A mixture of an organic liquid a and water distilled under one atmospheric pressure at 99.2 how many grams of steam
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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
Explanation:
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Explanation:
5.19g answer this is right answer
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