250 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)

$\begin{array}{l}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\to \text{ Glucose }\\ \text{ We know: }\frac{\text{ mass of }\mathrm{C}}{\text{ mass of glucose }}=\frac{72}{180}\\ \text{ Given: }\mathrm{\%}\mathrm{C}=10.8=\frac{\text{ mass of }\mathrm{C}}{\text{ mass of solution }}\times 100\\ \frac{10.8\times 250}{100}=\text{ mass of }\mathrm{C}\Rightarrow \text{ Mass of }\mathrm{C}=27\mathrm{gm}\\ \therefore \text{ mass of glucose }=67.5\mathrm{gm}\\ \therefore \text{ moles of glucose }=0.375\text{ moles }\\ \text{ Mass of solvent }=250-67.5\mathrm{gm}=182.5\mathrm{gm}\\ \therefore \text{ Molality }=\frac{0.375}{0.1825}=2.055\approx 2.06\end{array}$