Respuesta :
Answer : The correct option is, (C) 0.60 M [tex]Li_2CO_3[/tex]
Explanation :
According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.
Lowest vapor pressure ∝ i × m
where,
i = Van't Hoff factor
m = molarity
As we know that the vapor pressure depends on the molarity and the Van't Hoff factor.
So, the higher the product of Van't Hoff factor and molarity the lower the resultant vapor pressure.
(A) The dissociation of 0.120 M [tex]C_2H_6O[/tex] is not possible because it is a non-electrolyte solute. So, the Van't Hoff factor will be, 1.
Lowest vapor pressure = 1 × 0.120 = 0.120 m
(B) The dissociation of 0.40 M [tex](NH_4)_2SO_4[/tex] will be,
[tex](NH_4)_2SO_4\rightarrow 2NH_4^++SO_4^{2-}[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]2NH_4^++SO_4^{2-}[/tex] = 2 + 1 = 3
Lowest vapor pressure = 3 × 0.40 = 1.2 m
(C) The dissociation of 0.60 M [tex]Li_2CO_3[/tex] will be,
[tex]Li_2CO_3\rightarrow 2Li^{+}+CO_3^{2-}[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]2Li^{+}+CO_3^{2-}[/tex] = 2 + 1 = 3
Lowest vapor pressure = 3 × 0.60 = 1.8 m
(c) The dissociation of 0.30 M [tex]RbC_2H_3O_2[/tex] will be,
[tex]RbC_2H_3O_2\rightarrow Rb^{+}+C_2H_3O_2^{-}[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]Rb^{+}+C_2H_3O_2^{-}[/tex] = 1 + 1 = 2
Lowest vapor pressure = 2 × 0.30 = 0.6 m
From this we conclude that, 0.60 M [tex]Li_2CO_3[/tex] has the highest product of Van't Hoff factor and molarity which means that the solution will exhibit the lowest vapor pressure.
Hence, the correct option is, (C) 0.60 M [tex]Li_2CO_3[/tex]
The aqueous solution with the lowest vapor pressure : 0.060 m Li₂CO₃
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. mole fraction
the ratio of the number of moles of solute to the mole of solution
[tex] \large {\boxed {\bold {Xa = \frac {na} {na + nb}}} [/tex]
- 3. Vapor pressure
Vapor pressure depends on the mole fraction of the components in the solution
P = Xs. P °
P = vapor pressure solution
P ° = pure vapor pressure
Xs = mole fraction solvent
ΔP = P ° - P where
ΔP = change in vapor pressure
It is assumed that the solvent we use is water with an amount of 1 L or 1 kg, then the concentration of the solutions above is the mole of each solution
From the equation for determining the vapor pressure solution, shows a Proportional Comparisons / Directly proportional with the solvent mole fraction (in this case water)
mole solvent fraction:
[tex]\rm Xs=\dfrac{mol~solvent}{mol~solvent+mol~solute}[/tex]
Because the mole fraction used in the search for vapor pressure is the solvent mole fraction, the greater the mole solute, the smaller the solvent mole fraction, so the smaller the vapor pressure of solution
For electrolyte solutions, i = van't Hoff factor is calculated
formulated: i = 1 + (n-1) α
n = number of ions produced by the solution
a = degree of electrolyte ionization
So for electrolyte solutions, the solvent mole fraction becomes:
[tex]\rm Xs=\dfrac{mol~solvent}{mol~solvent+mol~solute\times i}[/tex]
So that the greater multiplication between mole solute x i , the larger the vapor pressure decreases
A. 0.120 m C₂H₆0₂
C₂H₆0₂: non electrolyte solution, i = 1
mol solute x i = 0.12 .1 = 0.12
For choices B, C and D, which are electrolyte solutions, because α is unknown, the value of i depends on the number of ions produced (n)
B. 0.040 m (NH₄)₂SO₄
(NH₄)₂SO₄ ---> 2NH₄ + SO₄²⁻, n = 3
mole solute x 3 = 0.040 x 3 = 0.12
C. 060 m Li₂CO₃
Li₂CO₃ ---> 2Li⁺ + CO₃²⁻, n = 3
mole solute x 3 = 0.06 x 3 = 0.18
D. 030 m RbC₂H₃O₂
RbC₂H₃O₂ ---> Rb⁺ + C₂H₃O²⁻, n = 2
mole solute x 2 = 0.03 x 2 = 0.06
The largest mole solute x i value is in Li₂CO₃ solution
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