Calculate the freezing point and boiling point of each of the following solutions:

Question

Calculate the freezing point and boiling point of each of the following solutions:

 

 

 

1)  the freezing point of the solution: 153 g of sucrose, C12H22O11 , a nonelectrolyte, dissolved in 1.35 kg of water (Kf =1.86°C )
 
2)  the boiling point of the solution: 153 g of sucrose, C12H22O11 , a nonelectrolyte, dissolved in 1.35 kg of water (Kb =0.52°C )

Answer 1

The freezing point is depressed proportional to the molality of molecules or ions in solution.

 

Sugar is a non electrolyte, so we calculate based on molality of sucrose *kf

 

MW of sucrose = 12*12 + 22*1 + 11*16 = 342 g/mole

Molality = moles of solute/kg of solvent =   (153g  /342g/M) /1.35kg =  0.331 M/kg

 

depression = 0.331 * 1.86 =  0.62 degrees so the freezing point is 0 - 0.62 = - 0.62 C

 

 

Similar for boiling point elevation  0.331 * 0.52 =   0.17 C  so the boiling point is 100 + 0.17 = 100.17 C