250\,g solution of D-glucose in water contains 10.8\% carbon by weight. The molality of the solution is nearest to:…

🧠 Carbon $\%$ tells you the glucose mass Glucose ($\ce{C6H12O6}$, $M = 180$) has $6 \times 12 = 72\,\text{g}$ of carbon per $180\,\text{g}$, i.e. $40\%$ carbon. So mass of glucose $= \text{mass of carbon} / 0.40$.

🗺️ Four steps Mass of C in $250\,\text{g}$ solution $= 0.108 \times 250 = 27\,\text{g}$. Moles of C $= 27 / 12 = 2.25\,\text{mol}$. Moles of glucose $= 2.25 / 6 = 0.375\,\text{mol}$. Mass of glucose $= 0.375 \times 180 = 67.5\,\text{g}$. Mass of water $= 250 - 67.5 = 182.5\,\text{g} = 0.1825\,\text{kg}$.

Final molality: $m = \frac{0.375}{0.1825} \approx 2.055.$ Nearest option is $2.06$.

$\boxed{\text{Answer: (b)}}$