Question

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

Answer 1

Molarity of the urea solution is 15/60 mol/litre = ¼ mol/litre

Now isotonic solutions have same osmotic pressure (π), which is given by
π = iMRT
M = molar mass
R = Universal gas constant
T = absolute temperature

For two solutions the ration of osmotic pressure at constant temperature is
π₁/π₂ = M₁/M₂
(the suffix 1 & 2 represent urea and unknown solution respectively)

As they have same osmotic pressure so π₁/π₂ = 1
So M₁ = M₂
So molarity of the unknown (glucose) solution is ¼ mol/litre.

Molar mass of glucose is 180 g/mol
¼ mol of glucose = ¼ X 180 g = 45 g
So the glucose solution contains 45g in one litre.