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At 40 ^(@) C, the vapour pressure (in to...

At `40 ^(@) C,` the vapour pressure (in torr) of methyl alcohol (A) and ethyl alcohol (B) solution is represented by :
`P = 120 X _(A) + 13 8 , ` where `X _(A)` is mole fraction of methyl alcohol. The valur of
`underset( X _(A) to 0) ( P _("Limit")) & underset( X _(B) to 0) ( P _( "limit"))` 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, P = P _(B)`
`P = 138`
If `X _(B) = 0 implies X _(A) = 1,`
`therefore P = P _(A) = 120 + 138 = 258.`
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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

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 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:

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 nd ethanol solution is P=119x+135 where x is the mole fraction of methanol. Hence

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 :

At 25^(@)C , the vapour pressure of pure methyl alcohol is 92.0 torr. Mol fraction of CH_(3)OH in a solution in which vapour pressure of CH_(3)OH is 23.0 torr at 25^(@)C , is:

The vapour pressure of methyl alcohol at 298 K is 0.158 bar. The vapour pressure of this liquid in solution with liquid B is 0.095 bar. Calculate the mole fraction of methyl alcohol in the solution if the mixture obeys Raoult's law.

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