+ + 2Cl-

 ⇒ Van't Hoff factor = 1+2 = 3

0.20 * 3 = 0.60

0.15m C6H12O6

⇒  for non-ionic compounds in solution, like glucose (C6H12O6) , the van't Hoff factor is 1. They do not dissociate in water.

0.15 * 1 = 0.15

0.15m CH3COOH.

CH3CO2H ⇄ CH3CO2− + H+

 ⇒ Van't hoff factor ≈ 1<x<2

0.15 * 2 = 0.30

Larger the concentration of the concentration of the particles, smaller the freezing point.

0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 >  0.20 m MgCl2 > 0.35 m NaCl

Histrionicus

The order of decreasing freezing point is -

0.15m C6H12O6 < 0.15m CH3COOH < 0.10 m Na3PO4 < 0.20 m MgCl2 < 0.35 m NaCl,

The Freezing Point Depression

It is calculated as:  

dT = i Kf m  

m = The concentration values in molalities  

Kf = the cryoscopic constant for water (-1.86, the same for all solutions)

i = the Van’t Hoff factor, which is the number of ion particles per individual molecule/formula of solute

Solution:

So you have to arrange in increasing order of the product i·m  

0.15 m C6H12O6        

im = 1 x 0.15 = 0.15     (undissociated, i=1)

0.15 m CH3COOH  

im =1.1 x 0.15 = 0.17 (partially dissociated) ( 1<i<2)

0.10 m Na3PO4

im = 4 x 0.1 = 0.4       (i=4)

0.20 m MgCl2

im = 3 x 0.2 = 0.6    (i=3)

0.35 m NaCl,

im = 2 x 0.35 = 0.7 (i=2)

Thus, the order of decreasing freezing point is -

0.15m C6H12O6 < 0.15m CH3COOH < 0.10 m Na3PO4 < 0.20 m MgCl2 < 0.35 m NaCl,

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