Ksp of CaF2 is 1.7  10-10 and its mol. wt. is 78 g mole-1. What volume of the saturated solution will contain 0.078 g of CaF2?

Sure, let's solve this step by step:

1. First, we need to find the molar concentration of CaF2 in the saturated solution. The Ksp of CaF2 is given as 1.7 x 10^-10. The dissolution of CaF2 can be represented as:

   CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)

   The Ksp expression for this reaction is:

   Ksp = [Ca2+][F-]^2

   Since the stoichiometric ratio of CaF2 to Ca2+ and F- is 1:1:2, the concentration of Ca2+ and F- in the saturated solution will be the same, let's call it 's'. So, the Ksp expression becomes:

   Ksp = s * (2s)^2 = 4s^3

   Solving for 's' gives:

   s = (Ksp / 4)^(1/3) = (1.7 x 10^-10 / 4)^(1/3) = 1.3 x 10^-3 M

2. Now, we need to find the volume of the saturated solution that will contain 0.078 g of CaF2. First, convert the mass of CaF2 to moles using its molar mass:

   Moles of CaF2 = 0.078 g / 78 g/mole = 0.001 mole

3. Then, use the molar concentration of CaF2 in the saturated solution to find the volume:

   Volume = moles / concentration = 0.001 mole / 1.3 x 10^-3 M = 0.77 L or 770 mL

So, 770 mL of the saturated solution will contain 0.078 g of CaF2.