Compute the boiling point of this solution:
21.6 g nis04 in 1.00 • 102 g h20
(assume 100% ionization)

The boiling point of the solution is 101,42ºC

Explanation:

Formula for boiling point elevation is:

ΔT = Kb · molality . i

Water boils at 100ºC, at 1 atm.

NiSO4 > Ni2+  +  SO4-2  (assume 100% ionization, the value of i, is 2)

Tº boiling point solution - Tº boiling point = Ebulloscopic constant . molality . Van't Hoff factor.

Tº bp solution - 100ºC = 0,512 ºC/molal . molality . 2

Molality means moles of solute in 1kg of solvent (1000g)

Molar mass NiSO4 = 154,75 g/m

Mass/ Molar mass = Moles --> 21,6 g / 154,75 g/m = 0,139 moles

These moles are in 100 g of H2O, so we have to make a rule of three to find, the moles in 1000g

100 g 0,139 moles

1000 g 1,39 moles (molality)

Tº bp solution - 100ºC = 0,512 ºC/molal . 1,39molal . 2

Tº bp solution = (0,512 ºC/molal . 1,39molal . 2) + 100ºC

Tº bp solution = 101,42ºC