Question

A solution containing 15 g urea (molar mass = 60 g mol\(^{-1}\)) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water. Calculate the mass of glucose present in one litre of its solution.

Hint:

When two solutions have the same osmotic pressure, the number of moles of solute is proportional to the molar mass of the solute.

Answer 1

To determine the mass of glucose in 1 litre of an isotonic solution relative to a solution containing 15 g of urea per litre, we use the provided molar masses: urea (60 g mol\(^{-1}\)) and glucose (180 g mol\(^{-1}\)).

1. Isotonic Solution Principle:
Isotonic solutions exhibit identical osmotic pressures. The osmotic pressure (\( \pi \)) is defined as \( \pi = CRT \). At constant temperature, for isotonic solutions, their molar concentrations must be equal: \( C_{\text{urea}} = C_{\text{glucose}} \).

2. Urea Solution Concentration Calculation:
Given: Urea mass = 15 g, Molar mass of urea = 60 g mol\(^{-1}\), Volume = 1 L.

Calculate moles of urea: \( \text{Moles of urea} = \frac{15 \, \text{g}}{60 \, \text{g mol}^{-1}} = 0.25 \, \text{mol} \).
Calculate concentration of urea: \( C_{\text{urea}} = \frac{0.25 \, \text{mol}}{1 \, \text{L}} = 0.25 \, \text{mol L}^{-1} \).

3. Equating Glucose and Urea Concentrations:
Due to isotonicity, the molar concentration of glucose is equal to that of urea: \( C_{\text{glucose}} = 0.25 \, \text{mol L}^{-1} \).

4. Glucose Mass Calculation:
Given: Molar mass of glucose = 180 g mol\(^{-1}\), Volume = 1 L.

From step 3, moles of glucose = 0.25 mol.
Calculate mass of glucose: \( \text{Mass of glucose} = 0.25 \, \text{mol} \times 180 \, \text{g mol}^{-1} = 45 \, \text{g} \).

Final Result:
The required mass of glucose in one litre of its solution is \( 45 \, \text{g} \).