Question

I take 20.00 grams of citric acid (C6H8O7) and dissolve it in water. The final solution has a volume of 110.0 mL. What is the boiling point of this new solution at 1 atm? (Be precise - overrounding will lead to the wrong answer) Density of the solution is = 1.080 g/ml i = 1.151 Kb = 0.5200 ºC kg mol-1

Answer 1

We know that ΔT b = iKbx m
Where
ΔT b= elevation in boiling point
= boiling point of solution - boilinging point of pure solvent
= Tb -100 oC
Kb = elevation in boiling point constant of water = 0.52 oC/m

i=1.151

Mass of solution= volume x density

= 110.0 mL * 1.080 g/ mL

= 118.8g

So mass of water = 118.8 g - 20.00 g= 98.8 g = 0.0988 kg
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg
= ( 20.00g/(192g/mol))/0.0988 kg

= 1.054 m
Plug the values we get Tb= 100.63oC