Consider a binary solution of two volatile liquid components 1 an… - Vidyadip

MCQ

Consider a binary solution of two volatile liquid components 1 and $2 x _1$ and $y _1$ are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as :

✓
$\frac{ P _1^0}{ P _2^0}, \frac{ P _2^0- P _1^0}{ P _2^0}$
B
$\frac{ P _2^0}{ P _1^0}, \frac{ P _1^0- P _2^0}{ P _2^0}$
C
$\frac{ P _1^0}{ P _2^0}, \frac{ P _1^0- P _2^0}{ P _2^0}$
D
$\frac{ P _2^0}{ P _1^0}, \frac{ P _2^0- P _1^0}{ P _2^0}$

✓

Answer

Correct option: A.

$\frac{ P _1^0}{ P _2^0}, \frac{ P _2^0- P _1^0}{ P _2^0}$

(A)
$\because$ For liquid solution of two liquids ' 1 ' and ' 2 '
$
\begin{array}{l}
P_1=P_{T y_1}=P_1^{o} x_1 \\
\therefore \frac{P_{T}}{x_1}=\frac{P_1^{o}}{y_1} \\
\therefore \frac{P_2^{o}+x_1\left(P_1^{o}-P_2^{o}\right)}{x_1}=\frac{P_1^{o}}{y_1}
\end{array}
$
$\begin{array}{l}\therefore \frac{ P _2^{ o }}{ x _1}+\left( P _1^{ o }- P _2^{ o }\right)=\frac{ P _1^{ o }}{ y _1} \\ \therefore \frac{1}{ x _1}=\left(\frac{ P _1^{ o }}{ P _2^{ o }}\right)\left(\frac{1}{ y _1}\right)+\left(\frac{ P _2^{ o }- P _1^{ o }}{ P _2^{ o }}\right) \\ \therefore \text { Slope }=\left(\frac{ P _1^{ o }}{ P _2^{ o }}\right) \\ \therefore \text { Intercept }=\left(\frac{ P _2^{ o }- P _1^{ o }}{ P _2^{ o }}\right)\end{array}$

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