Question

assuming ideal behaviour calculate the normal boiling point in degress celcius o a solution that contains 64.5 grams of the non volatile non electrolyte quinoline (MW=129) in 500 grams of benzene for the normal boiling point is 80.1o degress celcuus and KB =2.53 degress celcius

Answer 1

Answer:

82.63°C

Explanation:

We know,

∆Tb=Kb × m......[∆Tb= Elevation in boiling point]

We have 64.5 gm of quinoline. Which means we have 64.5/129 mol of quinoline.

Thus,

No. of moles of quinoline=63.5/129=0.5 mol

Molality=m= No. of moles of solute/ Mass of solvent

Thus,

m= 0.5mol/500g

= 1 mol/kg

Now,

∆Tb=Kb × 1

∆Tb=2.53

Thus,

Boiling point = 80.1 + 2.53

= 82.63°C