Consider a binary solution of two volatile li | vedclass

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(A) According to Raoult's law for a binary solution of two volatile liquids $1$ and $2$:$P_T = P_1^0 x_1 + P_2^0 x_2 = P_1^0 x_1 + P_2^0 (1 - x_1) = P

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$\frac{P_1^0}{P_2^0}, \frac{P_2^0-P_1^0}{P_2^0}$

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(A) According to Raoult's law for a binary solution of two volatile liquids $1$ and $2$:$P_T = P_1^0 x_1 + P_2^0 x_2 = P_1^0 x_1 + P_2^0 (1 - x_1) = P_2^0 + x_1 (P_1^0 - P_2^0)$Also,from Dalton's law,$P_1 = P_T y_1$,where $P_1 = P_1^0 x_1$.So,$P_T y_1 = P_1^0 x_1 \implies P_T = \frac{P_1^0 x_1}{y_1}$.Equating the two expressions for $P_T$:$P_2^0 + x_1 (P_1^0 - P_2^0) = \frac{P_1^0 x_1}{y_1}$Dividing both sides by $x_1 P_2^0$:$\frac{P_2^0}{x_1} + \frac{P_1^0 - P_2^0}{P_2^0} = \frac{P_1^0}{P_2^0 y_1}$Rearranging to the form $Y = mX + c$ where $Y = \frac{1}{x_1}$ and $X = \frac{1}{y_1}$:$\frac{1}{x_1} = \left( \frac{P_1^0}{P_2^0} \right) \frac{1}{y_1} + \left( \frac{P_2^0 - P_1^0}{P_2^0} \right)$Comparing with $Y = mX + c$,the slope is $\frac{P_1^0}{P_2^0}$ and the intercept is $\frac{P_2^0 - P_1^0}{P_2^0}$.

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 :

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