Raoult's Law

If ${\mathrm{P}}_{\mathrm{A}}^{°}=200 \mathrm{mm}$ of $\mathrm{Hg},{\mathrm{P}}_{\mathrm{B}}^{°}=100 \mathrm{mm}$ of $\mathrm{Hg}$ and ${\mathrm{X}}_{\mathrm{A}\left(\mathrm{I}\right)}=0.4$ then, find ${\mathrm{X}}_{\mathrm{A}\left(\mathrm{III}\right)}$ ?

Two liquid $\mathrm{A}$ and $\mathrm{B}$ form an ideal solution. The ratio of vapour pressures of pure liquids $\mathrm{A}$ and $\mathrm{B}$ are $3:5$. if the mole- fraction of $\mathrm{A}$ in the liquid solution at equilibrium with vapours, is $0.4$, then the mole fraction of $\mathrm{B}$ in the vapour is

The following graph represents variation of boiling point with composition of liquid and vapours of binary liquid mixture. The graph is plotted at constant pressure, for a solution of mole fraction $0.5$

Which of the following statement(s) is correct.

The vapour pressure of water is $12.3\mathrm{kPa}$ at $300 \mathrm{k}$. Now, calculate the vapour pressure of $1$ molar solution of a solute in it.

How we can prove that Raoult’s law as a special case of Henry’s law?

Raoult's law becomes a special case of Henry's law when ${\mathrm{K}}_{\mathrm{H}}$$=$${\mathrm{p}}^0$. In this equation, ${\mathrm{K}}_{\mathrm{H}}$ is 

Raoult's law becomes a special case of Henry's law when ${\mathrm{K}}_{\mathrm{H}}$ becomes equal to _____.

$\mathrm{A}. {\mathrm{p}}^0 \mathrm{B}. \mathrm{p} \mathrm{C}. {\mathrm{p}}^2$

"Enter the correct answer as $\mathrm{A}, \mathrm{B}$ or $\mathrm{C}$."

Condition at which Raoult's law becomes a special case of Henry's law?

Raoult's law becomes a special case of Henry's law when ${\mathrm{K}}_{\mathrm{H}}$ becomes equal to ${\mathrm{p}}^1$.

At $300 \mathrm{K}$, the vapour pressure of a solution containing one mole of $\mathrm{n}-$hexane and three moles of $\mathrm{n}-$heptane is $550 \mathrm{mm}$ of $\mathrm{Hg}$. At the same temperature, if one more mole of $\mathrm{n}-$heptane is added to this solution, the vapour pressure of the solution increases by $10 \mathrm{mm}$ of $\mathrm{Hg}$. What is the vapour pressure in $\mathrm{mm} \mathrm{Hg}$ of $\mathrm{n}-$heptane in its pure state?

A mixture of toluene and benzene forms a nearly ideal solution. Assume ${\mathrm{P}}_{\mathrm{B}}°$ and ${\mathrm{P}}_{\mathrm{T}}°$ to be the vapor pressures of pure benzene and toluene, respectively. The slope of the line obtained by plotting the total vapor pressure to the mole fraction of benzene is:

In the pressure vs. the mole fraction of benzene curves/lines shown below, the total vapour pressure of an ideal mixture of benzene and toluene will follow the curve/line.

Vapour pressure of benzene is $200 \mathrm{mm} \mathrm{Hg}$. When $2\mathrm{g}$ of non-volatile solute is dissolved in $78\mathrm{g}$ benzene, benzene solution has vapour pressure $\text{195 mm Hg}$. Calculate the molar mass of the solute in ${\mathrm{gmol}}^{-1}$( molar mass of benzene = $78{\mathrm{gmol}}^{-1}$).

Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

Vapour pressure of $\mathrm{A}$ and $\mathrm{B}$ mixture at $80°\mathrm{C}$ is given by $\mathrm{P}(\mathrm{in} \mathrm{mm} \mathrm{Hg})=100+200 {\mathrm{x}}_{\mathrm{A}},$ where ${\mathrm{x}}_{\mathrm{A}}$ is mole fraction of $\mathrm{A}.$ Solution is prepared by mixing $2 \mathrm{mol}$ of $\mathrm{A}$ and $3 \mathrm{mol}$ of $\mathrm{B}$ and if the vapours are removed and condensed into liquid and again brought to the temperature of $80°\mathrm{C},$ If the ratio of mole fraction of $\mathrm{A}$ in vapour state and mole fraction of $\mathrm{B}$ in vapour state above the condensate is $\mathrm{x}: 1,$ find value of $\text{'}\mathrm{x}\text{'}$ is -

At $40°\mathrm{C},$ the vapour pressure (in torr) of methyl alcohol (A) and ethyl alcohol solution is represented by: $\mathrm{P}=120{\mathrm{X}}_{\mathrm{A}}+138;$ where ${\mathrm{X}}_{\mathrm{A}}$ is mole fraction of
methyl alcohol. The value of ${\mathrm{P}}_{\mathrm{B}}°$ at ${\mathrm{limX}}_{\mathrm{A}}⟶0$ and ${\mathrm{P}}_{\mathrm{A}}°$ at ${\mathrm{limX}}_{\mathrm{B}}⟶0$ are :

The mass of a non-volatile solute of molar mass $40 \mathrm{g} {\mathrm{mol}}^{-1}$ that should be dissolved in $114 \mathrm{g}$ of octane to lower its vapour pressure by $20\%$ is :

The vapour pressure of benzene at $80°\mathrm{C}$ is lowered by $10 \mathrm{mm}$ by dissolving $2 \mathrm{g}$ of a non-volatile substance in $78 \mathrm{g}$ of benzene. The vapour pressure of pure benzene at $80°\mathrm{C}$ is $750 \mathrm{mm}$. The molecular mass of the substance will be :

Which condition is not satisfied by an ideal solution?

At $40°\mathrm{C}$, vapour pressure in torr of methanol and ethanol solution is $\mathrm{P}=119\mathrm{X}+135:$ where $\mathrm{X}$ is the mole fraction of methanol. Hence