Solution containing 320 gram of methanol to 30 gram of ethanol and 290 gram of the mole fraction of ethanol in the solution is
Answers
Answer:
Mol. mass of ethyl alcohol =C_2H_5OH=46=C
2
H
5
OH=46
No. of moles of ethyl alcohol =\dfrac{60}{46}=1.304=
46
60
=1.304
Mol. mass of methyl alcohol =CH_3OH=32=CH
3
OH=32
No. of moles of methyl alcohol =\dfrac{40}{32}=1.25=
32
40
=1.25
'X_A'
′
X
A
′
, mole fraction of ethyl alcohol =\dfrac{1.304}{1.304+1.25}=0.5107=
1.304+1.25
1.304
=0.5107
'X_B'
′
X
B
′
, mole fraction of methyl alcohol =\dfrac{1.25}{1.304+1.25}=0.4893=
1.304+1.25
1.25
=0.4893
Partial pressure ethyl alcohol =X_A\dot p^0_A=0.5107\times 44.5 = 22.73 mm Hg=X
A
p
˙
A
0
=0.5107×44.5=22.73mmHg
Partial pressure methyl alcohol =X_B\dot p^0_B=0.4893\times 88.7 = 73.40 mm Hg=X
B
p
˙
B
0
=0.4893×88.7=73.40mmHg
Total vapour pressure of solution =22.73+43.40=66.13 mm Hg=22.73+43.40=66.13mmHg
Mole fraction of methyl alcohol in the vapour
=\dfrac{\text{Partial pressure of CH}_3OH}{\text{Total vapour pressure}}=\dfrac{43.40}{66.13}=0.6563=
Total vapour pressure
Partial pressure of CH
3
OH
=
66.13
43.40
=0.6563