Question

Arrange the following aqueous solutions in order of decreasing freezing point: 0.10 m Na3PO4, 0.35 m NaCl, 0.20 m MgCl2, 0.15 m C6H12O6, 0.15 m CH3COOH. (Introduce only the formulas, not the concentrations, of each substance in the appropriate box, beginning with the highest freezing point. Click in the answer boxes to activate the palette.) > > > > highest freezing point lowest freezing point

Answer 1

Answer: 0.35 m
`NaCl` > 0.20 m
`MgCl_2` >0.10 m
`Na3PO_4` > 0.15 m
`CH_3COOH` > 0.15 m
`C_6H_(12)O_6`

Step-by-step explanation:

Depression in freezing point:

`T_f^0-T_f=i* k_b* m`

where,

`T_f`= freezing point of solution

`T^o_f` = freezing point of solvent

`k_f` = freezing point constant

m = molality

1. For 0.10 m
`Na3PO_4`

`Na_3PO_4\rightarrow 3Na^++PO_4^(3-)`

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be
`4* 0.10=0.40`

2. For 0.35 m
`NaCl`

`NaCl\rightarrow Na^++Cl^-`

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be
`2* 0.35=0.70`

3. For 0.20 m
`MgCl_2`

`MgCl_2\rightarrow Mg^(2+)+2Cl^-`

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be
`3* 0.20=0.60`

4. For 0.15 m
`C_6H_(12)O_6`

, i= 1 as it is a non electrolyte and does not dissociate to give ions.

5. For 0.15 m
`CH_3COOH`

`CH_3COOH\rightarrow CH_3COO^-+H^+`

, i= 2 as it is a electrolyte and dissociate to give 2 ions and thus have
`2* 0.15=0.30`

As concentration is highest for 0.35 m
`NaCl` , freezing point depression will be highest and thus has lowest freezing point. As concentration is lowest for 0.15 m
`C_6H_(12)O_6` , freezing point depression will be lowest and thus has highest freezing point