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 solve the problem, we need to calculate the mass of glucose in one litre of its solution that is isotonic with a solution containing 15 g of urea per litre, given the molar masses of urea (60 g mol\(^{-1}\)) and glucose (180 g mol\(^{-1}\)).

1. Understand Isotonic Solutions:
Isotonic solutions have the same osmotic pressure. Osmotic pressure (\( \pi \)) is given by \( \pi = CRT \), where \( C \) is the molar concentration, \( R \) is the gas constant, and \( T \) is the temperature. For isotonic solutions at the same temperature, \( C_{\text{urea}} = C_{\text{glucose}} \).

2. Calculate Molar Concentration of Urea:
Mass of urea = 15 g, molar mass = 60 g mol\(^{-1}\), volume = 1 L.

Moles of urea = \( \frac{15}{60} = 0.25 \, \text{mol} \).
Concentration of urea = \( \frac{0.25}{1} = 0.25 \, \text{mol L}^{-1} \).

3. Set Concentration of Glucose Equal to Urea:
Since the solutions are isotonic, \( C_{\text{glucose}} = C_{\text{urea}} = 0.25 \, \text{mol L}^{-1} \).

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

Moles of glucose = \( 0.25 \, \text{mol} \).
Mass of glucose = \( 0.25 \times 180 = 45 \, \text{g} \).

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