The only incorrect statement regarding Henry's law for a gas dissolving in water is

The values of $K_b$ and $K_f$ for water are $0.52 \, K \, kg \, mol^{-1}$ and $1.86 \, K \, kg \, mol^{-1}$ respectively. If the solution boils at $0.78 \, K$ above the boiling point of water,then the freezing point of the solution will be ........ $K$.

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 $K_{b}=0.76 \ K \ kg \ mol^{-1}$)

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.

An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:
Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.

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