1) You neeed to know and use the Ksp for BaF2.
At 25°C this Ksp is 1.0 * 10 ^ - 6
2) The solutibility of BaF2 is given by:
BaF2 ⇄ Ba(2+) + 2F(-)
x 2x
=> Ksp = x * (2x)^2 = 4x^3
3) When you have a NaF solution, you have to take into accout the concentration of the NaF solution
M = 0.1
Now the equilibrium species are:
BaF2 ⇄ Ba(2+) + 2F(-)
x 2x + 0.10
And Ksp = x* [2x + 0.10]^2 = 1.0 * 10 ^ -6
Given that the Ksp << 1 you may assume that 2x << 0.1 => 2x + 0.1 ≈ 0.1
=> 1.0 * 10 ^ - 6 ≈ x(0.1)^2 = 0.01x
=> x = 1.0 * 10^ -6 / 0.01 = 1.0 * 10^ - 4 M = 0.0001 M
That is the molar solubility.
4) Now, you calculate the number of moles from the molarity's formula:
M = n / v => n = M * v = 0.0001 M * 0.500 l = 0.00005 mol
And now convert to grams,
mass in grams = number of moles * molar mass
molar mass of BaF2 = 175.34 g/mol
mass in grams = 0.00005 moles * 175.34 g / mol = 0.0088 g
Answer: 0.0088 g
