"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

The vapour pressure of $30 \%$ $(w/v)$ aqueous solution of glucose is $...... \ mm \ Hg$ at $25^{\circ} \ C$. [Given : The density of $30 \%$ $(w/v)$ aqueous solution of glucose is $1.2 \ g \ cm^{-3}$ and vapour pressure of pure water is $24 \ mm \ Hg$.] (Molar mass of glucose is $180 \ g \ mol^{-1}$.)

The vapour pressure of pure benzene and methyl benzene at $27^{\circ} C$ is given as $80 \ Torr$ and $24 \ Torr$,respectively. The mole fraction of methyl benzene in vapour phase,in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is. . . . . . .$\times 10^{-2}$ (nearest integer)

What is the relationship between vapor pressure and temperature? What will be the vapor pressure at $373 \, K$?

Calculate the amount of solute dissolved in $612 \ g$ of water at $30^{\circ} C$ if molar mass of solute is $342 \ g \ mol^{-1}$ (Relative vapour pressure lowering is $0.025$ and molar mass of water $18 \ g \ mol^{-1}$ ). (in $g$)

If $x \ g$ of a non-volatile,non-electrolytic solute is dissolved in $114 \ g$ of octane to reduce its vapour pressure to $80 \%$ of that of pure octane,then $x$ is ........... $g$. (Molar mass of solute $= 40 \ g/mol$)