Arrange the following solutions in order of decreasing freezing point: 0.10m na3po4; 0.35m nacl; 0.20m mgcl2; 0.15m c6h12o6; 0.15m ch3cooh.

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,

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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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

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.