Question
Bromobenzene, C6H5Br, and chlorobenzene, C6H5Cl, form essentially ideal solutions in all proportions. At 100°C, the equilibrium vapour pressure of bromobenzene is 137mmHg and that of chlorobenzene is 285mmHg. Calculate the equilibrium vapour pressure above a 17.6% (w/w) solution of bromobenzene in chlorobenzene at a temperature of 100°C
16305 mmHg
163 mmHg
25895 mmHg
265 mmHg
157 mmHg
16305 mmHg
163 mmHg
25895 mmHg
265 mmHg
157 mmHg
Answers
Are you sure the 25895 choice is not 258.95?
Since the solution is 17.6% in bromobenene, that means 17.6 g/100 g soln OR
17.6 g bromobenzene in a solution made up of 17.6 g bromogenzene and (100-17.6 = 82.4 g chlorobenzene).
Determine moles bromo and moles chloro.
Calculate mole fraction bromo and chloro.
Then vapor pressure of each component is
P = X*P<sup>o</sup>
You will have one partial pressure for the bromo and one partial pressure for th chlor. Add them for the total pressure. Post your work if you get stuck.
17.6 g bromobenzene in a solution made up of 17.6 g bromogenzene and (100-17.6 = 82.4 g chlorobenzene).
Determine moles bromo and moles chloro.
Calculate mole fraction bromo and chloro.
Then vapor pressure of each component is
P = X*P<sup>o</sup>
You will have one partial pressure for the bromo and one partial pressure for th chlor. Add them for the total pressure. Post your work if you get stuck.
i have 0.112 mol of bromo,
and 0.732 mol chloro,
but i am stuck on the mole fraction.
Do i just do 1-0.112= 0.888 or am i missing a step?
and 0.732 mol chloro,
but i am stuck on the mole fraction.
Do i just do 1-0.112= 0.888 or am i missing a step?
ok, this is what I have.
Xbromo= 0.112/+0.732+0.112=0.133
Xchloro= 1-0.133= 0.867
P=Xbromo*p= 0.133*137= 18.221mmHg
P=Xchloro*p= 0.867*285= 247.095mmHg
so I add 18.221+247.095= 265.316mmHg
Is that right?
Xbromo= 0.112/+0.732+0.112=0.133
Xchloro= 1-0.133= 0.867
P=Xbromo*p= 0.133*137= 18.221mmHg
P=Xchloro*p= 0.867*285= 247.095mmHg
so I add 18.221+247.095= 265.316mmHg
Is that right?
correct