WHAT IS THE MOLARITY OF A SOLUTION WHICH FREEZES AT -0.192DEGREE CELSIUS ASSUMING NO CHANGE IN THE SOLUTE BY RAISING THE TEMPERATURE,AT WHAT TEMPERATURE WILL THE SAME SOLUTION BOIL?(Kf for water=1.86 degree celsius kg per mole and Kb for water is 0.515 degree clsius kg per mole)

Here, depression in freezing point Tf = Freezing point of pure solvent - freezing point of solution =
or, depression in freezing point =  00C (for water solvent) - (-0.1920C) = 0.1920C
Since, depression in freezing point = Kf x molality
Hence, molality = 0.192/ 1.86 = 0.103 
Since, for water 1L  equivalent to 1Kg so, M = m
Hence, molarity = 0.103
Now, elevation in boiling point for same amount of solute, i.e. 0.103molal solution,
Elevation in boiling point = m x Kb = 0.103 x 0.515 = 0.0530C
Hence, boiling point = 1000C (for water) + 0.053 0C = 100.0530C

  • 3
What are you looking for?