The vapour pressure of ethanol and methanol are44.5mm Hg and 88.7 mm Hg respectively. A solution is made by mixing 60 g of ethanol and 40 g of methanol. Assuming the solution tp be ideal. Calculate the vapour pressure of the solution
Answers
Answer:
Mol. mass of ethyl alcohol =C 2h5
OH=46
No. of moles of ethyl alcohol = 46/60
. =
Mol. mass of methyl alcohol =CH
3
OH=32
No. of moles of methyl alcohol =
32
40
=1.25
′
X
A
′
, mole fraction of ethyl alcohol =
1.304+1.25
1.304
=0.5107
′
X
B
′
, mole fraction of methyl alcohol =
1.304+1.25
1.25
=0.4893
Partial pressure ethyl alcohol =X
A
p
˙
A
0
=0.5107×44.5=22.73mmHg
Partial pressure methyl alcohol =X
B
p
˙
B
0
=0.4893×88.7=73.40mmHg
Total vapour pressure of solution =22.73+43.40=66.13mmHg
Mole fraction of methyl alcohol in the vapour
=
Total vapour pressure
Partial pressure of CH
3
OH
=
66.13
43.40
=0.6563
Explanation:
this is answer of your questionMol. mass of ethyl alcohol =C
2
H
5
OH=46
No. of moles of ethyl alcohol =
46
60
=1.304
Mol. mass of methyl alcohol =CH
3
OH=32
No. of moles of methyl alcohol =
32
40
=1.25
′
X
A
′
, mole fraction of ethyl alcohol =
1.304+1.25
1.304
=0.5107
′
X
B
′
, mole fraction of methyl alcohol =
1.304+1.25
1.25
=0.4893
Partial pressure ethyl alcohol =X
A
p
˙
A
0
=0.5107×44.5=22.73mmHg
Partial pressure methyl alcohol =X
B
p
˙
B
0
=0.4893×88.7=73.40mmHg
Total vapour pressure of solution =22.73+43.40=66.13mmHg
Mole fraction of methyl alcohol in the vapour
=
Total vapour pressure
Partial pressure of CH
3
OH
=
66.13
43.40
=0.6563