When 1.00 g. of MgCO3 is added to 2.0 L of water, some, but not all, will dissolve to form a saturated solution. Calculate the mass of solid that remains undissolved.

........MgCO3 ==> Mg^2+ + CO3^2-

I.......solid.....0........0
C...x dissolves...x........x
E.......solid......x.......x

Ksp = (Mg^2+)(CO3^2-)
Substitute for Ksp and the E line from the ICE chart and solve for x = solubility MgCO3 in mols/L.
You have 2L therefore, the solubility will be ?mols/L x 2L = ? then convert that to grams by g = mols x molar mass.
Subtract from 1.00g to find the amount not dissolved.

To calculate the mass of the solid that remains undissolved, we first need to determine the solubility of magnesium carbonate (MgCO3) in water.

The given problem tells us that the solution is saturated, which means it contains the maximum amount of solute (MgCO3) that can dissolve at a given temperature.

To find the solubility of MgCO3, we need to consult a solubility table or reference. Let's assume that the solubility of MgCO3 in water at the given temperature is 0.05 g/L.

Now, we can find the maximum amount of MgCO3 that can dissolve in 2.0 L of water (the volume given in the problem).

Maximum amount of MgCO3 that can dissolve = solubility of MgCO3 x volume of water
= 0.05 g/L x 2.0 L
= 0.1 g

Since the solution is saturated, it means that the 1.00 g of MgCO3 exceeds the maximum amount that can dissolve. Therefore, the mass of solid that remains undissolved is:

Mass of solid that remains undissolved = mass of MgCO3 initially added - maximum amount of MgCO3 that can dissolve
= 1.00 g - 0.1 g
= 0.90 g

Therefore, 0.90 grams of MgCO3 remains undissolved in the solution.