The correct statement(s) regarding sugars is(are) :

At a given temperature, 0.45 g of acetic acid in 50 mL of water is shaken with 1.0 g of charcoal and the pH of the resulting solution is 3.0 . Assume, the adsorption of acetic acid from the aqueous solution by charcoal follows Freundlich isotherm,

$$ \frac{x}{m}=k C^{1 / n} $$

If the plot of $\log _{10}(x / m)$ against $\log _{10} C$ gives a straight line with slope 1 , the value of $k$ in $\mathrm{L} \mathrm{mol}^{-1}$ is $\_\_\_\_$ .

Given: The molar mass of acetic acid is $60 \mathrm{~g} \mathrm{~mol}^{-1}$.

The acid dissociation constant of acetic acid is $1.0 \times 10^{-5}$ at the given temperature.

$x$ is the mass (in grams) of acetic acid adsorbed.

$m$ is the mass (in grams) of charcoal.

$C$ is the equilibrium concentration of acetic acid in the solution after the adsorption is complete.

$k$ and $n$ are constants for acetic acid-charcoal system at the given temperature.

In a solvent $\mathbf{S}$, a compound $\mathbf{B}$ is partially dissociated into $\mathbf{C}$ and $\mathbf{D}$ as given below :

$$ \mathbf{B} \rightleftharpoons 2 \mathbf{C}+2 \mathbf{D} $$

$\mathbf{B}, \mathbf{C}$ and $\mathbf{D}$ are non-volatile in nature. The molar mass of $\mathbf{B}$ is 10 times the molar mass of $\mathbf{S}$. The standard boiling point and the standard enthalpy of vaporization of $\mathbf{S}$ are 400 K and $10 R \mathrm{~J} \mathrm{~mol}^{-1}$, respectively ( $R$ is the gas constant in $\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ). A solution of $\mathbf{B}$ in $\mathbf{S}$ with an initial concentration of $\mathbf{B}$ as $0.25 \%$ (mass/mass) has a boiling point of 408 K at 1 bar pressure. In this solution, the mole percent of $\mathbf{B}$ that has been dissociated is $\_\_\_\_$ .