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Question: An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of m...

An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour
pressure of methanol and ethanol are 2.619 kPa and 4.556 kPa, respectively, the composition of
vapour (in terms of mole fraction) will be:
(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

Explanation

Solution

For ideal solution, we know partial pressure of solution is directly proportional to mole fraction
of individual substance.
To make proportionality sign into equal, we need to multiply a constant and that constant is total
pressure, so we can write:
PAXA PA=XA×PTotal  {P_A} \propto {X_A} \\\ {P_A} = {X_A} \times {P_{Total}} \\\
Here, PA{P_A} = partial pressure of gas A
XA{X_A} = Mole fraction of gas A
PTotal{P_{Total}} = Total pressure of all gases

Complete step by step answer:
Given:
Pmethanol=2.619kPa Pethanol=4.556kPa  {P_{methanol}} = 2.619kPa \\\ {P_{ethanol}} = 4.556kPa \\\
Now, we can calculate total pressure by adding both these partial pressures.
PTotal=2.619+4.556 =7.175kPa  {P_{Total}} = 2.619 + 4.556 \\\ = 7.175kPa \\\
We also know,
PAXA PA=XA×PTotal  {P_A}\propto {X_A} \\\ {P_A} = {X_A} \times {P_{Total}} \\\
Substitute the values for Methanol.
Pmethanol=Xmethanol×PTotal{P_{methanol}} = {X_{methanol}} \times {P_{Total}}
Substitute, partial pressure of methanol, and total pressure, we get

2.619=Xmethanol×7.1752.619 = {X_{methanol}} \times 7.175
Taking numerical values on one side,
2.6197.175=Xmethanol   \dfrac{{2.619}}{{7.175}} = {X_{methanol}} \\\ \\\
Xmethanol=0.365\therefore {X_{methanol}} = 0.365
Thus, mole fraction of Methanol is 0.365,
Similarly, when we have to find for ethanol,
Xethanol=PethanolPTotal{X_{ethanol}} = \dfrac{{{P_{ethanol}}}}{{{P_{Total}}}}
Now, substitute the values in above equation, we get
Xethanol=4.5567.175{X_{ethanol}} = \dfrac{{4.556}}{{7.175}}
Xethanol=0.635\therefore {X_{ethanol}} = 0.635
Thus mole fraction of Ethanol is 0.635.
So option (B) 0.365 MeOH, 0.635 EtOH is the correct answer.

Note:
First alternate method:
To find mole fraction of ethanol, after we know mole fraction of methanol, we can use this
alternative.
We know for binary system XA+XB=1{X_A} + {X_B} = 1

Xmethanol+Xethanol=1 Xethanol=1Xmethanol Xethanol=10.365 =0.635 {X_{methanol}} + {X_{ethanol}} = 1 \\\ {X_{ethanol}} = 1 - {X_{methanol}} \\\ {X_{ethanol}} = 1 - 0.365 \\\ = 0.635 \\\

Second Alternate method is hit and trial
We know, Partial pressure of gas is proportional to mole fraction of that substance.
So, we can take their ratio PAPB=XAXB\dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{{X_A}}}{{{X_B}}}
Here,

PmethanolPethanol=2.6194.556 =0.575 \dfrac{{{P_{methanol}}}}{{{P_{ethanol}}}} = \dfrac{{2.619}}{{4.556}} \\\ = 0.575 \\\

Now we can check each option one by one.
(E) 0.635 MeOH, 0.365 EtOH
XmethanolXethanol=0.6350.365=1.74\dfrac{{{X_{methanol}}}}{{{X_{ethanol}}}} = \dfrac{{0.635}}{{0.365}} = 1.74
(F) 0.365 MeOH, 0.635 EtOH
XmethanolXethanol=0.3650.635=0.575\dfrac{{{X_{methanol}}}}{{{X_{ethanol}}}} = \dfrac{{0.365}}{{0.635}} = 0.575
(G) 0.574 MeOH, 0.326 EtOH
XmethanolXethanol=0.5740.326=1.76\dfrac{{{X_{methanol}}}}{{{X_{ethanol}}}} = \dfrac{{0.574}}{{0.326}} = 1.76
(H) 0.173 MeOH, 0.827 EtOH
XmethanolXethanol=0.1730.827=0.21\dfrac{{{X_{methanol}}}}{{{X_{ethanol}}}} = \dfrac{{0.173}}{{0.827}} = 0.21
Thus, we can see that option (B) is same as the calculated ratio, thus right answer is (B)