$$250 \text{ g}$$ solution of D-glucose in water contains $$10.8\%$$ of carbon by weight. The molality of the solution is nearest to (Given: Atomic Weights are $$H = 1u; C = 12u; O = 16u$$)

We have a $$250 \text{ g}$$ solution of D-glucose in water containing $$10.8\%$$ carbon by weight. We need to find the molality.

The mass of carbon present is given by the percentage times the total mass: $$ \text{Mass of carbon} = 10.8\% \times 250 = \frac{10.8}{100} \times 250 = 27 \text{ g} $$.

The molecular formula of D-glucose is $$C_6H_{12}O_6$$, and its molecular weight is calculated as $$6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ g/mol}$$. Since there are 6 carbon atoms in one mole of glucose, the mass of carbon per mole is $$6 \times 12 = 72 \text{ g}$$, giving a mass fraction of carbon in glucose of $$\dfrac{72}{180} = \dfrac{2}{5}$$. Therefore, the mass of glucose in the solution is found by dividing the mass of carbon by this fraction: $$ \text{Mass of glucose} = \frac{\text{Mass of carbon}}{\text{Mass fraction of C}} = \frac{27}{2/5} = 27 \times \frac{5}{2} = 67.5 \text{ g} $$.

The mass of the solvent (water) is the difference between the total mass and the mass of glucose: $$ \text{Mass of water} = 250 - 67.5 = 182.5 \text{ g} = 0.1825 \text{ kg} $$.

The number of moles of glucose present is calculated as $$ n = \frac{67.5}{180} = 0.375 \text{ mol} $$.

The molality of the solution is then the moles of solute per kilogram of solvent: $$ m = \frac{n}{\text{mass of solvent in kg}} = \frac{0.375}{0.1825} = 2.055 \approx 2.06 $$.

The molality is nearest to $$2.06$$. The correct answer is Option B: $$2.06$$.