Question

1) Estimate the boiling point of the solution in degrees Celsius consisting of 15.9g of glucose (M=180.16 g/mol) dissolved in 140g of water; having K_f = 1.86 K kg mol^-1 and K_b = 0.51 K kg mol^-1 for water. (show work)

Answer 1

1)

Lets calculate molality first

Molar mass of C6H12O6,

MM = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass(C6H12O6)= 15.9 g

use:

number of mol of C6H12O6,

n = mass of C6H12O6/molar mass of C6H12O6

=(15.9 g)/(1.802*10^2 g/mol)

= 8.826*10^-2 mol

m(solvent)= 140 g

= 0.14 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(8.826*10^-2 mol)/(0.14 Kg)

= 0.6304 molal

lets now calculate ΔTb

ΔTb = Kb*m

= 0.51*0.6304

= 0.3215 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 0.3215

= 100.3215 oC

Answer: 100.32 oC