
The vapor pressure (in torr) of $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}} $ (A) and $ {{\text{C}}_2}{{\text{H}}_5}{\text{OH}} $ (B) solution is represented by $ {\text{P = 120}}{{\text{X}}_{\text{A}}}{\text{ + 138}} $ ; where $ {{\text{X}}_{\text{A}}} $ is mole fraction of $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}} $ . The value of $ {{\text{P}}_{\text{B}}} $ at lim $ {{\text{X}}_{\text{A}}} \to {\text{0}} $ and $ {{\text{P}}_{\text{A}}} $ at lim $ {{\text{X}}_{\text{B}}} \to {\text{0}} $ are:
(A)- \[138,258\]
(B)- $ 258,138 $
(C)- $ 120,138 $
(D)- $ 138,125 $
Answer
548.7k+ views
Hint Vapor pressure of the solution is calculated as the sum of the product of the mole fraction of the constituent and partial pressure of that constituent. And expression for vapor pressure is defined as: $ {\text{P = }}{{\text{P}}_{\text{A}}}{{\text{X}}_{\text{A}}}{\text{ + }}{{\text{P}}_{\text{B}}}{{\text{X}}_{\text{B}}} $ .
Complete step by step solution:
In the question vapour pressure equation for a solution of methyl alcohol ( $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}} $ ) and ethyl alcohol ( $ {{\text{C}}_2}{{\text{H}}_5}{\text{OH}} $ ) is shown as follow:
$ {\text{P = 120}}{{\text{X}}_{\text{A}}}{\text{ + 138}} $ ………… (i)
Let we compare the given equation to the general vapor pressure equation for a solution $ {\text{P = }}{{\text{P}}_{\text{A}}}{{\text{X}}_{\text{A}}}{\text{ + }}{{\text{P}}_{\text{B}}}{{\text{X}}_{\text{B}}} $ and we get,
$ {{\text{P}}_{\text{A}}} $ = initial vapor pressure of methyl alcohol = $ 120 $
$ {{\text{X}}_{\text{A}}} $ = mole fraction of methyl alcohol
$ {{\text{P}}_{\text{B}}} $ = initial vapor pressure of ethyl alcohol = $ 138 $
$ {{\text{X}}_{\text{B}}} $ = mole fraction of ethyl alcohol
As we know that, sum of the mole fraction of the constituents present in a solution is always equals to one i.e. $ {{\text{X}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{\text{ = 1}} $ .
-Now we will calculate the value of initial pressure of ethyl alcohol ( $ {{\text{P}}_{\text{B}}} $ ) at lim $ {{\text{X}}_{\text{A}}} \to {\text{0}} $ by following manner:
By equation (i) we get,
$ {\text{P = 120}}{{\text{X}}_{\text{A}}}{\text{ + 138}} $
Now we take lim $ {{\text{X}}_{\text{A}}} \to {\text{0}} $ on both side of the equation and we get,
$ {\text{lim}}{{\text{X}}_{\text{A}}} \to {\text{0(P) = lim}}{{\text{X}}_{\text{A}}} \to {\text{0(120}}{{\text{X}}_{\text{A}}}{\text{) + 138}} $
$ {\text{lim}}{{\text{X}}_{\text{A}}} \to {\text{0(P) = 138}} $
On putting the value of $ {{\text{X}}_{\text{A}}} = 0 $ , we will get the vapor pressure of the solution equals to the vapor pressure of ethyl alcohol which is equal to $ {\text{138torr}} $ i.e. $ {{\text{P}}_{\text{B}}}{\text{ = 138torr}} $ .
-Now we will calculate the value of initial pressure of methyl alcohol ( $ {{\text{P}}_{\text{A}}} $ ) at lim $ {{\text{X}}_{\text{B}}} \to {\text{0}} $ by following manner:
In the equation (i) mole fraction term of ethyl alcohol ( $ {{\text{X}}_{\text{B}}} $ ) is not given and we have to calculate the value of $ {{\text{P}}_{\text{A}}} $ at lim $ {{\text{X}}_{\text{B}}} \to {\text{0}} $ , and in this condition value of $ {{\text{X}}_{\text{A}}} $ is equals to one as value of $ {{\text{X}}_{\text{B}}} $ tends to zero and we know that for a solution $ {{\text{X}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{\text{ = 1}} $ . So equation (i) is written as:
$ {\text{lim}}{{\text{X}}_{\text{B}}} \to 0({\text{P) = lim}}{{\text{X}}_{\text{B}}} \to 0({\text{120 + 138)}} $
$ {\text{lim}}{{\text{X}}_{\text{B}}} \to 0({\text{P) = }}258 $
On putting the value of $ {{\text{X}}_{\text{B}}} = 0 $ , we will get the vapor pressure of the solution equals to the vapor pressure of methyl alcohol which is equal to $ {\text{258torr}} $ i.e. $ {{\text{P}}_{\text{A}}}{\text{ = 258torr}} $ .
