Raoult's Law

Two liquids $\mathrm{A}$ and $\mathrm{B}$ are miscible over the whole range of composition and may be treated as ideal (obeying Raoult's law.) At $350 \mathrm{K}$ the vapour pressure of pure $\mathrm{A}$ is $24.0 \mathrm{kPa}$ and of pure $\mathrm{B}$ is $12.0 \mathrm{kPa}.$ $\mathrm{A}$ mixture of $60\%$ A and $40\%$ $\mathrm{B}$ is distilled at this temperature. A small amount of the distillate is collected and and redistilled at $350 \mathrm{K} ;$ what is the mole percent of $\mathrm{B}$ in the second distillate ?

The vapor pressure of chlorobenzene and water at different temperatures are

At what temperature will $\mathrm{\varphi } \mathrm{Cl}$ steam-distillation under a total pressure of $800 \mathrm{mm} \mathrm{Hg}$?

Assuming the formation of an ideal solution, determine the boiling point of a mixture containing $1560 \mathrm{g}$ benzene (molar mass $=78$ ) and chlorobenzene (molar mass $=112.5$ ) using the following against an external pressure of $1000$ Torr.

Consider two liquids $\mathrm{A} \& \mathrm{B}$ having pure vapour pressures ${\mathrm{P}}_{\mathrm{A}}°\&{\mathrm{P}}_{\mathrm{B}}°$ forming an ideal solution. The plot of $\frac{1}{{\mathrm{X}}_{\mathrm{A}}}\mathrm{v}/\mathrm{s}\frac{1}{{\mathrm{Y}}_{\mathrm{A}}}$ (where ${\mathrm{X}}_{\mathrm{A}}$ and ${\mathrm{Y}}_{\mathrm{A}}$ are the mole fraction of liquid $\mathrm{A}$ in liquid and vapour phase respectively) is linear with slope and $\mathrm{Y}$ intercepts respectively:

A solution is prepared by mixing $8.5 \mathrm{g}$ of ${\mathrm{CH}}_2{\mathrm{Cl}}_2$ and $11.95 \mathrm{g}$ of ${\mathrm{CHCl}}_3.$ If vapour pressure of ${\mathrm{CH}}_2{\mathrm{Cl}}_2$ and ${\mathrm{CHCl}}_3$ at $298 \mathrm{K}$ are $415$ and $200 \mathrm{mm} \mathrm{Hg}$ respectively, the mole fraction of ${\mathrm{CHCl}}_3$ in vapour form is: 

The relative lowering of vapour pressure is equal to the mole fraction of the solute. This is the statement of:

For a dilute solution containing $2.5 \mathrm{g}.$of a non- electrolyte solute. in $100 \mathrm{g}$ of water, the elevation in boiling point at $1$ atm pressure pressure is $2°\mathrm{C}.$ Assuming concentration of  solute is much lower than the concentration of solvent, the vapour pressure $(\mathrm{mm}$ of $\mathrm{Hg}$ )  of the solution is (take ${\mathrm{K}}_{\mathrm{b}}=0.76 \mathrm{K} \mathrm{kg}$ ${\mathrm{mol}}^{-1} )$

At $323 \mathrm{K}$, the vapour pressure in millimeters of mercury of a methanol-ethanol solution is represented by the equation, $\mathrm{p}=120{\mathrm{X}}_{\mathrm{A}}+140$, where ${\mathrm{X}}_{\mathrm{A}}$ is the mole fraction of methanol. Then the value of $\underset{{\mathrm{x}}_{\mathrm{A}}\to 1}{\lim }\left(\frac{\mathrm{p}}{{\mathrm{X}}_{\mathrm{A}}}\right)$ is:

If ${\mathrm{P}}_0$ and $\mathrm{P}$ are the vapour pressures of a solvent and its solution, respectively, and ${\mathrm{N}}_1$ and ${\mathrm{N}}_2$ are the mole fractions of the solvent and non-volatile solute, respectively, then the correct relation is:

The vapour pressure of pure water at$20°\mathrm{C} \mathrm{is} 17.5 \mathrm{mm} \mathrm{Hg}$. If $18 \mathrm{g}$ of glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ is added to$178.2 \mathrm{g}$of water at $20° \mathrm{C}$, the vapour pressure of resulting solution will be

A very small amount of a non-volatile solute (non-associative, non-dissociative) is dissolved in $100 {\mathrm{cm}}^3$ of a solvent. At room temperature, vapour pressure of this solution is $98.7 \mathrm{mm}$ of $\mathrm{Hg}$ while that of pure solvent is $100 \mathrm{mm}$ of $\mathrm{Hg}$. If the freezing temperature of this solution is $0.72 \mathrm{K}$ lower than that of pure solvent, what is the value of cryoscopic constant of solvent (in $\mathrm{K} \mathrm{Kg}/\mathrm{mol}$) ? Round off your answer to the nearest whole number. Report your answer as $0$ (zero) if you find data insufficient. Given: Molar mass of solvent $= 78 \mathrm{g}/\mathrm{mol}$.

