The boiling point of a solution containing 20.0g C12H22O11 in 500.0g H2O is:
a. 100.037 degrees C
b. 100.060   "
c. 101.735   "
d.  10.468   "

I am here again teatea!!!

Some concepts:
The temperature at which the vapor pressure of the liquid is equal to
the atmospheric pressure is termed as the boiling point.
The boiling point of any liquid increases by adding a non-volatile
solute to it. This increases the vapor pressure.
Elevation in the boiling point ( DTb ) is directly proportional to the
amount of solute present.
Concentration of the solute is expressed in molalities (m) 
(molality = moles of solute / kilograms of solvent) 

DTb = m.Kb 
where
m = concentration in molalities 
Kb = molal boiling point elevation constant; for the water Kb = 0.512
ºC/mol.
Boiling point of the water is 100ºC.

Now we can solve the problem.

First of all we need to calculate how many moles of the solute we
have:
1 mole C12H22O11 = 12x12g + 22x1g + 11x16g = 342 grams.
We have 20g of C12H22O11 in the solution, that are 20g/342g = 0.058478
moles.

Now we can calculate the molality m:
m = moles of solute / kilograms of solvent = 
  = 0.058478 moles / 0.5 kg = 0.116956

Then the elevation in the boiling point ( DTb ) is:
DTb = m.Kb = 0.116956 x 0.512 = 0.06 ºC

The boiling point of the solution is:
SolTb = 100ºC + 0.06ºC = 100.06 ºC.

The correct answer is b.

Hope this helps.

Regards.
livioflores-ga