Question

A 6% solution of glucose (molar mass = 180 g mol\(^{-1}\)) is isotonic with a 2.5% solution of an unknown organic substance. Calculate the molecular weight of the unknown organic substance.

Hint:

If two solutions are isotonic, their osmotic pressures and molar concentrations must be equal.

Answer 1

Since the two solutions are isotonic, their osmotic pressures are equal, meaning the concentrations in terms of molarity are the same: \[ C_G = C_U \] where: - \( C_G = \frac{6}{180} \) (Glucose solution molarity) - \( C_U = \frac{2.5}{M_U} \) (Unknown substance molarity) Since, \[ \frac{6}{180} = \frac{2.5}{M_U} \] Rearranging to solve for \( M_U \): \[ M_U = \frac{2.5 \times 180}{6} \] \[ M_U = 75 \, g \, mol^{-1} \]