Calculate the molar mass of nonelectrolyte solute when 6 gram of it is dissolved in 1 dm ^3 water has osmotic pressure 2.4 atm at 300 K (R = 0.0821 atm dm ^3 K ⁻¹ mol ⁻¹)

Step-by-step solution

The osmotic pressure is given by the formula: = C R T = W R T M V Rearranging the formula to solve for the molar mass M : M = W R T V Substituting the given values W = 6 g , R = 0.0821 atm dm ^3 K ⁻¹ mol ⁻¹ , T = 300 K , = 2.4 atm , and V = 1 dm ^3 : M = 6 0.0821 300 2.4 1 M = 147.78 2.4 M = 61.575 g mol ⁻¹ Rounding to two decimal places, the molar mass is 61.58 g mol ⁻¹ . Answer: 61.58 g mol ⁻¹