$1 \, \text{mole}$ of liquid $A$ and $2 \, \text{moles}$ of liquid $B$ make a solution having an observed vapour pressure of $42 \, \text{torr}$. The vapour pressures of pure $A$ and pure $B$ are $45 \, \text{torr}$ and $36 \, \text{torr}$ respectively. The described solution:

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

Two elements $A$ and $B$ form compounds having formula $AB_{2}$ and $AB_{4}$. When dissolved in $20 \ g$ of benzene $(C_{6}H_{6})$,$1 \ g$ of $AB_{2}$ lowers the freezing point by $2.3 \ K$ whereas $1.0 \ g$ of $AB_{4}$ lowers it by $1.3 \ K$. The molar depression constant for benzene is $5.1 \ K \ kg \ mol^{-1}$. Calculate atomic masses of $A$ and $B$.

At $80^o C$,the vapor pressure of pure liquid $A$ is $520 \ mm \ Hg$ and that of pure liquid $B$ is $1000 \ mm \ Hg$. If a mixture of $A$ and $B$ boils at $80^o C$ and $1 \ atm$ pressure,the mole percentage of $A$ in the mixture is ........... $(1 \ atm = 760 \ mm \ Hg)$.

Elevation in the boiling point for $1 \ m$ solution of glucose is $2 \ K$. The depression in the freezing point for $2 \ m$ solution of glucose in the same solvent is $2 \ K$. The relation between $K_b$ and $K_f$ is

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