When a solid solute is dissolved in a volatile solvent to form a solution,then .......

$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:

Assertion : If a liquid solute more volatile than the solvent is added to the solvent,the vapour pressure of the solution may increase i.e.,$p_s > p^o$.
Reason : In the presence of a more volatile liquid solute,only the solute will form the vapours and solvent will not.

If $P^o$ and $P$ are the vapour pressure of a solvent and its solution respectively and $N_1$ and $N_2$ are the mole fractions of the solvent and solute respectively,then the correct relation is:

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

Which solution has the highest vapour pressure?