Question

An aqueous solution of glucose is 10% (w/w). What would be the molality and mole fraction of each component in the solution? (Molar mass of glucose = 180 g mol⁻¹)

Hint:

10% w/w means 10 g glucose in 90 g water; use molality = mol solute / kg solvent and mole fraction = component mol / total mol.

Answer 1

Concept: 10% (w/w) glucose means 10 g of glucose is present in 100 g of solution, so the solvent (water) is \(100 - 10 = 90\) g.

Step 1: Moles of each component.
\[ n_{glucose} = \frac{10}{180} = 0.0556\ \text{mol} \]
\[ n_{water} = \frac{90}{18} = 5\ \text{mol} \]

Step 2: Molality.
Molality \(= \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}\). Mass of water \(= 90\ g = 0.090\ kg\).
\[ m = \frac{0.0556}{0.090} = 0.617\ \text{mol kg}^{-1} \]
\[\boxed{m = 0.617\ m}\]

Step 3: Mole fractions.
Total moles \(= 0.0556 + 5 = 5.0556\).
\[ x_{glucose} = \frac{0.0556}{5.0556} = 0.011 \]
\[ x_{water} = \frac{5}{5.0556} = 0.989 \]
Check: \(0.011 + 0.989 = 1.000\).
\[\boxed{x_{glucose} = 0.011,\quad x_{water} = 0.989}\]