Respuesta :
Answer:
A. 0525.
B. 0.3483 atm.
C. 0.3890.
Explanation:
A.
Molar mass of Benzene =
(12*6) + (1*6)
= 78 g/mol.
Number of moles of Benzene = mass/molar mass
= 50/78
= 0.641 moles.
Molar mass of n-hexane =
(12*6) + (1*14)
= 86 g/mol.
Number of moles of n-hexane = mass/molar mass
= 50/86
= 0.581 moles.
Total moles of the solution = number of moles of Benzene + number of moles of n-hexane
= 0.641 + 0.581
= 1.222 moles
Mole fraction is the number of moles of a particular substance in a solution divided by the total number of moles of substances in the solution.
Mole fraction of Benzene = 0.641/1.222
= 0.525.
B.
Vapour Pressure of a solution is the amount of pressure that the vapour exert on the liquid solvent when they are in equilibrium and at a certain temperature.
Uaing Raouit's law and Dalton's law,
Raoult's law states that the the partial pressure of a solution is directly proportional to the mole fraction of the solute component.
Dalton's law states that the total pressure of a solution is the sum of its individual partial pressure.
Uaing Dalton's law,
Psol = Pbenzene + Pn-hexane
= 0.1355 + 0.2128
= 0.3483 atm
C.
Using Raoult's equation, Mole fraction of Benzene = P°benzene/Psol
= 0.1355/0.3483
= 0.3890
A. The mole fraction of benzene in the given solution has been 0.525.
B. The total pressure of the solution at 300 K has been 0.3483 atm.
C. The mole fraction of benzene in vapor has been 0.3890.
The moles of the substance can be given as the mass of the sample with respect to the molar mass. The expression for moles of sample has been given as:
[tex]\rm Moles\;=\;\dfrac{Mass}{Molecular\;mass}[/tex]
The moles in 50 grams benzene has been:
[tex]\rm Moles\;(Benzene)\;=\;\dfrac{50\;g}{78\;g/mol}\\Moles\;(Benzene)\;=\;0.641\;mol[/tex]
The moles in 50 grams n-hexane has been:
[tex]\rm Moles\;(n-hexane)\;=\;\dfrac{50\;g}{86\;g/mol}\\Moles\;(n-hexane)\;=\;0.581\;mol[/tex]
The solution has been the mixture of n-hexane and benzene. The total moles of solution has been:
[tex]\rm Total \;moles = n-hexane + benzene\\Total\;moles=0.581+0.641\;mol\\Total\;moles=1.222\;mol[/tex]
A. The mole fraction of a compound has been given by:
[tex]\rm Mole\;fraction\;=\;\dfrac{Moles\;of\;compound}{Total\;moles}[/tex]
The mole fraction of benzene (m) in the given solution has been given by substituting the values as:
[tex]m=\dfrac{0.641}{1.222}\\\\m=0.525[/tex]
The mole fraction of benzene in the given solution has been 0.525.
B. The total pressure of the solution been the sum of the partial pressure of the individual component.
The given partial pressure has been:
Partial pressure of benzene, [tex]P_B\;=\;0.1355\; \rm atm[/tex]
Partial pressure of n-hexane, [tex]P_H\;=\;0.2128\; \rm atm[/tex]
The total pressure, [tex]P_T_o_t_a_l[/tex] of the sample has been given as:
[tex]P_T_o_t_a_l=P_B\;+\;P_H[/tex]
Substituting the values for total pressure:
[tex]P_T_o_t_a_l=0.1355\;\rm atm\;+\;0.2128\;atm\\P_T_o_t_a_l=0.3483\;atm[/tex]
The total pressure of the solution at 300 K has been 0.3483 atm.
C. The mole fraction (m) for the vapor benzene has been given by:
[tex]m=\dfrac{P_B}{P_t_o_t_a_l}[/tex]
Substituting the values:
[tex]m=\dfrac{0.1355}{0.3483} \\\\m=0.3890[/tex]
The mole fraction of benzene in vapor has been 0.3890.
For more information about the vapor pressure, refer to the link:
https://brainly.com/question/25356241
