$2.5 \ g$ 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 $= 18 \ g \ mol^{-1}]$

The boiling point of water in a $0.1 \ m$ molal silver nitrate solution (solution $A$) is $x \ ^{\circ}C$. To this solution $A$,an equal volume of $0.1 \ m$ molal aqueous barium chloride solution is added to make a new solution $B$. The difference in the boiling points of water in the two solutions $A$ and $B$ is $y \times 10^{-2} \ ^{\circ}C$. (Assume: Densities of the solutions $A$ and $B$ are the same as that of water and the soluble salts dissociate completely. Use: Molal elevation constant,$K_b = 0.5 \ K \ kg \ mol^{-1}$; Boiling point of pure water as $100 \ ^{\circ}C$.) $(1)$ The value of $x$ is $(2)$ The value of $|y|$ is

Drinking water contains some salt impurities dissolved in it. When this solution is heated in an open container,vapours are formed and separated slowly. In this process,the freezing point and osmotic pressure of the remaining solution will continuously

An aqueous solution of a non-volatile solute boils at $100.15\,^{\circ}C$. If the solution is diluted with an equal volume of water,the freezing point of the resulting solution will be ...... $^{\circ}C$. (Given: $K_b = 0.512\,K\,kg\,mol^{-1}$ and $K_f = 1.86\,K\,kg\,mol^{-1}$ for water)

The number of pairs of solutions having the same value of osmotic pressure from the following is:
(Assume $100\%$ ionization)
$A.$ $0.500 \ M \ C_2H_5OH \ (aq)$ and $0.25 \ M \ KBr \ (aq)$
$B.$ $0.100 \ M \ K_4[Fe(CN)_6] \ (aq)$ and $0.100 \ M \ FeSO_4(NH_4)_2SO_4 \ (aq)$
$C.$ $0.05 \ M \ K_4[Fe(CN)_6] \ (aq)$ and $0.25 \ M \ NaCl \ (aq)$
$D.$ $0.15 \ M \ NaCl \ (aq)$ and $0.1 \ M \ BaCl_2 \ (aq)$
$E.$ $0.02 \ M \ KCl \cdot MgCl_2 \cdot 6H_2O \ (aq)$ and $0.05 \ M \ KCl \ (aq)$

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 $x_L=1$