Question

The vapour pressure of a mixture(PM) for liquid-liquid
system is given by PM = P0A XA + P0B XB where as for
solid- liquid system the vapour pressure of solution PS
is given by PS = P0 x mole fraction of solvent. When
36 g. water mixed with 362 g. of ethanol, the vapour
pressure of ethanol-water system is given by PM
(mm) = 32.80 + 11.20 x Xethanol
1. What is the law used in the above discussion.
2. The vapour pressure of pure ethanol is
3. The vapour pressure of pure water is
4. What is the mole fraction of ethanol in the vapour
phase
5. The vapour pressure of a solution having 0.7
mole fraction of water in ethanol is

Answer 1

Answer:

by PM = P0A XA + P0B XB where as for

solid- liquid system the vapour pressure of solution PS

is given by PS = P0 x mole fraction of solvent. When

36 g. water mixed with 362 g. of ethanol, the vapour

pressure of ethanol-water system is given by PM

(mm) = 32.80 + 11.20 x Xethanol