molar mass problem

Question

If 50 g of an unknown electrolyte raise the boiling point of 650 g naphthalene to 220 C, find the unknown electrolyte’s molar mass. Assume naphthalene has a Kb of 5.8 C/m, a normal tB of 217.7 C, and that the van’t Hoff factor of the electrolyte is 4.

Answer 1

ΔT = imK

ΔT = change in BP = 220 - 217.7 = 2.3^o

i = vant Hoff factor = 4

m = molality = mols / kg = ?

K = boiling constant = 5.8^o/m

Solving for m…

m = ΔT/(i)(K) = (2.3)/(4)(5.8)

m = 0.0991 moles X / kg naphthalene

There are 650 g naphthalene = 0.650 kg

moles X = 0.0981 moles/kg x 0.650 kg = 0.0638 moles

molar mass X = grams/moles = 50 g/0.0638 moles

molar mass = 784 g/mole = 780 g/mol (2 sig figs)

Answer 2

The increase in boiling point after the addition of electrolyte is expressed by the following equation:

T_N-T_O= i K_bW₂x1000/(M₂x650)

here T_N and T_O is new and old boiling point, respectively. W₂ is mass of electrolyte added and M₂ is molar mass of electrolyte.

So, putting given values,

220-217.7= 4x5.8x50x1000/(M₂X650)

2.3=1784.61/M₂

M₂= 1784.61/(2.3)

M₂= 775.92