Hence the value of $ {{\text{P}}_{\text{B}}} $ at lim $ {{\text{X}}_{\text{A}}} \to {\text{0}} $ and $ {{\text{P}}_{\text{A}}} $ at lim $ {{\text{X}}_{\text{B}}} \to {\text{0}} $ are $ {\text{138torr}} $ and $ {\text{258torr}} $ respectively.
Note:
In this question some of you may do wrong calculation if you are not familiar with the relation between the mole fractions of the constituents of the solution i.e. $ {{\text{X}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{\text{ = 1}} $ . Always keep in mind if one of the mole fraction gets zero then the other will get the value equals to one.
Complete step by step solution:
In the question vapour pressure equation for a solution of methyl alcohol ( $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}} $ ) and ethyl alcohol ( $ {{\text{C}}_2}{{\text{H}}_5}{\text{OH}} $ ) is shown as follow:
$ {\text{P = 120}}{{\text{X}}_{\text{A}}}{\text{ + 138}} $ ………… (i)
Let we compare the given equation to the general vapor pressure equation for a solution $ {\text{P = }}{{\text{P}}_{\text{A}}}{{\text{X}}_{\text{A}}}{\text{ + }}{{\text{P}}_{\text{B}}}{{\text{X}}_{\text{B}}} $ and we get,
$ {{\text{P}}_{\text{A}}} $ = initial vapor pressure of methyl alcohol = $ 120 $
$ {{\text{X}}_{\text{A}}} $ = mole fraction of methyl alcohol
$ {{\text{P}}_{\text{B}}} $ = initial vapor pressure of ethyl alcohol = $ 138 $
$ {{\text{X}}_{\text{B}}} $ = mole fraction of ethyl alcohol
As we know that, sum of the mole fraction of the constituents present in a solution is always equals to one i.e. $ {{\text{X}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{\text{ = 1}} $ .
-Now we will calculate the value of initial pressure of ethyl alcohol ( $ {{\text{P}}_{\text{B}}} $ ) at lim $ {{\text{X}}_{\text{A}}} \to {\text{0}} $ by following manner:
By equation (i) we get,
$ {\text{P = 120}}{{\text{X}}_{\text{A}}}{\text{ + 138}} $
Now we take lim $ {{\text{X}}_{\text{A}}} \to {\text{0}} $ on both side of the equation and we get,
$ {\text{lim}}{{\text{X}}_{\text{A}}} \to {\text{0(P) = lim}}{{\text{X}}_{\text{A}}} \to {\text{0(120}}{{\text{X}}_{\text{A}}}{\text{) + 138}} $
$ {\text{lim}}{{\text{X}}_{\text{A}}} \to {\text{0(P) = 138}} $
On putting the value of $ {{\text{X}}_{\text{A}}} = 0 $ , we will get the vapor pressure of the solution equals to the vapor pressure of ethyl alcohol which is equal to $ {\text{138torr}} $ i.e. $ {{\text{P}}_{\text{B}}}{\text{ = 138torr}} $ .
-Now we will calculate the value of initial pressure of methyl alcohol ( $ {{\text{P}}_{\text{A}}} $ ) at lim $ {{\text{X}}_{\text{B}}} \to {\text{0}} $ by following manner:
In the equation (i) mole fraction term of ethyl alcohol ( $ {{\text{X}}_{\text{B}}} $ ) is not given and we have to calculate the value of $ {{\text{P}}_{\text{A}}} $ at lim $ {{\text{X}}_{\text{B}}} \to {\text{0}} $ , and in this condition value of $ {{\text{X}}_{\text{A}}} $ is equals to one as value of $ {{\text{X}}_{\text{B}}} $ tends to zero and we know that for a solution $ {{\text{X}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{\text{ = 1}} $ . So equation (i) is written as:
$ {\text{lim}}{{\text{X}}_{\text{B}}} \to 0({\text{P) = lim}}{{\text{X}}_{\text{B}}} \to 0({\text{120 + 138)}} $
$ {\text{lim}}{{\text{X}}_{\text{B}}} \to 0({\text{P) = }}258 $
On putting the value of $ {{\text{X}}_{\text{B}}} = 0 $ , we will get the vapor pressure of the solution equals to the vapor pressure of methyl alcohol which is equal to $ {\text{258torr}} $ i.e. $ {{\text{P}}_{\text{A}}}{\text{ = 258torr}} $ .
Hence the value of $ {{\text{P}}_{\text{B}}} $ at lim $ {{\text{X}}_{\text{A}}} \to {\text{0}} $ and $ {{\text{P}}_{\text{A}}} $ at lim $ {{\text{X}}_{\text{B}}} \to {\text{0}} $ are $ {\text{138torr}} $ and $ {\text{258torr}} $ respectively.
Note:
In this question some of you may do wrong calculation if you are not familiar with the relation between the mole fractions of the constituents of the solution i.e. $ {{\text{X}}_{\text{A}}}{\text{ + }}{{\text{X}}_{\text{B}}}{\text{ = 1}} $ . Always keep in mind if one of the mole fraction gets zero then the other will get the value equals to one.
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