Respuesta :
Answer:
We have 2.58 moles toluene in solution
Explanation:
Step 1: Data given
Mass of toluene = 801 grams
Boiling point of the solution = 90.3 °C
Boiling point elevation constant = 2.92 °C/m
Boiling point temperature = 80.9 °C
Step 2: Calculate molality
ΔT = i*Kb*m
⇒with ΔT = the boiling point elevation = 90.3-80.9 = 9.4 °C
⇒with i = the van't Hoff factor = 1
⇒with Kb = the Boiling point elevation constant = 2.92 °C/m
⇒with m = the molality
m = 9.4 °C / 2.92
m = 3.22 molal = moles / kg
Step 3: Calculate moles toluene
Molality = moles tolune / mass cyclohexane
3.22 molal = moles toluene / 0.801 kg
Moles toluene = 2.58 moles
We have 2.58 moles toluene in solution
The moles of toluene in 801 grams of the solution have been 2.578 mol.
For heating, a solution, the change in temperature can be given by:
[tex]\Delta[/tex]T = i [tex]\rm \times\;k_b\;\times[/tex] m
[tex]\Delta[/tex]T = Change in temperature = Final temperature - Initial temperature
[tex]\Delta[/tex]T = 90.3[tex]\rm ^\circ C[/tex] - 80.9[tex]\rm ^\circ C[/tex]
[tex]\Delta[/tex]T = 9.4 [tex]\rm ^\circ C[/tex]
i = von't Hoff factor = 1
[tex]\rm k_b[/tex] = 2.92[tex]\rm ^\circ C[/tex]/m
m = molality
9.4 = 1 [tex]\times[/tex] 2.92 [tex]\times[/tex] m
m = 3.21 m
The molality of the solution of toluene has been 3.21 mol.
The molality can be given as moles per kg solvent.
The mass of solvent = 801 g. = 0.801 kg
molality = [tex]\rm \dfrac{moles}{solvent}[/tex]
3.21 = [tex]\rm \dfrac{moles}{0.801}[/tex]
Moles of toluene = 3.21 [tex]\times[/tex] 0.801
Moles of toluene = 2.578 mol.
The moles of toluene in 801 grams of the solution have been 2.578 mol.
For more information about moles in a solution, refer to the link:
https://brainly.com/question/12399522
