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Q1.5:
A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g \(mL^{-1}\), then what shall be the molarity of the solution?

Solution :

Step 1: Establishing the Initial Parameters

The label "$10\% \text{ w/w}$ glucose solution" implies that there are $10\text{ g}$ of glucose ($C_6H_{12}O_6$) dissolved in $90\text{ g}$ of water ($H_2O$), making a total solution mass of $100\text{ g}$.

  • Mass of solute (glucose) ($w_2$) = $10\text{ g}$
  • Mass of solvent (water) ($w_1$) = $90\text{ g} = 0.09\text{ kg}$
  • Molar mass of glucose ($M_2$) = $(6 \times 12) + (12 \times 1) + (6 \times 16) = 180\text{ g mol}^{-1}$
  • Molar mass of water ($M_1$) = $(2 \times 1) + 16 = 18\text{ g mol}^{-1}$

Step 2: Calculating Moles of Components

Using the formula $n = \frac{\text{mass}}{\text{molar mass}}$ [By definition of mole]:

  • $n_{\text{glucose}} = \frac{10\text{ g}}{180\text{ g mol}^{-1}} \approx 0.0556\text{ mol}$
  • $n_{\text{water}} = \frac{90\text{ g}}{18\text{ g mol}^{-1}} = 5.00\text{ mol}$

Step 3: Calculating Molality ($m$)

Molality is defined as the number of moles of solute per kilogram of solvent [Per IUPAC standard]:

$m = \frac{n_2}{w_1 \text{ (in kg)}} = \frac{0.0556\text{ mol}}{0.09\text{ kg}} \approx 0.617\text{ mol kg}^{-1}$

Step 4: Calculating Mole Fraction ($x$)

The mole fraction of a component is the ratio of its moles to the total moles in the mixture [Dalton’s Law of Partial Pressures context]:

Total moles ($n_{\text{total}}$) = $0.0556 + 5.00 = 5.0556\text{ mol}$

  • $x_{\text{glucose}} = \frac{n_{\text{glucose}}}{n_{\text{total}}} = \frac{0.0556}{5.0556} \approx 0.011$
  • $x_{\text{water}} = 1 - 0.011 = 0.989$

Step 5: Calculating Molarity ($M$)

Given density ($d$) = $1.2\text{ g mL}^{-1}$. First, determine the volume of the solution:

$V = \frac{\text{Mass of solution}}{\text{Density}} = \frac{100\text{ g}}{1.2\text{ g mL}^{-1}} \approx 83.33\text{ mL} = 0.08333\text{ L}$

Molarity is the moles of solute per liter of solution [By definition of molar concentration]:

$M = \frac{n_{\text{glucose}}}{V \text{ (in L)}} = \frac{0.0556\text{ mol}}{0.08333\text{ L}} \approx 0.667\text{ mol L}^{-1}$

Solution: 100g 10g Glucose 90g Water Density: 1.2 g/mL V = 83.33 mL
ParameterResult
Molality$0.617\text{ m}$
Mole Fraction (Glucose)$0.011$
Mole Fraction (Water)$0.989$
Molarity$0.667\text{ M}$

Final Solution: Molality = $0.617\text{ mol kg}^{-1}$, Mole Fraction (glucose) = $0.011$, Mole Fraction (water) = $0.989$, Molarity = $0.667\text{ mol L}^{-1}$.


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