Respuesta :
Answer:
1. 0.27 m [tex]NH_4I[/tex] : Highest boiling point
2. 0.11 m
: Lowest boiling point
3. 0.16 m
: Third highest boiling point
4. 0.5 m sucrose : Second highest boiling point
Explanation:
[tex]\Delta T_b=i\times k_b\times m[/tex]
[tex]\Delta T_b[/tex] = elevation in boiling point
i = Van'T Hoff factor
[tex]k_b[/tex] = boiling point constant
m = molality
1. For 0.27 m [tex]NH_4I[/tex]
[tex]NH_4I\rightarrow NH_4^{+}+I^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]2\times 0.27=0.54m[/tex]
2. For 0.11 m [tex]FeCl_3[/tex]
[tex]FeCl_3\rightarrow Fe^{3+}+3Cl^{-}[/tex]
i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]4\times 0.11=0.44m[/tex]
3. For 0.16 m [tex]CaCl_2[/tex]
[tex]CaCl_2\rightarrow Ca^{2+}+2Cl^{-}[/tex]
, i= 3 as it is a electrolyte and dissociate to give 3 ions, concentration of ions will be [tex]3\times 0.16=0.48m[/tex]
4. For 0.5 m sucrose
, i= 1 as it is a non electrolyte and does not dissociate to give ions, concentration will be 0.5 m
Thus as concentration of solute follows the order : [tex]NH_4I[/tex] > sucrose > [tex]CaCl_2[/tex] > [tex]FeCl_3[/tex], the boiling point will also follow the same order.
