He solubility of ba(no3)2is 130.5 grams per liter at 0°c. how many moles of dissolved salt are present in 4.0 liters of a saturated solution of ba(no3)2at 0ºc

A saturated solution is when there is a dynamic equilibrium between the solute dissolved, the dissociated ions, the undissolved and the compound in the solution. The concentration of a solution that is saturated is equal to the solubility of the solute in solvent at a certain temperature since solubility is the maximum amount of solute that can be dissolved in the solvent to form a solution. For this case, the solubility of barium nitrate is 130.5 g per liter. To determine the moles of barium nitrate, we do as follows:

130.5 g / L = x g / 4.0 L      where x is the mass of barium nitrate in the solution.

x = 522 g Ba(NO3)2

522g Ba(NO3)2 ( 1 mol / 261.34 g ) = 2.00 moles Ba(NO3)2