(B) In the vapor phase,the mole fraction of a component is given by the ratio of its partial pressure to the total pressure of the solution.Total pres

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(B) In the vapor phase,the mole fraction of a component is given by the ratio of its partial pressure to the total pressure of the solution.Total pressure $P_{total} = P_{MeOH} + P_{EtOH} = 2.619 \ kPa + 4.556 \ kPa = 7.175 \ kPa$.Mole fraction of $MeOH$ in vapor phase $(y_{MeOH}) = \frac{P_{MeOH}}{P_{total}} = \frac{2.619}{7.175} = 0.365$.Mole fraction of $EtOH$ in vapor phase $(y_{EtOH}) = \frac{P_{EtOH}}{P_{total}} = \frac{4.556}{7.175} = 0.635$.

Class 12 Chemistry Solutions Ideal and Non-ideal solution

Medium

$A$ mixture of methanol $(MeOH)$ and ethanol $(EtOH)$ forms an ideal solution. If the partial pressures of methanol and ethanol are $2.619 \ kPa$ and $4.556 \ kPa$ respectively,what will be the mole fractions of the components in the vapor phase?

A
$0.635 \ MeOH, 0.365 \ EtOH$
B
$0.365 \ MeOH, 0.635 \ EtOH$
C
$0.574 \ MeOH, 0.326 \ EtOH$
D
$0.173 \ MeOH, 0.827 \ EtOH$

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Ideal and Non-ideal solution

In a binary ideal solution of liquids $A$ and $B$ at a constant temperature,the mole fraction of $A$ in the liquid phase is $0.7$ and in the vapor phase is $0.4$. If $P_A^o + P_B^o = 90 \, mm$,then the vapor pressures of pure liquids $A$ and $B$ at that temperature are respectively:

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Liquids $A$ and $B$ form an ideal solution. If $X_A$ and $Y_A$ are the mole fractions of $A$ in the solution and in vapour phase in equilibrium,respectively,then (Given $P_A^{\circ} > P_B^{\circ}$):

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$1 \, \text{mole}$ of liquid $A$ and $2 \, \text{mole}$ of liquid $B$ form a solution with a vapor pressure of $38 \, \text{torr}$. If the vapor pressures of pure $A$ and pure $B$ are $45 \, \text{torr}$ and $36 \, \text{torr}$ respectively,the solution will be:

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Which of the following is true when components forming an ideal solution are mixed?

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$p_A$ and $p_B$ are the vapour pressure of pure liquid components,$A$ and $B,$ respectively of an ideal binary solution. If $x_A$ represents the mole fraction of component $A,$ the total pressure of the solution will be

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$A$ mixture of methanol $(MeOH)$ and ethanol $(EtOH)$ forms an ideal solution. If the partial pressures of methanol and ethanol are $2.619 \ kPa$ and $4.556 \ kPa$ respectively,what will be the mole fractions of the components in the vapor phase?

Similar Questions

In a binary ideal solution of liquids $A$ and $B$ at a constant temperature,the mole fraction of $A$ in the liquid phase is $0.7$ and in the vapor phase is $0.4$. If $P_A^o + P_B^o = 90 \, mm$,then the vapor pressures of pure liquids $A$ and $B$ at that temperature are respectively:

Liquids $A$ and $B$ form an ideal solution. If $X_A$ and $Y_A$ are the mole fractions of $A$ in the solution and in vapour phase in equilibrium,respectively,then (Given $P_A^{\circ} > P_B^{\circ}$):

$1 \, \text{mole}$ of liquid $A$ and $2 \, \text{mole}$ of liquid $B$ form a solution with a vapor pressure of $38 \, \text{torr}$. If the vapor pressures of pure $A$ and pure $B$ are $45 \, \text{torr}$ and $36 \, \text{torr}$ respectively,the solution will be:

Which of the following is true when components forming an ideal solution are mixed?

$p_A$ and $p_B$ are the vapour pressure of pure liquid components,$A$ and $B,$ respectively of an ideal binary solution. If $x_A$ represents the mole fraction of component $A,$ the total pressure of the solution will be

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