Respuesta :
Answer:
0.13 m of [tex]Mn(NO_3)_2[/tex] → Highest boiling point
0.19 m of [tex]AgNO_3[/tex] → Second Highest boiling point
0.17 m of [tex]CrSO_4[/tex] → Third highest boiling point
0.31 m Sucrose (nonelectrolyte) → Lowest boiling point
Explanation:
Elevation in boiling is given by :
[tex]\Delta T_b=i\times k_b\times m[/tex]
Where :
i = van't Hoff factor
[tex]k_b[/tex]= Molal Elevation constant of solvent
m = molaity of the solution
1) 0.19 m of [tex]AgNO_3[/tex]
[tex]AgNO_3\rightarrow Ag^++NO_3^{-}[/tex]
i = 2 (electrolyte)
Molality of the solution = 0.19
Elevation is boiling point of solution:
[tex]\Delta T_b=2\times k_b\times 0.19 m[/tex]
[tex]\Delta T_b=0.38 m\times k_b[/tex]
2) 0.17 m of [tex]CrSO_4[/tex]
[tex]CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}[/tex]
i = 2 (electrolyte)
Molality of the solution = 0.17
Elevation is boiling point solution :
[tex]\Delta T_b=2\times k_b\times 0.17 m[/tex]
[tex]\Delta T_b=0.34 m\times k_b[/tex]
3) 0.13 m of [tex]Mn(NO_3)_2[/tex]
[tex]Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}[/tex]
i = 3 (electrolyte)
Molality of the solution = 0.13
Elevation is boiling point solution :
[tex]\Delta T_b=3\times k_b\times 0.13 m[/tex]
[tex]\Delta T_b=0.39 m\times k_b[/tex]
4) 0.31 m Sucrose (nonelectrolyte)
i = 1 ( non electrolyte)
Molality of the solution = 0.31 m
Elevation is boiling point solution :
[tex]\Delta T_b=1\times k_b\times 0.31 m[/tex]
[tex]\Delta T_b=0.31 m\times k_b[/tex]
Higher the value of elevation in temperature higher will be the boiling point of the solution .
The decreasing order of solution from highest boiling point to lowest boiling point is :
[tex]0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b[/tex]
0.13 m of [tex]Mn(NO_3)_2[/tex] → Highest boiling point
0.19 m of [tex]AgNO_3[/tex] → Second Highest boiling point
0.17 m of [tex]CrSO_4[/tex] → Third highest boiling point
0.31 m Sucrose (nonelectrolyte) → Lowest boiling point
