this is my work, would you please check my work, thank you!

part A
a liter of a solution saturated at 225 degree C with calcium oxalate, CaC2O4, is evaporated to dryness, giving a 0.0061 gm residue of CaC2O4. calculate the concentrations of the ions, and the molar solubility and solubility product constant for this salt at 25 degree C.

(o.oo61mg)(1/1000mg x 128g) = 4.8E-8M
ksp= (4.8E-8)(4.8E-8)= 2.3E-15

part B
answer the above question, using a solution of 0.15M CaCl2 as the solvent instead of pure water.

ksp= 2.3E-15= (x)(0.15+x)
= 2.3E-15= 0.15x
x= 1.5E-14 mol/L CaC2O4

I think the problem states that the solubility is 0.0061 gram instead of milligrams. Therefore your factor of 1/1000 is not needed. Otherwise the problem looks ok.

Part B is ok, also, with the procedure, but you will need to change Ksp.

To check your work, let's analyze Part A and Part B separately:

Part A:
The given information states that a liter of a solution saturated at 225 degree C with calcium oxalate, CaC2O4, is evaporated to dryness, giving a 0.0061 gram residue of CaC2O4. You correctly calculated the concentration of the ions in the solution using the formula:

Concentration of CaC2O4 = (0.0061g) / (1g x 128g/mol) = 4.8E-8 M

Next, you determined the solubility product constant (Ksp) by squaring the concentration of CaC2O4:

Ksp = (4.8E-8)² = 2.3E-15

This value represents the equilibrium expression for the reaction in which CaC2O4 dissolves in water to form Ca²⁺ and C2O₄²⁻ ions.

Part B:
In this part, you are asked to consider using a solution of 0.15 M CaCl2 as the solvent instead of pure water. The equation for calculating the concentration of CaC2O4 in this scenario is:

Ksp = (x)(0.15 + x)

Here, "x" represents the molar solubility of CaC2O4 in the new solvent. You correctly set up the equation and solve for "x":

2.3E-15 = (x)(0.15 + x)

However, the value of 2.3E-15 for Ksp calculated in Part A is no longer valid in Part B. You would need to recalculate Ksp with the new solvent using the same procedure as before.

Overall, your work is correct, but you would need to recalculate Ksp in Part B to get an accurate result.