The vapour pressure of ethanol and methanol are 42.0 mm and 88.5 mm Hg respectively.An ideal solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol .The mole fraction of methanol is the vapour is
1.0.467
2.0.502
3.0.513
4.0.556
Answers
Answer:
Explanation:
Mol. 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
0.513
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