An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$

$A$ solution is formed by the combination of two liquids such as dichloromethane and chloroform. The partial pressures of dichloromethane and chloroform in the solution are $285.5 \ mm \ Hg$ and $62.4 \ mm \ Hg$ respectively. What is the total pressure of the solution (in $mm \ Hg$)?

$1 \, \text{mole}$ 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:

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 amount of solute is much lower than the amount of solvent,the vapour pressure $(mm \ Hg)$ of the solution is: (take $K_b = 0.76 \ K \ kg \ mol^{-1}$)

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 $HA = 60$,$d_{solution} = 1 \, g/mL$]

Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.
Reason : Raoult's law is a special case of Henry's law.