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 $C_6H_{12}O_6$

Explanation:

Depression in freezing point:

$T_f^0-T_f=i\times k_b\times m$

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 $4\times 0.10=0.40$

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 $2\times 0.35=0.70$ 

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 $3\times 0.20=0.60$ 

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 $2\times 0.15=0.30$ 

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