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

Framing: Fix a convenient sample of exactly 100 g of solution so the percentages become masses directly.

Step 1: Assign masses.
In 100 g solution: glucose \(= 10\) g, water \(= 90\) g.

Step 2: Convert to moles.
Glucose: \(10 \div 180 = 0.0556\) mol. Water: \(90 \div 18 = 5.0\) mol.

Step 3: Molality.
Molality counts moles of solute per kilogram of solvent. With \(0.090\) kg water:
\(m = 0.0556 / 0.090 = 0.617\) mol/kg.
\(\boxed{0.617\ m}\)

Step 4: Mole fraction of water first.
\(x_{water} = 5.0 / (5.0 + 0.0556) = 5.0 / 5.0556 = 0.989\).
Since the mole fractions add to 1, \(x_{glucose} = 1 - 0.989 = 0.011\).
\(\boxed{x_{water} = 0.989,\ x_{glucose} = 0.011}\)

Both routes give the same molality and mole fractions, confirming the answer.