Respuesta :
Answer: 0.100 m [tex]Li_2SO_{4}[/tex]
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_f[/tex] = boiling point constant
m = molality
1. For 0.100 m [tex]Li_2SO_{4}[/tex]
[tex]Li_2SO_4\rightarrow 2Li^{+}+SO_4^{2-}[/tex]
i= 3 as it is a electrolyte and dissociate to give 3 ions.
Thus concentration of ions =[tex]3\times 0.100=0.300[/tex]
2. For 0.100 m [tex]KNO_{2}[/tex]
[tex]KNO_2\rightarrow K^{+}+NO_2^{-}[/tex]
i= 2 as it is a electrolyte and dissociate to give 2 ions.
Thus concentration of ions =[tex]2\times 0.100=0.200[/tex]
3. For 0.200 m [tex]C_3H_8O_3[/tex]
i= 1 as it is a non electrolyte and do not dissociate to give ions.
4. For 0.060 m [tex]Li_3PO_4[/tex]
[tex]Li_3PO_4\rightarrow 3Li^{+}+PO_4^{3-}[/tex]
i= 4 as it is a electrolyte and dissociate to give 4 ions.
Thus concentration of ions =[tex]4\times 0.060=0.24[/tex]
Thus as concentration of ions is highest for [tex]Li_2SO_{4}[/tex] and the boiling point will be highest.
0.060 m Li₃PO₄ has the highest boiling point
Further explanation
Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.
Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles because electrolytes break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes.
The term is used in the Solution properties
- 1. molal
that is, the number of moles of solute in 1 kg of solvent
[tex] \large {\boxed {\bold {m = mole. \frac {1000} {mass \: of \: solvent (in \: grams)}}} [/tex]
- 2. Boiling point and freezing point
Solutions from volatile substances have a higher boiling point and lower freezing points than the solvent
ΔTb = Tb solution - Tb solvent
ΔTb = boiling point elevation
[tex] \rm \Delta T_f = T_fsolvent-T_fsolution [/tex]
[tex] \large {\boxed {\boxed {\bold {\Delta Tb \: = \: Kb.m}}} [/tex]
[tex] \rm \Delta T_f = K_f \times m [/tex]
Kb = molal boiling point increase
Kf = molal freezing point constant
m = molal solution
For electrolyte solutions there is a van't Hoff factor = i
i = 1 + (n-1) α
n = number of ions from the electrolyte
α = degree of ionization, strong electrolyte α = 1, for non electrolytes i = 1
so the boiling point formula becomes:
[tex] \ rm \ Delta T_f = K_b \ times m \ times i [/ tex]
All solutions in the problem have the same solvent -> assuming water (The same [tex] \ rm K_b [/ tex]) so that what affects the value of [tex] \ rm \ Delta T_b [/ tex] is the value of i and m
Assuming the degree of electrolyte ionization α = 1, the magnitude i is determined by the number of ions produced by the electrolyte (n)
a. 0.100 m Li₂SO₄
Li₂SO₄ ---> 2Li ++ SO₄²⁻ → 3 ions
m x i = 0.1 x 3 = 0.3
b. 0.100 m KNO₂
KNO₂ ---> K⁺ + NO₂⁻ → 2 ions
m x i = 0.1 x 2 = 0.2
c. 0.200 m C₃H₈O₃
Non-electrolyte solution, i = 1
ΔTb = Kb .m (based only on concentration m)
m x i = 0.2 x 1 = 0.2 or m only = 0.2
d. 0.060 m Li₃PO₄
Li₃PO₄ ---> 3Li⁺ + PO₄ ³⁻ → 4 ions
m x i = 0.06 x 4 = 0.24
Li₃PO₄ has the highest number of m x i, so it has the highest [tex] \ rm \ Delta T_b [/ tex] and the highest boiling point.
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