Following is false when in a volatile solvent $\mathrm{A}$ and a non volatile solute $\mathrm{B}$ is mixed (where symbols have their usual meaning):

The vapour pressure of two pure isomeric liquids $\mathrm{X}$ and $\mathrm{Y}$ are $200 \mathrm{torr}$ and $100 \mathrm{torr}$, respectively, at a given temperature. Assuming a solution of these components to obey Raoult’s law. The mole fraction of component $\mathrm{X}$ in the vapour phase in equilibrium with the solution containing equal amounts of $\mathrm{X}$ and $\mathrm{Y}$, at the same temperature, is

Mole fraction of the toluene in the vapor phase which is in equilibrium with a solution of benzene $(\mathrm{p}° = 120 \mathrm{Torr})$ and toluene $(\mathrm{p}° = 80 \mathrm{Torr})$ having $2.0 \mathrm{mol}$ of each is:

The vapor pressure of a pure liquid $\mathrm{A}$ is $40 \mathrm{mmHg} \mathrm{at} 310 \mathrm{K}$. The vapor pressure of this liquid in a solution with liquid $\mathrm{B}$ is $32 \mathrm{mmHg}$. Mole fraction of $\mathrm{A}$ in the solution, if it obeys Raoult’s law is:

A solution at $20°\mathrm{C}$ is composed of $1.5 \mathrm{mol}$ of benzene and $3.5 \mathrm{mol}$ of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are $74.7$ torr and $22.3$ torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively:

On mixing heptane and octane form an ideal solution at $373 \mathrm{K}$, the vapour pressures of the two liquid components (heptane and octane) are $105 \mathrm{kPa}$ and $45 \mathrm{kPa}$, respectively. The vapour pressure of the solution obtained by mixing $25.0 \mathrm{g}$ of heptane and $35 \mathrm{g}$ of octane will be (molar mass of heptane$=100 \mathrm{g} {\mathrm{mol}}^{–1}$ and of octane$=114 \mathrm{g} {\mathrm{mol}}^{–1}$)

Two liquids $\mathrm{X}$ and $\mathrm{Y}$ form an ideal solution. At $300 \mathrm{K}$, the vapour pressure of the solution containing $1 \mathrm{mol}$ of $\mathrm{X}$ and $3 \mathrm{mols}$ of $\mathrm{Y}$ is $550 \mathrm{mm} \mathrm{Hg}$. At the same temperature, if $1 \mathrm{mol}$ of $\mathrm{Y}$ is further added to the solution, the vapour pressure of the solution increases by $10 \mathrm{mm} \mathrm{Hg}$. Vapour pressure (in $\mathrm{mm} \mathrm{Hg}$) of $\mathrm{X}$ and $\mathrm{Y}$ in their pure states will be, respectively

At $80°\mathrm{C}$, the vapour pressure of a pure liquid $\text{'}\mathrm{A}\text{'}$ is $520 \mathrm{mm} \mathrm{Hg}$ and that of a pure liquid $\text{'}\mathrm{B}\text{'}$ is $1000 \mathrm{mm} \mathrm{Hg}$. If a mixture solution of $\text{'}\mathrm{A}\text{'}$ and $\text{'}\mathrm{B}\text{'}$ boils at $80°\mathrm{C}$, at $1 \mathrm{atm}$ pressure, the amount of $\text{'}\mathrm{A}\text{'}$ in the mixture is $(1 \mathrm{atm}=760 \mathrm{mm} \mathrm{Hg})$

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $290 \mathrm{mm}$ at $300 \mathrm{K}$. The vapour pressure of propyl alcohol is $200 \mathrm{mm}$. If the mole fraction of ethyl alcohol is $0.6$, its vapour pressure (in $\mathrm{mm}$) at the same temperature will be