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A solution containing 6 gram of a solute dissolved into 250 cm cube of water gave an osmotic pressure of 4.5 atmosphere at 27° celsius calculate the boiling point of the solution the moral elevation constant for water is 0.52 degree celsius 1000 gram

A solution containing 6 gram of a solute dissolved into 250 cm cube of water gave an osmotic pressure of 4.5 atmosphere at 27° celsius calculate the boiling point of the solution the moral elevation constant for water is 0.52 degree celsius 1000 gram

Grade:12

1 Answers

ayush agrawal
24 Points
5 years ago
π=CRTπ=n/v * RTPutting π=4.5 atm R=0.0821 l atm mol-1 and T=300K V=0.25l; we get,n=0.04∆tb = kb * mm = 0.04/0.25 = 0.182Kb= 0.52∆tb = 0.52*0.182= 0.095Boiling point= 100+0.095 = 100.095

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