Answer to Question #168267 in General Chemistry for Shein

Question

Calculate the boilng point of a solution made by dissolving 17.5 g analine, C6H5NH2, in 113g of water.

Answer 1

molar weight of aniline = 93 g/mol

113 g water contain 17.5 g aniline

1 g water contain = (17.5/113) g aniline

1000 g water contain

= [(17.5×1000)/113] g aniline

= 154.87 g

=[(154.87 g) ÷(93 g/mol)]

= 1.698 mol

Since, 1 kg = 1000 g

So, molality of the solution of aniline, m = 1.698 mol/kg

= 1.698 m

As we know, for a solution

If boiling point elevation = ∆Tb

= (Ti – Tf)

Ti = boiling point of solvant

Tf = boiling point of solution

And molality of the solution = m

boiling point elevation constant = Kf

Then, ∆T = Kf.m

Or, (Ti – Tf) = Kf.m

Or, Tf = Ti – Kf.m

For water, Kf = 0.51K/m

And Ti = 373K

Or, Tf = 373 – (0.51×1.698)

= 372.134 K

So, boiling point of the solution = 372.134 K