Respuesta :
Answer:
56.4 mmHg
Explanation:
Given:
Vapor pressure of the solution, P solution = 55 mmHg
The mass of sucrose (C₆H₁₂O₆) = 10 g
Also, Molar mass of sucrose (C₆H₁₂O₆) = 180 g/mol
So, moles = Given mass/ molar mass
Hence, moles of sucrose in the solution = 10 g / 180 g/mol = 0.05556 mol
Given that: Mass of ethanol = 100 g
Molar mass of ethanol = 46 g/mol
Hence, moles of ethanol = 100 g / 46 g/mol = 2.174 mol
Mole fraction of solvent, ethanol is:
X ethanol = 2.174 mol / (2.174 + 0.05556) mol = 0.975
Applying Raoult's Law
P solution = X ethanol*P° ethanol
=> P° ethanol = P solution / X ethanol = 55 mmHg / 0.975 = 56.4 mm Hg
Explanation:
The given data is as follows.
Vapor pressure of the solution [tex](P_{solution})[/tex] = 55 mm Hg
Mass of sucrose = 10 g
Molar mass of sucrose = 180 g/mol
Therefore, moles of sucrose present into the solution will be calculated as follows.
No. of moles = [tex]\frac{mass}{molar mass}[/tex]
= [tex]\frac{10 g}{180 g/mol}[/tex]
= 0.055 mol
Mass of ethanol is given as 100 g and its molar mass is 46 g/mol.
Hence, number of moles of ethanol will be calculated as follows.
No. of moles = [tex]\frac{mass}{molar mass}[/tex]
= [tex]\frac{100 g}{46 g}[/tex]
= 2.174 mol
As mole fraction = [tex]\frac{no. of moles}{total number of moles}[/tex]
Hence, mole fraction of etahnol will be calculated as follows.
[tex]X_{ethanol}[/tex] = [tex]\frac{no. of moles}{total number of moles}[/tex]
= [tex]\frac{2.174}{2.174 + 0.055}[/tex]
= 0.975
Now, using Raoult's Law as follows.
[tex]P_{solution} = X_{ethanol} \times P_{ethanol}[/tex]
[tex]P_{ethanol}[/tex] = [tex]\frac{P_{solution}}{X_{ethanol}}[/tex]
= [tex]\frac{55 mm Hg}}{0.975}}[/tex]
= 56.4 mm Hg
Thus, we can conclude that the vapor pressure of the pure solvent is 56.4 mm Hg.
