At a certain temperature,the mole fraction of a solute is $0.15$ and the vapor pressure of the pure solvent is $120 \, torr$. If the solute is non-volatile,what will be the relative lowering in vapor pressure?

$X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ at different concentrations:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is:

For two volatile liquids $A$ and $B$,if the vapour pressure ratio of $P_{A}^{0} : P_{B}^{0} = 1 : 2$ and the mole fraction ratio in the liquid phase is $X_{A} : X_{B} = 1 : 2$,then find the mole fraction of component $A$ in the vapour state $(Y_{A})$?

The vapour pressure of pure liquid solvent is $0.50 \ atm$. When a non-volatile solute $B$ is added to the solvent,its vapour pressure drops to $0.30 \ atm$. The mole fraction of the solute $B$ is $......$

Among acetone,ether,and ethanol,which one will have the maximum and which one will have the minimum vapor pressure?

$p_A$ and $p_B$ are the vapour pressures of pure liquid components $A$ and $B$ respectively of an ideal binary solution. If $x_A$ represents the mole fraction of component $A,$ the total pressure of the solution will be