b
\(C _{6} H _{12} O _{6} \rightarrow\) Glucose

We know: \(\frac{\text { mass of } C }{\text { mass of glucose }}=\frac{72}{180}\)

Given: \(\% C =10.8=\frac{\text { mass of } C }{\text { mass of solution }} \times 100\)

\(\frac{10.8 \times 250}{100}=\) mass of \(C \Rightarrow\) Mass of \(C =27\,gm\)

\(\therefore\) mass of glucose \(=67.5\,gm\)

\(\therefore\) moles of glucose \(=0.375\,moles\)

Mass of solvent \(=250-67.5\,gm =182.5\,gm\)

\(\therefore\) Molality \(=\frac{0.375}{0.1825}=2.055 \approx 2.06\)