Final answer:
The change in the boiling point after dissolving 14.7 g of glucose into 150.0 ml of water is calculated as 0.279°C using the molal boiling point elevation constant for water (Kb) and the molality of the solution (m). None of the provided answer choices match this result.
Step-by-step explanation:
The boiling point elevation (ΔTb) of a solution can be calculated using the formula ΔTb = Kb · m, where Kb is the molal boiling point elevation constant and m is the molality of the solution. The molality is the number of moles of solute per kilogram of solvent. First, we need to calculate the moles of glucose (C₆H₁₂O₆) in the 14.7 g sample. The molar mass of glucose is approximately 180.16 g/mol.
Therefore, moles of glucose,
= 14.7 g / 180.16 g/mol
= 0.0816 mol.
Since we have 150.0 ml of water and the density is 1.00 g/mL, the mass of the water is 150.0 g, which is 0.150 kg. Thus, the molality (m) of the solution is,
= 0.0816 mol / 0.150 kg
= 0.544 mol/kg. Now, using the given Kb value of 0.512°C/m, the change in boiling point,
ΔTb = 0.512°C/m · 0.544 m
= 0.279 degree C.
Therefore, none of the provided answers (a, b, c, d) are correct.