Question 16 of 50intermediate🔧 ApplyNumerical3 marks

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?

Correct Answer

For a 10% w/w solution, we consider 10 g of glucose dissolved in 90 g of water. The molar mass of glucose (C6H12O6C_6H_{12}O_6) is 180 g mol−1^{-1} and water (H2OH_2O) is 18 g mol−1^{-1}. Moles of glucose = 10/180=0.055510 / 180 = 0.0555 mol. Moles of water = 90/18=5.090 / 18 = 5.0 mol.

Molality is calculated as moles of solute per kilogram of solvent. Molality = 0.0555 mol/0.09 kg=0.617 m0.0555 \text{ mol} / 0.09 \text{ kg} = 0.617 \text{ m}. Mole fraction of glucose (xglucosex_{glucose}) = 0.0555/(0.0555+5.0)=0.0110.0555 / (0.0555 + 5.0) = 0.011. Mole fraction of water (xwaterx_{water}) = 1−0.011=0.9891 - 0.011 = 0.989.

To calculate Molarity, we determine the volume of the solution using the density of 1.2 g mL−1^{-1}. Volume = 100 g/1.2 g mL−1=83.33 mL=0.08333 L100 \text{ g} / 1.2 \text{ g mL}^{-1} = 83.33 \text{ mL} = 0.08333 \text{ L}. Molarity = 0.0555 mol/0.08333 L=0.67 M0.0555 \text{ mol} / 0.08333 \text{ L} = 0.67 \text{ M}.

Exercise: EXERCISES | Q: 1.5 | (Page 27)
For More Understanding

Explanation

The solution follows the standard method for concentration calculations as illustrated in the provided context. The textbook explicitly states that a 10% solution by mass contains 10 g of solute in 90 g of solvent. The mole fraction calculation mirrors the method in Example 1.1, and the molar masses are derived from the text (180 for glucose, 18 for water).

Solution Steps

  1. Step 1: Define the composition based on percentage (10 g glucose, 90 g water) and calculate moles. Moles of glucose = 0.0555 mol; Moles of water = 5.0 mol.

  2. Step 2: Calculate Molality using mass of solvent in kg. Molality = 0.0555 mol / 0.09 kg = 0.617 m.

  3. Step 3: Calculate Mole Fractions. x_glucose = 0.011; x_water = 0.989.

  4. Step 4: Calculate Molarity using density to find volume. Volume = 83.33 mL. Molarity = 0.0555 mol / 0.08333 L = 0.67 M.

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