Respuesta :
Explanation:
(a) It is known that relation between vapor pressure and mole fraction is as follows.
vapor pressure = [tex]x1 \times P_{1} + x_{2} \times P_{2}[/tex]
where, [tex]x_{1}[/tex] = mole fraction of component one
[tex]P_{1}[/tex] = vapor pressure of component one when pure
Also, [tex]x_{1} + x_{2}[/tex] = 1
Now, putting the given values into the above formula as follows.
vapor pressure = [tex]x1 \times P_{1} + x_{2} \times P_{2}[/tex]
33 = [tex]x_{1} \times 75 + (1 - x_{1}) \times 22[/tex]
11 = [tex](75 - 22) \times x_{1}[/tex]
[tex]x_{1}[/tex] = 0.20754717
= 0.21 (approx)
And, [tex]x_{2} = 1 - x_{1}[/tex]
= (1 - 0.21)
= 0.79
Hence, the composition in mole fractions of a solution that has a vapor pressure of 33 torr at [tex]20^{o}C[/tex] is 0.79.
(b) According to Raoult's law,
[tex]x_{1} \times P_{1} = (P_{vapor}) \times y_{1 }[/tex]
where, [tex]y_{1}[/tex] = composition in gas phase
Therefore, calculate the value of [tex]y_{1}[/tex] (mole fraction of benzene) as follows.
[tex]y_{1} = {x_{1} \times \frac{P_{1}}{P_{vap}}[/tex]
= [tex]\frac{0.21 \times 75}{33}[/tex]
= 0.471698114
= 0.47 (approx)
Thus, mole fraction of benzene in the given vapor is 0.47.
