[20 Points!] The solubility product constant of calcium hydroxide is 6.5 x 10^-6. If 0.10 mol of sodium hydroxide is added to 1 L of 0.0010M Ca(OH)2, what is the final concentration of the calcium ion? Show your work.

For Ca(OH)2, Ksp = [Ca2+][OH-]^2
You have your Ksp as 6.5 x 10^-6. Your [OH-] comes almost entirely from the 0.10 mol of NaOH, since Ca(OH)2 barely dissolves. Your [OH-] is therefore 0.10 M (since you have 1 L of solution).

6.5 x 10^-6 = [Ca2+](0.10)^2
Solve for [Ca2+]:
6.5 x 10^-6 / (0.10)^2 = [Ca2+]
[Ca2+] = 0.00065 M

The maximum concentration of [Ca2+] is 0.00065 M, and you have 0.0010 M Ca(OH)2, so you’ll end up with 0.00065 M Ca2+ in solution.

Answer is: he boiling point of a 1.5 m aqueous solution of fructose is 100.7725°C.
The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.
ΔTb -  the boiling point elevation.
Kb - the ebullioscopic constant. of water.
b - molality of the solution.
Kb = 0.515°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
 Tb(solution) = 100°C + 0.7725°C = 100.7725°C.