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?

$x \ g$ of a solute is dissolved in $y \ g$ of two different liquids $A$ and $B$. The relative lowering of vapor pressure of the solution in $A$ is twice that of the solution in $B$. If the moles of solute are negligible compared to the moles of solvent,which of the following statements regarding the molar masses $M_A$ and $M_B$ of the solvents is correct?

Calculate the mole fraction of solute in a solution if the vapour pressure of the solution and the pure solvent are $27 \text{ mm Hg}$ and $30 \text{ mm Hg}$ respectively at $298 \text{ K}$.

The graph between vapour pressure and mole fraction of the solvent for an ideal solution is:

At $25\, ^oC$,benzene and toluene form an ideal solution. What will be the vapor pressure (in $mm \, of \, Hg$) of a solution containing benzene and toluene in a weight ratio of $1:2$? The vapor pressures of pure benzene and pure toluene are $75\, mm \, of \, Hg$ and $30\, mm \, of \, Hg$,respectively.

The vapour pressure of pure benzene at a certain temperature is $0.850 \ bar$. $A$ non-volatile,non-electrolyte solid weighing $0.5 \ g$ is added to $39.0 \ g$ of benzene (molar mass $78 \ g/mol$). The vapour pressure of the solution then is $0.845 \ bar$. What is the molecular mass of the solid substance?