At $T \ K$,the vapour pressure of pure benzene $(C_6H_6)$ and toluene $(C_7H_8)$ are $75 \ mm \ Hg$ and $22 \ mm \ Hg$ respectively. $23.4 \ g$ of benzene and $64.4 \ g$ of toluene are mixed to form an ideal solution. If the vapours are in equilibrium with the liquid mixture,the mole fraction of toluene in the vapour phase is (Atomic weight: $C=12, H=1$).

An aqueous solution contains $6\%$ and $4\%$ by weight of a weak acid $(HA)$ and urea respectively. If the depression in freezing point is $\frac{13.6}{3} \, ^{\circ}C$,calculate the $K_a$ of $HA$. [Given: $K_f = 1.8 \, K \, kg/mol$,Molecular weight of $HA = 60$,$d_{solution} = 1 \, g/mL$]

When a solute is added to a solvent,the freezing point of the solution decreases to $1.86 \ K$. What is the value of $\Delta T_b$? $[K_f = 1.86, K_b = 0.512]$

Elevation in boiling point for $1.5 \, m$ solution of glucose in water is $4 \, K$. The depression in freezing point for $4.5 \, m$ solution of glucose in water is $4 \, K$. The ratio of molal elevation constant to molal depression constant $(K_{b} / K_{f})$ is $....$

$W \ g$ 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 $= 373 \ K$):

For a dilute solution containing $2.5 \ g$ of a non-volatile non-electrolyte solute in $100 \ g$ of water,the elevation in boiling point at $1 \ atm$ pressure is $2^{\circ} C$. Assuming the concentration of solute is much lower than the concentration of solvent,the vapour pressure ($mm$ of $Hg$) of the solution is (take $K_{b}=0.76 \ K \ kg \ mol^{-1}$)