Answer to Question #269359 in General Chemistry for leo

Question

A solution of glycerol, C3H5(OH)3, in 735 g of water has a boiling point of 104.4 °C at a

pressure of 760 mm Hg. What is the mass of glycerol in the solution? What is the

mole fraction of the solute?

Answer 1

We know that ΔTb = Kb x m

Where

ΔT b = elevation in boiling point

= boling point of solution - boiling point of water

= 104.4 oC - 100 oC

= 4.4 oC

K b = elevation in boiling point constant of water = 0.512 oC / m

m = molality of the solution

= ( mass / Molar mass ) / weight of the solvent in Kg

= ( w / 92 g/mol ) / 0.735 Kg

= 0.0148 w

Plug the values we get 4.4 = 0.512 x 0.0148 w

---> w = 581.1 g

No.of moles of solute , n = mass/molar mass

= 581.1 g / 92 g/mol

= 6.32 mol

no.of moles of water , n' = 735 g / 18 g/mol

= 40.83 mol

Mole fraction of solute , X = n / (n+n')

= 6.32 / ( 6.32 + 40.83 )

= 0.134