Question

The concentration of the solution of glucose in water is 10% (w/w). If the density of this solution is 1.20 g mL$^{-1}$, then calculate – 
(i) Molality 
(ii) Molarity 
(iii) Mole fraction of each component in solution 
 

Hint:

Always remember: w/w percentage is based on 100 g of solution. Use density to convert mass into volume when calculating molarity.

Answer 1

Given:
10% (w/w) solution $\implies$ 10 g glucose in 100 g solution.
Mass of solution = 100 g.
Mass of solute (glucose) = 10 g.
Mass of solvent (water) = 90 g = 0.090 kg.
Molar mass of glucose = 180 g/mol.
Density = 1.20 g/mL = 1200 g/L.
(i) Molality:
\[ \text{Moles of glucose} = \frac{10}{180} = 0.0556 \, \text{mol} \] \[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.0556}{0.090} = 0.617 \, \text{mol/kg} \] (ii) Molarity:
\[ \text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density}} = \frac{100}{1200} = 0.0833 \, L \] \[ \text{Molarity} = \frac{0.0556}{0.0833} = 0.667 \, M \] (iii) Mole Fraction:
\[ \text{Moles of water} = \frac{90}{18} = 5.0 \, \text{mol} \] \[ \text{Total moles} = 0.0556 + 5.0 = 5.0556 \] \[ X_{\text{glucose}} = \frac{0.0556}{5.0556} = 0.011, \quad X_{\text{water}} = \frac{5.0}{5.0556} = 0.989 \] Final Answer:
Molality = 0.617 mol/kg
Molarity = 0.667 M
Mole fraction of glucose = 0.011
Mole fraction of water = 0.989