Question
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 )
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 )
C. 61.58\ g mol ^ -1
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
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