Respuesta :
1)
b(solution) = 1,80 m = 1,80 mol/kg..
If we use 1000 g of water to make solution:
m(H₂O) =
1000 g ÷ 1000 g/kg = 1 kg.
n(sucrose - C₁₂O₂₂O₁₁) = b(solution) · m(H₂O).
n(C₁₂O₂₂O₁₁) =
1,8 mol/kg · 1 kg.
n(C₁₂O₂₂O₁₁) =
1,8 mol.
n(H₂O)
= 1000 g ÷ 18 g/mol.
n(H₂O)
= 55,55 mol.
Mole fraction of solvent = 55,55 mol ÷
(55,55 mol + 1,8 mol) = 0,968.
Raoult's Law: p(solution) = mole fraction of solvent · p(solvent).
p(solution) = 0,968 · 17,54 torr = 16,99 torr.
Δp = 17,54 torr - 16,87 torr = 0,55 torr.
2) b(solution) = 1,80 m = 1,80 mol/kg..
If we use 1000 g of water to make solution:
m(H₂O) =
1000 g ÷ 1000 g/kg = 1 kg.
n(NaCl) = b(solution) · m(H₂O).
n(NaCl) = 1,8 mol/kg · 1 kg.
n(NaCl) = 1,8 mol.
n(H₂O) =
1000 g ÷ 18 g/mol.
n(H₂O) =
55,55 mol.
i(NaCl) = 2; Van 't Hoff factor. Because dissociate on one
cation and one anions.
Mole fraction of solvent = 55,55 mol ÷ (55,55 mol + 1,8 mol · 2) = 0,94.
Raoult's Law: p(solution) = mole fraction of
solvent · p(solvent)
p(solution) = 0,94 · 17,54 torr = 16,47 torr.
Δp = 17,54 torr - 16,47 torr = 1,06 torr.
The lowering of vapor pressure with sucrose solution addition has been 0.56 torr. The lowering of vapor pressure with NaCl addition has been 1.07 torr.
Suppose the mass of water in the given solution has been 1000 grams.
Mole = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of water = [tex]\rm \dfrac{1000}{18}[/tex]
Moles of water (solvent) = 55.55 mol
Given molality of the solution = 1.80 m
Molality = moles of solute per kg of solvent.
Since the amount of solvent (water) is 1 kg (1000 grams), the moles of solute = molality of the solution.
Moles of solute = 1.80 moles.
Mole fraction of solvent = [tex]\rm \dfrac{moles\;of\;solvent}{total\;moles\;of\;solution}[/tex]
Mole fraction of solvent = [tex]\rm \dfrac{55.55}{55.55\;+\;1.80}[/tex]
Mole fraction of solvent = 0.968
According to Raoult's law:
Vapor pressure of solution = Mole fraction of solvent [tex]\times[/tex] Vapor pressure of the solvent.
Given, the vapor pressure of solvent (water) = 17.54 torr.
Vapor pressure of solution = 0.968 [tex]\times[/tex] 17.54 torr
Vapor pressure of solution = 16.978 torr.
Lowering of vapor pressure = 17.54 - 16.978 torr
Lowering of vapor pressure = 0.56 torr.
The lowering of vapor pressure with sucrose solution addition has been 0.56 torr.
The NaCl has been dissociated in the form of 1 cation and 1 anion. The van't Hoff factor (i) for NaCl has been 2.
The mole fraction of solvent has been:
Mole fraction = [tex]\rm \dfrac{moles\;of\;solvent}{total\;moles\;of\;solution\;\times\;van't\;Hoff\;factor}[/tex]
Mole fraction of solvent (water) = [tex]\rm \dfrac{55.55}{55.55\;+\;1.80\;\times\;2}[/tex]
Mole fraction of solvent (water) = 0.94
According to Raoult's law:
Vapor pressure of solution = Mole fraction of solvent [tex]\times[/tex] Vapor pressure of the solvent.
Vapor pressure of solution = 0.94 [tex]\times[/tex] 17.54 torr
Vapor pressure of solution = 16.47 torr
Lowering of vapor pressure = 17.54 - 16.47 torr
Lowering of vapor pressure = 1.07 torr.
The lowering of vapor pressure with NaCl addition has been 1.07 torr.
For more information about the lowering of vapor pressure, refer to the link:
https://brainly.com/question/8646601
