If 6 moles of sugar was added to a kilogram of water, the new boiling point would be degrees c.

Using the equation for boiling point elevation Δt
     Δt = i Kb m 
we can find the new boiling point T for the solution:
     Δt = T - 100∘C 
since we know that pure water boils at 100 °C.

We know that the van't Hoff Factor i is equal to 1 because sugar does not dissociate in water.

Also, the value of Ebullioscopic constant Kb for water is listed as 0.512 °C·kg/mol.

The molality m of the solution of 6 moles of sugar dissolved in a kilogram of water can be calculated as 
     m = 6 moles / 1 kg
         = 6 mol/kg

Therefore the new boiling point T would be
     T - 100 °C = i Kb m 
     T = i Kb m + 100 °C.
        = (1) (0.512 °C·kg/mol) (6 mol/kg) + 100 °C
        = 3.072 °C + 100 °C 
        = 103.072 °C

Idon't see the pic?