Respuesta :
Explanation:
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] =Elevation in boiling point
[tex]K_b[/tex] = boiling point constant of solvent
i = van't Hoff factor
m = molality
As we can see that molality is directly proportional ti elevation in boiling point, so higher the molality of the solution at more high temperature it will boil.
1) 0.16 m [tex]Pb(CH_3COO)_2[/tex]
i = 3
[tex]\Delta T_b=3K_b\times 0.16 m=K_b\times 0.48 m[/tex]
Third highest boiling point
2) 0.17 m [tex]NiBr_2[/tex]
i = 3
[tex]\Delta T_b=3K_b\times 0.17 m=K_b\times 0.51 m[/tex]
Second highest boiling point
3) [tex]8.8\times 10^{-2} m[/tex] of [tex]Al_2(SO_4)_3[/tex]
i = 5
[tex]\Delta T_b=5\times K_b\times 8.8\times 10^{-2} m=K_b\times 0.44 m[/tex]
Lowest boiling point
4) 0.53 m Urea
i = 1
[tex]\Delta T_b=1\times K_b\times 0.53 m =K_b\times 0.53 m[/tex]
Highest boiling point
