Chemistry, asked by anshnagpal2443, 19 days ago

A mixture of an organic liquid a and water distilled under one atmospheric pressure at 99.2 how many grams of steam

Answers

Answered by clever61
0

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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Answered by princesharma96951
0

Explanation:

5.19g answer this is right answer

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