If 4.27 grams of sucrose, C12H22O11, are dissolved in 15.2 grams of water, what will be the boiling point of the resulting solution?

Answer is: the boiling point of the resulting solution of sucrose is 100.42°C.
m(H₂O) = 15.2 g ÷ 1000 g/kg = 0.0152 kg.
m(C₁₂H₂₂O₁₁) = 4.27 g.
n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 4.27 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0125 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0125 mol ÷ 0.0152 kg.
b(solution) = 0.82 m.
ΔT = b(solution) · Kb(H₂O).
ΔT = 0.82 m · 0.512°C/m.
ΔT = 0.42°C.
Tb = 100°C + 0.42°C = 100.42°C.
Hope this helps!

The gas would begin to expand, and eventually the can would burst or explode.