Respuesta :
Answer:
[tex]P=361.2torr[/tex]
Explanation:
Hello,
In this case, considering that the formed liquid is solution is ideal, we can relate the vapor pressure and molar fraction of water with the total vapor pressure of the solution by using the Dalton's equilibrium-based law:
[tex]P=x_{H_2O}P_{H_2O}^v[/tex]
In such a way, we compute the molar fraction of water by computing its moles as well as diglyme:
[tex]n_{H_2O}=100mL*\frac{1g}{1mL} *\frac{1mol}{18g} =5.56molH_2O\\\\n_{C_6H_{14}O_3}=150g*\frac{1mol}{134g}=1.12molC_6H_{14}O_3[/tex]
Thus, the mole fraction of water:
[tex]x_{H_2O}=\frac{5.56mol}{5.56mol+1.12mol}=0.832[/tex]
Thereby, the vapor pressure of the solution:
[tex]P=0.832*434torr\\\\P=361.2torr[/tex]
Regards.
The vapor pressure of the pure water with diglyme has been 361.28 torrs.
Vapor pressure of a solution can be given by :
Vapor pressure = Mole fraction of solvent [tex]\times[/tex] Vapor pressure of the pure solvent
The pure solvent has been pure water.
The vapor pressure of pure water = 434 torr.
The mole fraction of the solvent can be given as:
Mole fraction of solvent = [tex]\rm \dfrac{Moles\;of\;solvent}{Moles\;of\;solution}[/tex]
The solvent has been water. The density of water has been 1g/ml. It states that the mass of the 1 ml sample is 1 gram.
Mass of 100 ml water sample = 100 grams
Moles can be expressed as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of solvent (water) = [tex]\rm \dfrac{100}{18}[/tex]
Moles of solvent (water) = 5.55 mol.
The moles of solute (Diglyme) can be given as:
Moles of solute (Diglyme) = [tex]\rm \dfrac{150}{134.17}[/tex]
Moles of solute (Diglyme) = 1.117 mol.
The moles of solution = Solvent + solute
The moles of solution = 5.55 + 1.117 mol
The moles of solution = 6.667 mol.
The mole fraction of the solvent has been;
Mole fraction of solvent = [tex]\rm \dfrac{5.55}{6.667}[/tex]
Mole fraction of solvent = 0.832
The vapor pressure of the solution will be:
Vapor pressure = 0.832 [tex]\times[/tex] 434 torr
The vapor pressure of the solution = 361.28 torrs.
The vapor pressure of the pure water with diglyme has been 361.28 torrs.
For more information about the vapor pressure, refer to the link:
https://brainly.com/question/13423145
