Final answer:
The boiling point of a 0.875 m aqueous solution of lead(II) nitrate, Pb(NO3)2, is approximately 100.44625 °C.
Step-by-step explanation:
The boiling point of a solution depends on the concentration and nature of the solute. In this case, we have a 0.875 m aqueous solution of lead(II) nitrate, Pb(NO3)2. To determine the boiling point of this solution, we need to consider the boiling point elevation equation: ΔT = Kb * m, where ΔT is the boiling point elevation, Kb is the molal boiling point elevation constant, and m is the molality of the solution.
Given that the molal boiling point elevation constant for water is 0.51 °C/m, we can use this value to calculate the boiling point elevation. Since the concentration of the solution is given as 0.875 m, we can plug these values into the equation to find the boiling point elevation:
ΔT = (0.51 °C/m) * (0.875 m) = 0.44625 °C
To find the boiling point of the solution, we need to add the boiling point elevation to the boiling point of pure water. The boiling point of pure water is 100 °C, so:
Boiling point = 100 °C + 0.44625 °C = 100.44625 °C