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At 40°C, the vapour pressure (in torr) o...

At 40°C, the vapour pressure (in torr) of methyl alcohol (A) and ethyl alcohol solution is represented by:
`P= 120X_A + 138`, where `X_A` is mole fraction of methyl alcohol. The value of `P_B^@` at
`limX_A to 0 and P_A`° at
`limX_B to 0` are :

A

138, 258

B

258, 138

C

120, 138

D

138, 125

Text Solution

Verified by Experts

The correct Answer is:
A
If `X_A=0,` then pure
`B :. P_B^@=138`
If `X_A=1`, thn pure
`A :. P_A^@=120+138=258`
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At 40^(@)C , the vapour pressure in torr of methyl and ethyl alcohol solutions is represented by P = 119 X_(A)+135 , where X_(A) is mole fraction of methyl alcohol. The value of (P_(B)^(@))/(X_(B)) at lim X_(A) rarr 0) , and (P_(A)^(@))/(X_(A)) at lim X_(B) rarr 0 are:

Vapour pressure of methyl alcohol and ethyl alcohol solutions is represented by P = 115 x_(A)+140 where x_(A) is the mole fraction of methyl alcohol. The value of underset(x_(A)rarr0)(lim)(P_(B)^(0))/(x_(B)) is:

At 40^(@)C , the vapour pressures in torr, of methyl alcohol and ethyl alcohol solutions is represented by the equation. P = 119X_(A) + 135 where X_(A) is mole fraction of methyl alcohol, then the value of underset(x_(A)rarr1)(lim)(P_(A))/(X_(A)) is

At 323K , the vapour pressure in millimeters of mercury of a methanol-ethanol solution is represented by the equation p=120X_(A)+140 , where X_(A) is the mole fraction of methanol. Then the value of underset(x_(A) to 1)(lim) (p_(A))/(X_(A)) is:

At 323 K, the vapour pressure in millimeters of mercury of a methanol - ethanol solution is represented by the equation p=120X_(A)+140, where X_(A) is the mole fraction of methanol. Then the value of underset(x_(A)rarr1)"lim"(p_(A))/(p_(B)) is :

At 40^(@)C , vapour pressure in torr of methanol and ethanol solution is P=119x+135, where x is the mole fraction of methanol. Hence :

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Mole fraction of ethyl alcohol in aqueous ethyl alcohol solution is 0.25 . Hence percentage of ethyl alcohol by weight is :

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