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 inversely proportional to the molar mass of the solute.

Answer 1

Osmotic pressure (\( \pi \)) is given by the formula: \[ \pi = \frac{nRT}{V} \] where:
\( n \) is the number of moles of solute,
\( R \) is the gas constant,
\( T \) is the temperature,
\( V \) is the volume of the solution.
Since both solutions have the same osmotic pressure, the number of moles of solute in both solutions must be equal. For urea, the number of moles \( n_1 \) is calculated as: \[ n_1 = \frac{15 \, \text{g}}{60 \, \text{g/mol}} = 0.25 \, \text{mol}. \] Let the mass of glucose be \( m_2 \). The number of moles of glucose \( n_2 \) is: \[ n_2 = \frac{m_2}{180 \, \text{g/mol}}. \] Since the osmotic pressures are identical, the number of moles of urea must equal the number of moles of glucose: \[ 0.25 = \frac{m_2}{180}. \] Solving for \( m_2 \): \[ m_2 = 0.25 \times 180 = 45 \, \text{g}. \] Therefore, the mass of glucose present in one litre of its solution is \( 45 \, \text{g} \). \bigskip