Calculate the boiling point of the solution

Question

A certain liquid X has a normal boiling point of 129.00°C

 and a boiling point elevation constant K_b = 0.54·°C·kgmol^-1

.Calculate the boiling point of a solution made of 0.20 kg of iron(III) nitrate (FeNO33) dissolved in 600. g of X. Round your answer to 4

 significant digits.

Answer 1

Hi Amani, I can help out with this one.

First, calculate the molality.

The molar mass of Fe(NO3)3 = 55.85 + (14.01 + 48.00)*3 = 241.88 g/mol

Next, we are given the mass of Fe(NO3)3 as 0.2 kg, or 200 g.

The number of moles of Fe(NO3)3, n = mass of Fe(NO3)3/molar mass of Fe(NO3)3

= 200 g/241.88 g/mol = 0.827 mol.

Next, the mass of the solvent is given as 600 g, or 0.6 kg. Use molality, m = number of mol/mass of solvent in Kg

= 0.827 mol/0.6 kg = 1.378 molal

Next, we need to use the formula Change in Tb = i*Kb*m

i for Fe(NO3)3 is 4, because we have 1 Fe ion and 3 NO3 ions.

Change in Tb = i*Kb*m = 4*0.54*1.378 = 2.976 °C

The normal boiling point of liquid X is given in the problem as 129.00 °C. So, the new boiling point = 129.00°C + 2.976 °C = 131.976°C

The boiling point is 132 °C.