Chemistry - Concentration Terms Question with Solution | TestHub

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\to \mathrm{Glucose}$

Given: $\%\mathrm{C}=10.8=\frac{\mathrm{mass} of \mathrm{C}}{\mathrm{mass} \mathrm{of} solution}\times 100$

$\frac{10.8\times 250}{100}=$ mass of $\mathrm{C}\Rightarrow$ Mass of $\mathrm{C}=27\mathrm{gm}$

$$\begin{gathered} \frac{\mathrm{mass} of \mathrm{C}}{\mathrm{mass} \mathrm{of} \mathrm{glucose}}=\frac{72}{180} \\ \mathrm{mass} \mathrm{of} \mathrm{glucose} = 27\times \frac{180}{72} \end{gathered}$$

$\therefore$ mass of glucose $=67.5\mathrm{gm}$

$\therefore$ moles of glucose $=\frac{67.5}{180} = 0.375$ moles

Mass of solvent $=250-67.5\mathrm{gm}=182.5\mathrm{gm}$

$\therefore$ Molality $=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solvent} \left(\mathrm{kg}\right)} = \frac{0.375}{0.1825}=2.055\approx 2.06$