What is the vapour pressure of a solution containing $0.1 \ mol$ of non-volatile solute dissolved in $16.2 \ g$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

The mass of glucose that should be dissolved in $100 \ g$ of water in order to produce the same relative lowering of vapour pressure as is produced by dissolving $1 \ g$ of urea (molar mass $= 60 \ g/mol$) in $50 \ g$ of water is ........... $g$. (Assume dilute solution in both cases):-

When a non-volatile solute is dissolved in a solvent,the relative lowering of vapour pressure is equal to

Vapour pressure of a solution of $5 \ g$ of non-electrolyte in $100 \ g$ of water at a particular temperature is $2985 \ N/m^2$. The vapour pressure of pure water is $3000 \ N/m^2$. The molecular weight of the solute is

Vapour pressure of a pure solvent is $550 \ mm$ of $Hg$. By addition of a non-volatile solute,it decreases to $510 \ mm$ of $Hg$. Calculate the mole fraction of the solute in the solution.

At $T$ $(K)$,the vapour pressure of pure benzene is $0.85$ bar. $A$ non-volatile,non-electrolyte substance weighing $0.5$ $g$ when added to $39$ $g$ of benzene,the vapour pressure of the solution is $0.845$ bar. The molar mass (in $g$ $mol^{-1}$) of the substance is