Solid Lead nitrate is dissolved in $1 \, L$ of water. The solution was found to boil at $100.15^{\circ}C$. When $0.2 \, mol$ of $NaCl$ is added to the resulting solution,it was observed that the solution froze at $-0.8^{\circ}C$. The solubility product of $PbCl_2$ formed is $........... \times 10^{-6}$ at $298 \, K$. (Nearest integer) Given: $K_b = 0.5 \, K \, kg \, mol^{-1}$ and $K_f = 1.8 \, K \, kg \, mol^{-1}$. Assume molality to be equal to molarity in all cases.

$1 \, \text{mole}$ of each of $A$ and $B$ form an ideal solution of vapour pressure $100 \, \text{mm Hg}$. Addition of $2 \, \text{moles}$ of $B$ to it decreases the vapour pressure by $20 \, \text{mm Hg}$. The vapour pressure of $A$ and $B$ in pure state are,respectively:

At $35^{\circ} C$,the vapour pressure of $CS_{2}$ is $512 \; mm \; Hg$ and that of acetone is $344 \; mm \; Hg$. $A$ solution of $CS_{2}$ in acetone has a total vapour pressure of $600 \; mm \; Hg$. The false statement amongst the following is

The specific heat of a certain substance is $0.86 \,J \,g^{-1} \,K^{-1}$. Assuming ideal solution behaviour,the energy required (in $J$) to heat $10 \,g$ of $1 \,molal$ of its aqueous solution from $300 \,K$ to $310 \,K$ is closest to $.... \,J$
[Given: Molar mass of the substance $= 58 \,g \,mol^{-1}$; specific heat of water $= 4.2 \,J \,g^{-1} \,K^{-1}$]

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 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 $x_L=1$