The vapour pressure of a pure liquid $A$ is $70 \ torr$ at $300 \ K$. It forms an ideal solution with another liquid $B$. The mole fraction of $B$ is $0.2$ and the total vapour pressure of the solution is $84 \ torr$ at the same temperature. The vapour pressure of pure liquid $B$ (in $torr$) is:

If vapour pressure of pure solvent and solution are $240 \ mm Hg$ and $216 \ mm Hg$ respectively,then the mole fraction of solvent in the solution is:

According to Raoult's law for a non-volatile solute,which of the following is correct?

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?

Lowering of vapour pressure due to a solute in $1 \, molal$ aqueous solution at $100 \, ^\circ C$ is ........ $torr$.

Two $5 \ molal$ solutions are prepared by dissolving a non-electrolyte,non-volatile solute separately in the solvents $X$ and $Y$. The molecular weights of the solvents are $M_X$ and $M_Y$,respectively,where $M_X = \frac{3}{4} M_Y$. The relative lowering of vapour pressure of the solution in $X$ is $m$ times that of the solution in $Y$. Given that the number of moles of solute is very small in comparison to that of solvent,the value of $m$ is