$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

The relative lowering of vapor pressure is equal to the mole fraction of the non-volatile solute. This statement was given by whom?

The vapour pressure of an aqueous solution of methanol is:

If two substances $A$ and $B$ have $P_{A}^{\circ} : P_{B}^{\circ} = 1 : 2$ and have mole fraction in solution in the ratio $1 : 2$,then mole fraction of $A$ in vapour phase is:-

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 $CH_2Cl_2 = 85 \ u$)

The vapour pressure of pure benzene at a certain temperature is $200 \ mm \ Hg$. At the same temperature,the vapour pressure of a solution containing $2 \ g$ of a non-volatile,non-electrolyte solid in $78 \ g$ of benzene is $195 \ mm \ Hg$. What is the molecular weight of the solid?