At 300 K,the vapour pressures of two pure li | vedclass

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(D) According to Raoult's law,the total vapour pressure $p_T$ is given by: $p_T = x_A p_A^{\circ} + x_B p_B^{\circ}$ Since $x_B = 1 - x_A$,we have: $p

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$1 / 2$ and $1 / 6$

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(D) According to Raoult's law,the total vapour pressure $p_T$ is given by: $p_T = x_A p_A^{\circ} + x_B p_B^{\circ}$ Since $x_B = 1 - x_A$,we have: $p_T = x_A p_A^{\circ} + (1 - x_A) p_B^{\circ}$ Substituting the given values: $300 = x_A(100) + (1 - x_A)(500)$ $300 = 100x_A + 500 - 500x_A$ $400x_A = 200$ $x_A = 1 / 2$ Thus,the mole fraction of $A$ in the liquid phase is $1 / 2$. In the vapour phase,the mole fraction of $A$ $(y_A)$ is given by: $y_A = \frac{p_A}{p_T} = \frac{x_A p_A^{\circ}}{p_T}$ $y_A = \frac{(1 / 2) 	imes 100}{300} = \frac{50}{300} = 1 / 6$ Therefore,the mole fractions of $A$ in the liquid and vapour phases are $1 / 2$ and $1 / 6$ respectively.

At $300 \ K$,the vapour pressures of two pure liquids,$A$ and $B$ are $100 \ mm \ Hg$ and $500 \ mm \ Hg$,respectively. If in a mixture of $A$ and $B$,the total vapour pressure is $300 \ mm \ Hg$,the mole fractions of $A$ in the liquid and in the vapour phase,respectively,are:

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