Respuesta :
Answer: Thus the boiling point of a solution is [tex]102.68^0C[/tex]
Explanation:
Elevation in boiling point is given by:
[tex]\Delta T_b=i\times K_b\times m[/tex]
[tex]\Delta T_b=T_b-T_b^0=(T_b-100)^0C[/tex] = elevation in boiling point
[tex]K_b[/tex] = boiling point constant = [tex]0.51K/kgmol[/tex]
m = molality = ?
i = vant hoff factor = 1 (for non electrolyte)
[tex]\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (water)= 250.0 g = 0.250 kg
Molar mass of solute (sucrose) = 342 g/mol
Mass of solute (sucrose) = 450.0 g
[tex](T_b-100)^0C=1\times 0.51\times \frac{450.0g}{342g/mol\times 0.250kg}[/tex]
[tex](T_b-100)^0C=2.68[/tex]
[tex]T_b=102.68^0C[/tex]
Thus the boiling point of a solution is [tex]102.68^0C[/tex]
