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
581.1k+ 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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

