Question

An aqueous solution is made by dissolving 14.2g of AL(NO3)3 to 655g of water Kb H2O= 0.52 C/m

Molarity of solution

Boiling Point of elevation

Boiling point of solution

Answer 1

Molality = (weight / molecular weight) x (1000 / moass of the solute in grams)

weight = 14.2 grams

molar mass of AL(NO3)3 = 213 g/mol (taken from google)

weight of the solvent = 655 grams

put all these in the above equation

molality = (14.2 / 213) x (1000/655)

molality = 14200 / 139515m

molality = 0.102m

Boiling point elevation = deltaTb = Kb x m

Kb = 0.52 ºC/m

m = 0.102

deltaTb = 0.52 ºC/m x 0.102m

delta Tb = 0.0529 ºC is the boiling point elevation

now DeltaTb = boiloing point of the solution - boiling point of the solvent

0.0529 = boiling point of the solution - 100 ºC (here solvent is water and its boiling point = 100 ºC)

boiling point of the solution = 0.0529 +100

= 100.0529 ºC