An aqueous solution of glycerol, c3h8o3, is 48.0% glycerol by mass and has a density of 1.120 g ml-1. calculate the molality of the glycerol solution.

Answer is: the molality of the glycerol solution is 10.03 m.
V(solution) = 100 mL.
m(solution) = V(solution) · d(solution).
m(solution) = 100 mL ·1.120 g/mL.
m(solution) = 112 g.
m(glycerol) = ω(glycerol) · m(solution).
m(glycerol) = 0.48 · 112 g.
m(glycerol) = 53.76 g.
m(water) = 112 g - 53.76 g.
m(water) = 58.24 g ÷ 1000 g/kg = 0.05824 kg.
n(C₃H₈O₃) = m(C₃H₈O₃) ÷ M(C₃H₈O₃).
n(C₃H₈O₃) = 53.76 g ÷ 92 g/mol.
n(C₃H₈O₃) = 0.584 mol.
b(C₃H₈O₃) = n(C₃H₈O₃) ÷ m(H₂O).
b(C₃H₈O₃) = 0.584 mol ÷ 0.05824 kg.
m(C₃H₈O₃) = 10.027 mol/kg.

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c.  and at that point will start to turn into steam (gas)