$25 \ mL$ of an aqueous solution of $KCl$ was found to require $20 \ mL$ of $1 \ M \ AgNO_3$ solution when titrated using $K_2CrO_4$ as an indicator. What is the depression in freezing point of $KCl$ solution of the given concentration? (Nearest integer). Given: $K_f = 2.0 \ K \ kg \ mol^{-1}$. Assume: $(1)$ $100 \%$ ionization and $(2)$ density of the aqueous solution as $1 \ g \ mL^{-1}$.

Consider the following aqueous solutions.
$I$. $2.2 \ g$ Glucose in $125 \ mL$ of solution.
$II$. $1.9 \ g$ Calcium chloride in $250 \ mL$ of solution.
$III$. $9.0 \ g$ Urea in $500 \ mL$ of solution.
$IV$. $20.5 \ g$ Aluminium sulphate in $750 \ mL$ of solution.
The correct increasing order of boiling point of these solutions will be:
[Given: Molar mass in $g \ mol^{-1}$: $H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27, S=32$]

$P$ and $Q$ combine to form two compounds $PQ_2$ and $PQ_3$. If $1 \ g$ of $PQ_2$ is dissolved in $51 \ g$ of benzene, the depression of freezing point is $0.8^{\circ} C$. If $1 \ g$ of $PQ_3$ is dissolved in $51 \ g$ of benzene, the depression of freezing point is $0.625^{\circ} C$. Given $K_f$ of benzene $= 5.1 \ K \ kg \ mol^{-1}$, calculate the atomic masses of $P$ and $Q$.

The osmotic pressure of $0.1 \ M$ monobasic acid of $pH \ 3$ at $27^{\circ} C$ is (in $atm$)

$W \ g$ of a non-volatile electrolyte solid solute of molar mass $M \ g \ mol^{-1}$ when dissolved in $100 \ mL$ water, decreases vapor pressure of water from $640 \ mm \ Hg$ to $600 \ mm \ Hg$. If aqueous solution of the electrolyte boils at $375 \ K$ and $K_b$ for water is $0.52 \ K \ kg \ mol^{-1}$, then the mole fraction of the electrolyte solute $(X_2)$ in the solution can be expressed as (Given density of water $= 1 \ g/mL$ and boiling point of water $= 373 \ K$):

$1 \, \text{mole}$ 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: