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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1 / 2$ and $1 / 6$

Vedclass · Answer

(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:

Similar Questions

If the vapour pressure of a solution containing a non-volatile solute is $2\%$ less than the vapour pressure of pure water,find the molality of the solution.

What is the vapour pressure of a solution containing $0.1 \text{ mol}$ of non-volatile solute dissolved in $16.2 \text{ g}$ water (in $\text{ mmHg}$)? ($P_1^0 = 24 \text{ mmHg}$, molar mass of water $18 \text{ g mol}^{-1}$)

The vapour pressure of water at $20\, ^oC$ is $17.5\, mm\, Hg$. If $18\, g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2\, g$ of water at $20\, ^oC$,the vapour pressure of the resulting solution will be .......... $mm\, Hg$.

$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

"Relative lowering in vapour pressure of a solution containing a non-volatile solute is directly proportional to the mole fraction of the solute". The above statement is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module