If the vapor pressure of a solution formed by dissolving $1 \ mol$ of a non-volatile solute in $4 \ mol$ of a solvent is $24.0 \ kPa$,what will be the vapor pressure of the pure solvent in $kPa$?

An aqueous solution of methanol in water has a vapour pressure:

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 120 - 80 \, X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and $B$ at the same temperature are respectively:

At $300 \ K$,the vapour pressure of an ideal solution containing $3 \ mol$ of $A$ and $2 \ mol$ of $B$ is $600 \ torr$. At the same temperature,if $1.5 \ mol$ of $A$ and $0.5 \ mol$ of $C$ (non-volatile) are added to this solution,the vapour pressure of the solution increases by $30 \ torr$. What is the value of $p_B^o$?

Vapour pressure of chloroform $(CHCl_3)$ and dichloromethane $(CH_2Cl_2)$ at $25^\circ C$ are $200 \ mm \ Hg$ and $41.5 \ mm \ Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \ g$ of $CHCl_3$ and $40 \ g$ of $CH_2Cl_2$ at the same temperature will be (Molecular mass of $CHCl_3 = 119.5 \ u$ and molecular mass of $CH_2Cl_2 = 85 \ u$)

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.