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 ........... $(1 \ atm = 760 \ mm \ Hg)$.

At $T(K)$,the vapour pressure of pure benzene (molar mass $= 78 \ g \ mol^{-1}$) is $0.85 \ bar$. When $2.0 \ g$ of a non-volatile,non-electrolyte solute is added to $39 \ g$ of benzene,the vapour pressure of the solution at $T(K)$ is $0.83 \ bar$. The elevation in boiling point (in $K$) of the same solution is: ($K_b$ of benzene is $2.6 \ K \ kg \ mol^{-1}$)

If a $1 \ m$ solution of benzoic acid in benzene has a freezing point depression of $2.56 \ ^{\circ}C$ $(K_f = 5.12 \ ^{\circ}C \ kg \ mol^{-1})$ and a boiling point elevation of $2.53 \ ^{\circ}C$ $(K_b = 2.53 \ ^{\circ}C \ kg \ mol^{-1})$,select the correct statement$(s)$:
Statement $I$: There is dimer formation when undergoing freezing.
Statement $II$: There is no change when undergoing boiling.
Statement $III$: Reverse of $I$ and $II$.
Statement $IV$: Dimer formation in freezing and boiling state.

At $300 \ K$,an ideal solution is formed by mixing $460 \ g$ of toluene with $390 \ g$ of benzene. If the vapour pressure of pure toluene and pure benzene at $300 \ K$ are $32 \ mm$ and $40 \ mm$ respectively,the mole fraction of toluene in the vapour phase is:

$1$ molal $K_{4}[Fe(CN)_{6}]$ solution has a degree of dissociation of $0.4$. Its boiling point is equal to that of another solution which contains $18.1$ weight percent of a non-electrolytic solute $A$. The molar mass of $A$ is $.......\, u$. (Round off to the Nearest Integer). [Density of water $= 1.0\, g\, cm^{-3}$]

$1 \, \text{mole}$ 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: