Question

A solution has `1:4` mole ratio of pentane to hexane . The vapour pressure of pure hydrocarbons at `20^@C`are `440` mmHgfor pentane and `120`mmHg for hexane .The mole
A. `0.200`
B. `0.478`
C. `0.549`
D. `0.786`

Answer 1

Correct Answer - 2
Mole fraction of pentane in the vapor phase is
`Y_("pentane")=P_("pentane")/P_("total")`
`P_("pentane")= P_("pentane")^(0)chi_("pentane") ...(1)`
`P_("total")=P_("pentane")+P_("hexane")=P_("pentane")^(0)chi_("pentane")+P_("hexane")^(0)chi_("hexane") …(2)`
Considering 1 mole of pentane, we have 4 moles of hexane.
Thus
`chi_("pentane")=n_("pentane")/(n_("pentane")+n_("hexane"))`
`=1/(1+4)=1/5=0.2`
`chi_("hexane")=1-chi_("pentane")`
`=1-0.2`
`=0.8`
we are given
`P_("pentane")^(0)=440 mmHg`
`P_("hexane")^(0)=120 mmHg`
substituting thses results into Eqs (1) and (2), we obtain
`P_("pentane")=(440 mmHg)(0.2)`
`=88 mmHg`
`P_("total")=(440 mmHg)(0.2)+(120 mmHg)(0.8)`
`=88 mmHg+96 mmHg`
`=184 mmHg`
Using these results, we have
`Y_("pentane")=(88 mm Hg)/(184 mm Hg)`
`=0.478`