Respuesta :
Answer:
One mole of the jonic compound [tex]Na_3PO_4[/tex] dissolved in 1000 g of [tex]H_2O[/tex] → lowest freezing point
One mole of the jonic compound [tex]CuSO_4[/tex] dissolved in 1000 g of [tex]H_2O[/tex] → intermediate freezing point
One mole of the jonic compound [tex]C_6H{12}O_6[/tex] dissolved in 1000 g of [tex]H_2O[/tex] → highest freezing point
Explanation:
[tex]\Delta T_f=iK_f\times m[/tex]
[tex]\Delta T_f=T-T_f[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
[tex]K_f[/tex] = freezing point constant
m = molality
i = van't Hoff factor
[tex]T_f[/tex]= freezing point of solution
T = freezing point of pure solvent
Since the moles of solute and mass of solvent is same in all cases.So [tex]T_f[/tex] is directly related to [tex]\Delta T_f[/tex] and
Higher the value of 'i' more will be depression in freezing and lowest will be freezing point of the solution.
1) One mole of the jonic compound [tex]Na_3PO_4[/tex] dissolved in 1000 g of [tex]H_2O[/tex]
[tex]Na_3PO_4\rightarrow 3Na^++PO_4^{3-}[/tex]
i = 4
This solution will have the lowest freezing point.
2) One mole of the jonic compound [tex]CuSO_4[/tex] dissolved in 1000 g of [tex]H_2O[/tex]
[tex]CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}[/tex]
i = 2
This solution will have the intermediate freezing point.
3)One mole of the jonic compound [tex]C_6H{12}O_6[/tex] dissolved in 1000 g of [tex]H_2O[/tex]
Glucose is a non electrolyte substance.
i = 1
This solution will have the highest freezing point.
