Given below are two statements.
Statement-$I$: Liquids $A$ and $B$ form a non-ideal solution with negative deviation. The interactions between $A$ and $B$ are weaker than $A-A$ and $B-B$ interactions.
Statement-$II$: In reverse osmosis,the applied pressure must be higher than the osmotic pressure of solution.
The correct answer is

For a dilute solution containing $2.5 \ g$ of a non-volatile non-electrolyte solute in $100 \ g$ of water,the elevation in boiling point at $1 \ atm$ pressure is $2 \ ^\circ C$. Assuming the amount of solute is much lower than the amount of solvent,the vapour pressure $(mm \ Hg)$ of the solution is: (take $K_b = 0.76 \ K \ kg \ mol^{-1}$)

For a solution formed by mixing liquids $L$ and $M$,the vapour pressure of $L$ plotted against the mole fraction of $M$ in solution is shown in the following figure. Here $x_L$ and $x_M$ represent mole fractions of $L$ and $M$,respectively,in the solution. The correct statement$(s)$ applicable to this system is(are)
$A$. Attractive intermolecular interactions between $L-L$ in pure liquid $L$ and $M-M$ in pure liquid $M$ are stronger than those between $L-M$ when mixed in solution
$B$. The point $Z$ represents vapour pressure of pure liquid $M$ and Raoult's law is obeyed when $x_L \rightarrow 0$
$C$. The point $Z$ represents vapour pressure of pure liquid $L$ and Raoult's law is obeyed when $x_L \rightarrow 1$
$D$. The point $Z$ represents vapour pressure of pure liquid $M$ and Raoult's law is obeyed from $x_L=0$ to $x_L=1$

$2.5 \ g$ of a non-volatile,non-electrolyte is dissolved in $100 \ g$ of water at $25^{\circ} C$. The solution showed a boiling point elevation by $2^{\circ} C$. Assuming the solute concentration is negligible with respect to the solvent concentration,the vapour pressure of the resulting aqueous solution is . . . . . . $mm$ of $Hg$ (nearest integer).
[Given : Molal boiling point elevation constant of water $(K_b) = 0.52 \ K \ kg \ mol^{-1}$,
$1 \ atm$ pressure $= 760 \ mm$ of $Hg$,molar mass of water $= 18 \ g \ mol^{-1}]$

$A$ concentrated solution of copper sulphate,which is dark blue in colour,is mixed at room temperature with a dilute solution of copper sulphate,which is light blue. For this process:

$A$ solution has a $1:4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20 \ ^oC$ are $440 \ mm \ Hg$ for pentane and $120 \ mm \ Hg$ for hexane. The mole fraction of pentane in the vapour phase would be