A mixture of an organic liquid a and water distilled under one atmospheric pressure at 99.2 how many grams of steam
Answers
Answer:
Total pressure of the solution(A+Water)=1 atm=760mm Hg
Vapour pressure of water=739mm Hg
Vapour pressure of organic liquid A is=760−739=21mm Hg
Moleucular mass of A=123
W
water
W
A
=
P
B
×MM
w
ater
P
A
×MM
A
Where W
A
&W
water
represents the mass of organic liquid A and mass of water respectively.
MM
A
&MM
w
ater represents the molar mass of A and molar mass of water respectively
W
water
1
=
739×18
21×123
=5.19 g
Answer:
total pressure of the solution(A+water=1atm=760mmHg
vapour pressure of water= 739 mmhg
vapour pressure of organic liquid A is= 760 - 739= 21 mm HG
moleucular mass of A is=123
Wa/water=Pa×MMa/Pb×MMw ater
where w a n w water represent the mass of organic liquid a and mass of water respectively
MMa&MMw ater represent the molar mass ofA and molar mass of water respectively
1/Wwater=21×123/739x18=5.19gram
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